Answer: option A. volume
Explanation:
Potassium nitrate has a lattice energy of -163.8 kcal/mol and a heat of hydration of -155.5 kcal/mol.
How much potassium nitrate has to dissolve in water to absorb 101 kJ of heat?
Answer:
293.99 g
OR
0.293 Kg
Explanation:
Given data:
Lattice energy of Potassium nitrate (KNO3) = -163.8 kcal/mol
Heat of hydration of KNO3 = -155.5 kcal/mol
Heat to absorb by KNO3 = 101kJ
To find:
Mass of KNO3 to dissolve in water = ?
Solution:
Heat of solution = Hydration energy - Lattice energy
= -155.5 -(-163.8)
= 8.3 kcal/mol
We already know,
1 kcal/mol = 4.184 kJ/mole
Therefore,
= 4.184 kJ/mol x 8.3 kcal/mol
= 34.73 kJ/mol
Now, 34.73 kJ of heat is absorbed when 1 mole of KNO3 is dissolved in water.
For 101 kJ of heat would be
= 101/34.73
= 2.908 moles of KNO3
Molar mass of KNO3 = 101.1 g/mole
Mass of KNO3 = Molar mass x moles
= 101.1 g/mole x 2.908
= 293.99 g
= 0.293 kg
293.99 g potassium nitrate has to dissolve in water to absorb 101 kJ of heat.
Final answer:
To absorb 101 kJ of heat, 7.63 grams of potassium nitrate must dissolve in water, calculated based on the lattice energy and heat of hydration provided.
Explanation:
The question asks how much potassium nitrate needs to dissolve in water to absorb 101 kJ of heat, given the lattice energy and heat of hydration for potassium nitrate. First, we convert the given heat into kcal because the energies are provided in kcal/mol: 101 kJ × (1 kcal / 4.184 kJ) = 24.1 kcal. The total heat involved in dissolving potassium nitrate in water can be found by adding the lattice energy to the heat of hydration: -163.8 kcal/mol + (-155.5 kcal/mol) = -319.3 kcal/mol. This value represents the heat released when 1 mole of potassium nitrate dissolves in water.
To find the amount of potassium nitrate that needs to dissolve to absorb 101 kJ (24.1 kcal), we set up a proportion, knowing that -319.3 kcal is released per mole of potassium nitrate dissolved: (1 mole / -319.3 kcal) = (x moles / 24.1 kcal). Solving for x gives x = 24.1 kcal / -319.3 kcal/mol = 0.0755 moles. Finally, to find the mass of potassium nitrate, we multiply the moles by the molar mass of potassium nitrate (KNO3), which is approximately 101.1 g/mol: 0.0755 moles ×101.1 g/mol = 7.63 grams.
Therefore, 7.63 grams of potassium nitrate need to dissolve in water to absorb 101 kJ of heat.
There are some exceptions to the trends of first and successive ionization energies. For each of the following pairs, explain which ionization energy would be higher:
(a) IE₁ of Ga or IE₁ of Ge
(b) IE₂ of Ga or IE₂ of Ge
(c) IE₃ of Ga or IE₃ of Ge
(d) IE₄ of Ga or IE₄ of Ge
Explanation:
a) IE1 of Germenium. ionization energy decreases down the group.
b) Here IE2 of Ga is higher because second valence electron of gallium is in s orbital whereas second valence electron of germenium is in p orbital.
c) IE3 of Ge is higher as this follows the trend.
d)IE4 of Ga is higher because the 4th valence electron of Ga is a core electron and the for Ge it is in s orbital and it takes higher energy to break a core electron than the orbital ones.
When comparing Ga and Ge, IE₁ and IE₂ of Ga are lower due to being earlier in the periodic table, but IE₄ of Ga is expected to be higher as it involves removing an electron from a full orbital. IE₃ for both may be similar due to both elements having three valence electrons. The correct answer is d) IE₄ of Ga or IE₄ of Ge.
The subject in question involves comparing first and successive ionization energies (IE) for elements in the periodic table, particularly gallium (Ga) and germanium (Ge). Ionization energy refers to the energy required to remove an electron from a gaseous atom or ion.
As general rules, (a) the first IE tends to increase across a period as the nuclear charge increases, and (b) IE increases for successive ionizations as the atom or ion becomes progressively more positively charged.
(a) IE₁ of Ga would be lower than IE₁ of Ge because Ga is to the left of Ge in the periodic table, and thus Ge has a higher nuclear charge and a stronger attraction to its valence electron.
(b) IE₂ of Ga would also be lower than IE₂ of Ge because the additional electron from Ga would come from the same valence shell as the first electron, while Ge's second ionization would remove an electron that experiences a stronger nuclear charge.
(c) As both Ga and Ge have three valence electrons, the IE₃ of Ga and Ge will both involve removing an electron from a similarly charged ion, but because Ga has a higher energy 4p subshell, its IE₃ might be slightly lower than Ge's, which is removing an electron from the 4s subshell (which after removing two electrons is now full and lower in energy).
(d) At the IE₄ level, we see a large jump in ionization energy for both elements as electrons are being removed from an energy level closer to the nucleus; however, IE₄ of Ga will generally be higher than IE₄ of Ge due to of an electron from a full orbital in Ga, which has higher energy compared to the removal from a half-filled orbital in Ge.
These comparisons are based on the understanding that ionization energies increase both with increasing positive charge and with the removal of electrons closer to the nucleus.
Ammonia is oxidized to nitric oxide in the following reaction:
4NH3 + 5O2 --------> 4NO + 6H2O
a. Calculate the ratio (lb-mole O2 react/lb-mole NO formed).
b. If ammonia is fed to a continuous reactor at a rate of 100.0 kmol NH3/h, what oxygen feed rate (kmol/h) would correspond to 40.0% excess O2?
c. If 50.0 kg of ammonia and 100.0 kg of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.
Final answer:
a) The ratio of lb-mole O2 react/lb-mole NO formed is 1.25. b) The oxygen feed rate corresponding to 40% excess O2 is 175.0 kmol/h. c) Ammonia is the limiting reactant, with the oxygen being in excess by 5.85%. The extent of reaction is 36.17 mol NO produced, with a mass of 1085.73 g.
Explanation:
a. To calculate the ratio (lb-mole O2 react/lb-mole NO formed), we can use the stoichiometry of the reaction. From the balanced equation, we can see that for every 5 moles of O2, we get 4 moles of NO. So, the ratio is 5/4 or 1.25 lb-moles O2 react/lb-mole NO formed.
b. To find the oxygen feed rate corresponding to 40% excess O2, we need to calculate the stoichiometric amount of O2 required and then add 40% of that amount. Since the reaction requires 5 moles of O2 for every 4 moles of NO, the stoichiometric amount of O2 required is (5/4) * 100.0 kmol NH3/h = 125.0 kmol O2/h. Adding 40% of that amount gives 125.0 kmol O2/h + (40/100) * 125.0 kmol O2/h = 175.0 kmol O2/h.
c. To determine the limiting reactant, we need to compare the amounts of ammonia and oxygen given. The molecular weight of ammonia (NH3) is 17 g/mol and oxygen (O2) is 32 g/mol. So, the mass of 50.0 kg of ammonia is 50.0 kg * (1000 g/kg) / (17 g/mol) = 2941.18 mol. The mass of 100.0 kg of oxygen is 100.0 kg * (1000 g/kg) / (32 g/mol) = 3125.0 mol. Comparing these amounts, we can see that ammonia is the limiting reactant as there is less of it. The percentage by which the other reactant (oxygen) is in excess can be calculated as [(3125.0 mol - 2941.18 mol) / 3125.0 mol] * 100 = 5.85 %. The extent of reaction and mass of NO produced can be determined using the stoichiometry of the reaction. From the balanced equation, we can see that for every 4 moles of NO produced, 5 moles of O2 are consumed. So, the extent of reaction is (3125.0 mol - 2941.18 mol) / 5 = 36.17 mol NO produced. The mass of NO produced can be calculated as 36.17 mol * (30.01 g/mol) = 1085.73 g.
a) Ratio (lb-mole O2 react/lb-mole NO formed) is 1.25 lb-mole O₂ / lb-mole NO.
b) Oxygen feed rate is 175.0 kmol O₂/h.
c) Limiting reactant is NH₃; Percentage excess of O₂ is 10.1%; Extent of reaction is 2.94 kmol NH₃; Mass of NO produced is 88.23 kg
a. Calculate the ratio (lb-mole O₂ react/lb-mole NO formed).
From the given reaction, we can see that 5 moles of O₂ react to form 4 moles of NO. Therefore, the ratio of lb-mole O₂ react to lb-mole NO formed is:5 moles O₂ / 4 moles NO = 1.25 lb-mole O₂ / lb-mole NOb. If ammonia is fed to a continuous reactor at a rate of 100.0 kmol NH₃/h, what oxygen feed rate (kmol/h) would correspond to 40.0% excess O₂?
From the reaction, we know that 4 moles of NH₃ react with 5 moles of O₂. Therefore, the stoichiometric oxygen feed rate would be: 5 moles O₂ / 4 moles NH₃ = 1.25 moles O₂ / mole NH₃Since 100.0 kmol NH₃/h is fed to the reactor, the stoichiometric oxygen feed rate would be:1.25 moles O₂ / mole NH₃ × 100.0 kmol NH₃/h = 125.0 kmol O₂/hTo achieve 40.0% excess O₂, we need to add 40.0% of the stoichiometric oxygen feed rate:125.0 kmol O₂/h × 1.4 (1 + 0.4) = 175.0 kmol O₂/hc. If 50.0 kg of ammonia and 100.0 kg of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.
First, let's convert the given masses to moles:50.0 kg NH₃ × (1 kmol / 17.03 kg) = 2.94 kmol NH₃ 100.0 kg O₂ × (1 kmol / 32.00 kg) = 3.13 kmol O₂From the reaction, we know that 4 moles of NH₃ react with 5 moles of O₂. Therefore, the limiting reactant is NH₃, since 2.94 kmol NH₃ is less than the stoichiometric amount required to react with 3.13 kmol O₂.The percentage excess of O₂ can be calculated as:(3.13 kmol O₂ - 2.94 kmol NH₃ × 5/4) / 2.94 kmol NH₃ × 5/4 × 100% ≈ 10.1%Now, let's calculate the extent of reaction and mass of NO produced:The reaction proceeds to completion, so the extent of reaction is equal to the limiting reactant (NH₃):Extent of reaction = 2.94 kmol NH₃The mass of NO produced can be calculated as:4 moles NO / 4 moles NH₃ × 2.94 kmol NH₃ × 30.01 g/mol NO ≈ 88.23 kg NOA city water district wants to encourage local businesses and homeowners to landscape with drought-tolerant plants. After disappointing results from a publicity campaign, the water district decides to subsidize local plant nurseries so they can offer the plants at a lower price.
Suppose the graph shows the supply and demand curves for a drought-tolerant plant, such as purple sage. Drag the appropriate curve to show the impact of the water district subsidy.
Answer:
the complete question is found in the attachment
Explanation:
the complete explanation is found in the attachment
A subsidy from the water district to local plant nurseries would lower the cost of production, enabling them to offer more drought-tolerant plants. This would shift the supply curve to the right on a graph, representing an increase in supply. Assuming demand remains steady, this could result in a lower price and increased quantity of plants.
Explanation:In the economics of supply and demand, a subsidy given to local plant nurseries would likely increase the supply of drought-tolerant plants like purple sage. As the cost of production decreases due to the subsidy, nurseries can afford to produce and sell more plants, shifting the supply curve to the right.
On a graph showing supply and demand curves, you would drag the supply curve to the right to represent this change. This movement should ideally cause a decrease in the price of the plants and increase in quantity, assuming demand stays consistent.
This scenario illustrates the basic economic principle that if you reduce the cost of production (in this case by providing a subsidy), producers are able to supply more of the product, leading to increases in quantity and potential decreases in price. This is an effective tool used often by governments and organizations to influence market dynamics and promote specific goods or services.
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For many purposes we can treat nitrogen as an ideal gas at temperatures above its boiling point of - 196.°C.
Suppose the temperature of a sample of nitrogen gas is raised from -98.0 °C to -89.0 °C, and at the same time the pressure is increased by 10.0%.
A) Does the volume of the sample increase, decrease, or stay the same?
Answer: the volume of the sample decreased
Explanation:
T1 = -98°C = - 98 + 273 = 175K
T2 = -89°C = -89 +273 = 184K
P1 = P
P2 = 110%P = 1.1P
V1 = V
V2 =?
P1V1/T1 = P2V2/T2
PxV/175 = 1.1PxV2/184
175x1.1PxV2 = PVx 184
V2 = (PVx 184) /(175x1.1P)
V2 = 0.96V = 96%V
Therefore, the final volume is 96% of the initial volume. This means that the final volume decreased by 96%
The volume of the sample will increase.
• Based on the given information,
• Let us assume that we have constant number of moles of nitrogen gas at -98 degree C, and the initial pressure is P1.
• It is given that the pressure is increased by 10%, and the temperature is increased to -89 degree C.
Now, the final pressure (P2) will be,
P1 + P1*10/100 = 1.10 P1
T1 = -98 degree C = -98 + 273 K = 175 K
T2 = -89 degree C = -89 + 273 K = 184 K
At constant no of moles, the ideal gas equation is,
PV = nRT
Here, n and R are constant, So, P1V1/T1 = P2V2/T2
P1 * V1/T1 / 175 K = 1.10 P1 * V2/184K
V2/V1 = 184/175 * 1.10 = 1.15
V2 = 1.15 V1
Thus, the volume of sample increase by 1.15 times from the initial volume.
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You need to purchase an expensive chemical for a series of enzyme assays. Sigma Chemical Company sells the chemical in 1g ($30), 5g ($120), 10g ($200), and 25g ($400) vials. You will perform 40 assays. Each assay requires 3 ml of a 250mM solution of the chemical (M.W. = 156 g/mole). Which vial(s) is the best choice for your limited budget?
Answer:
5g ($120)
Explanation:
The amount of chemical needed for assay is 3 ml of 250 mM solution of a chemical
Molarity (M) = number of moles / volume in L
number of mole = M × volume in L
M of the chemical = 250 mM = 250 / 1000 = 0.25 M
number of moles = 0.25 × ( 3 / 1000) in L = 0.00075 m
mass of the chemical needed = 0.00075 × molar mass = 0.00075 × 156 = 0.117 g for each assay.
mass needed for 40 assay = 40 × 0.117 = 4.68 g
It is therefore wise for him to buy the 5 g ( $ 120), though the 10 g and 25 g yields better prices per gram, they much more than what he needed for the assay.
Calculate the speed of sound at 288 K in hydrogen, helium, and nitrogen. Under what conditions will the speed of sound in hydrogen be equal to that in helium?
Explanation:
The sped of sound is given as follows.
C = [tex]\sqrt{\gamma RT}[/tex]
It is known that for hydrogen,
R = 4124 J/kg K
T = 288 k
[tex]\gamma[/tex] = 1.41
Therefore, calculate the value of [tex]C_{hydrogen}[/tex] as follows.
[tex]C_{hydrogen} = \sqrt{\gamma RT}[/tex]
= [tex]\sqrt{1.41 \times 4124 J/kg K \times 288}[/tex]
= 1294.1 m/s
For helium,
R = 2077 J/kg K
T = 288 k
[tex]\gamma[/tex] = 1.66
Therefore, calculate the value of [tex]C_{helium}[/tex] as follows.
[tex]C_{helium} = \sqrt{\gamma RT}[/tex]
= [tex]\sqrt{1.66 \times 2077 J/kg K \times 288}[/tex]
= 996.48 m/s
For nitrogen,
R = 296.8 J/kg K
T = 288 k
[tex]\gamma[/tex] = 1.4
Therefore, calculate the value of [tex]C_{hydrogen}[/tex] as follows.
[tex]C_{hydrogen} = \sqrt{\gamma RT}[/tex]
= [tex]\sqrt{1.4 \times 296.8 J/kg K \times 288}[/tex]
= 345.93 m/s
So, speed of sound in hydrogen is calculated as follows.
= [tex]\sqrt{1.41 \times 4124 \times T_{H}}[/tex]
= [tex]76.26 \sqrt{T_{H}}[/tex]
Speed of sound in helium is as follows.
= [tex]\sqrt{1.66 \times 2077 \times T_{He}}[/tex]
= [tex]58.72 \sqrt{T_{He}}[/tex]
For both the speeds to be equal,
[tex]76.26 \sqrt{T_{H}}[/tex] = [tex]58.72 \sqrt{T_{He}}[/tex]
[tex]\frac{T_{H}}{T_{He}}[/tex] = 0.593
Therefore, we can conclude that the temperature of hydrogen is 0.593 times the temperature of helium.
Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely? Hint: Write a balanced chemical reaction first, then find which one is the limiting reactant.
Answer:
16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:
[tex]Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)[/tex]
Moles of calcium nitrate = [tex]\frac{31.3 g}{164 g/mol}=0.1908 mol[/tex]
Moles of ammonium fluoride = [tex]\frac{38.7 g}{37 g/mol}=1.046 mol[/tex]
According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.
Then 1.046 moles of ammonium fluoride will react with :
[tex]\frac{1}{2}\times 1.046 mol=0.523 mol[/tex] calcium nitarte .
This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.
Hence, calcium nitrate is a limiting reactant.
So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.
So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .
Then 0.1908 moles of calcium nitrate will give:
[tex]\frac{2}{1}\times 0.1908 mol=0.3816 mol[/tex]of dinitrogen monoxide gas.
Mass of 0.03816 moles of dinitrogen monoxide gas:
0.03816 mol × 44 g/mol = 16.79 g
16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.
After the reaction is complete the mass of dinitrogen monoxide N2O 8.39 grams.
let's follow these steps.
Step 1. Write the balanced chemical equation.
The reaction given is between calcium nitrate [tex]\( \text{Ca(NO}_3\text{)}_2 \)[/tex] and ammonium fluoride [tex]\( \text{NH}_4\text{F} \)[/tex], producing calcium fluoride [tex]\( \text{CaF}_2 \),[/tex] dinitrogen monoxide nitrous oxide, [tex]\ \text{N}_2\text{O} \)[/tex], and water vapor [tex]\( \text{H}_2\text{O} \)[/tex]
The balanced chemical equation is.
[tex]\[ \text{Ca(NO}_3\text{)}_2 + 2 \text{NH}_4\text{F} \[/tex] right arrow [tex]\text{CaF}_2 + \text{N}_2\text{O} + 4 \text{H}_2\text{O} \][/tex]
Step 2. Calculate the molar masses.
- Molar mass of. [tex]\( \text{Ca(NO}_3\text{)}_2 \)[/tex]
[tex]\[ \text{Ca}[/tex] = [tex]40.08 \, \text{g/mol} \][/tex]
[tex]\[ \text{N} = 14.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{O} = 16.00 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } \text{Ca(NO}_3\text{)}_2 = 40.08 + 2 \cdot (14.01 + 3 \cdot 16.00) = 164.10 \, \text{g/mol} \][/tex]
- Molar mass of [tex]\( \text{NH}_4\text{F} \):[/tex]
[tex]\[ \text{N} = 14.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{H} = 1.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{F} = 19.00 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } \text{NH}_4\text{F} = 14.01 + 4 \cdot 1.01 + 19.00 = 37.05 \, \text{g/mol} \][/tex]
- Molar mass of [tex]\( \text{N}_2\text{O} \)[/tex].
[tex]\[ \text{N} = 14.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{O} = 16.00 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } \text{N}_2\text{O} = 2 \cdot 14.01 + 16.00 = 44.02 \, \text{g/mol} \][/tex]
Step 3. Determine the limiting reactant.
Calculate the number of moles for each reactant.
- Moles of[tex]\( \text{Ca(NO}_3\text{)}_2 \).[/tex]
[tex]\[ \text{Moles} = \frac{31.3 \, \text{g}}{164.10 \, \text{g/mol}} \approx 0.1908 \, \text{mol} \][/tex]
- Moles of[tex]\( \text{NH}_4\text{F} \):[/tex]
[tex]\[ \text{Moles} = \frac{38.7 \, \text{g}}{37.05 \, \text{g/mol}} \approx 1.0451 \, \text{mol} \][/tex]
According to the balanced equation,[tex]\( \text{Ca(NO}_3\text{)}_2 \) and \( \text{NH}_4\text{F} \)[/tex] react in a 1:2 molar ratio. Therefore, [tex]\( \text{Ca(NO}_3\text{)}_2 \)[/tex] is the limiting reactant because it produces fewer moles of products compared to [tex]\( \text{NH}_4\text{F} \).[/tex]
Step 4. Calculate the mass of [tex]\( \text{N}_2\text{O} \)[/tex] produced.
From the balanced equation,[tex]\( 1 \) mole of \( \text{Ca(NO}_3\text{)}_2 \) produces \( 1 \) mole of \( \text{N}_2\text{O} \).[/tex]
- Moles of [tex]\( \text{N}_2\text{O} \)[/tex] produced.
[tex]\[ \text{Moles} = 0.1908 \, \text{mol} \][/tex]
- Mass of [tex]\( \text{N}_2\text{O} \).[/tex]
[tex]\[ \text{Mass} = \text{Moles} \times \text{Molar mass of } \text{N}_2\text{O} \][/tex]
[tex]\[ \text{Mass} = 0.1908 \, \text{mol} \times 44.02 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass} = 8.39 \, \text{g} \][/tex]
A West Virginia coal is burned at a rate of 8.02 kg/s. The coal has a sulfur content of 4.40 % and the bottom ash contains 2.80 % of the input sulfur. What is the annual rate (in kg/yr) of stack emissions of SO2
Answer: The annual emission rate of SO2 is 1.08 × [tex]10^{7}[/tex] kg/yr
Explanation:
The rate r at which the coal is been burnt is 8.02 kg/s.Amount of sulphur in the burning coal is given as 4.40 %i.e., 4.4/100 × 8.02 = 0.353 kg/s. Which is equivalent to the rate at which the sulphur is been burnt.
Since the burning of sulphur oxidizes it to produce SO2, it follows that the non-oxidized portion of the sulphur will go with the bottom ash.The bottom ash is said to contain 2.80 % of the input sulphur.Hence the portion of the SO2 produced is 100 — 2.80 = 97.20 %.The rate of the SO2 produced is percentage of SO2 × rate at high sulphur is been burnt.= 97.20/100 × 0.353 kg/s.
= 0.343 kg/s.
To get the annual emission rate of SO2, we convert the kg/s into kg/yr.
1 kg/s = 1 kg/s × (60 × 60 × 24 × 365) s/yr
1 kg/s = 31536000 kg/yr
Therefore, 0.343 kg/s = 0.343 × 31536000 kg/yr
= 10816848 kg/yr
= 1.08 × 10^7 kg/yryr.
If a reaction of 5.0 g of hydrogen with 5.0 g of carbon monoxide produced 4.5 g of methanol, what was the percent yield?
Answer:
The percentage yield of methanol is 78.74%.
Explanation:
[tex]2H_2+CO\rightarrow CH_3OH[/tex]
Theoretical yield of methanol ;
Moles of hydrogen gas = [tex]\frac{5.0 g}{2 g/mol}=2.5 mol[/tex]
Moles of carbon monoxide = [tex]\frac{5.0 g}{28 g/mol}=0.1786 mol[/tex]
According to reaction ,1 mole of CO reacts with 2 moles of hydrogen gas. Then 0.1786 moles of CO will :
[tex]\frac{2}{1}\times 0.1786 mol=0.0893 mol[/tex]
As we can see, that moles of hydrogen gas are in excess and CO are in limiting amount, so amount of methanol will depend upon moles of CO.
According to recation 1 mole of CO gives 1 mole of methanol, then 0.1786 moles of CO will give:
[tex]\frac{1}{1}\times 0.1786 mol=0.1786 mol[/tex]
Mass of 0.1786 moles of methanol =
= 0.1786 mol × 32 g/mol = 5.7152 g
Theoretical yield of methanol = 5.7152 g
Experimental yield of methanol = 4.5 g
Percentage yield of methanol:
To calculate the percentage yield , we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
[tex]=\frac{4.5 g}{5.7152 g}\times 100=78.74\%[/tex]
The percentage yield of methanol is 78.74%.
The percent yield of the reaction is approximately 78.53%.
First, we need to write the balanced chemical equation for the reaction between hydrogen [tex](H_2)[/tex] and carbon monoxide (CO) to produce methanol [tex](CH_3OH)[/tex]:
[tex]\[ 2H_2 + CO \rightarrow CH_3OH \][/tex]
From the stoichiometry of the balanced equation, we can see that 2 moles of hydrogen react with 1 mole of carbon monoxide to produce 1 mole of methanol.
Now, we calculate the moles of hydrogen and carbon monoxide that reacted:
Molar mass of hydrogen [tex](H_2)[/tex] = 2 [tex]\times[/tex] 1.008 g/mol = 2.016 g/mol
Moles of hydrogen = mass / molar mass = 5.0 g / 2.016 g/mol = 2.48 moles
Molar mass of carbon monoxide (CO) = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Moles of carbon monoxide = mass / molar mass = 5.0 g / 28.01 g/mol = 0.179 moles
Since the reaction requires a 2:1 ratio of hydrogen to carbon monoxide, and we have fewer moles of carbon monoxide, carbon monoxide is the limiting reactant.
The balanced equation tells us that 1 mole of CO produces 1 mole of methanol. Therefore, the theoretical yield of methanol is equal to the moles of CO that reacted:
Theoretical yield in moles = 0.179 moles
Now, we convert the moles of methanol to grams using its molar mass:
Molar mass of methanol [tex](CH_3OH)[/tex] = 12.01 g/mol (C) + 4 \times 1.008 g/mol (H) + 16.00 g/mol (O) = 32.042 g/mol
Theoretical yield in grams = theoretical yield in moles \times molar mass = 0.179 moles \times 32.042 g/mol = 5.73 g
The actual yield of the reaction is given as 4.5 g of methanol.
Now, we can calculate the percent yield:
[tex]\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \][/tex]
[tex]\[ \text{Percent Yield} = \left( \frac{4.5 \text{ g}}{5.73 \text{ g}} \right) \times 100\% \][/tex]
[tex]\[ \text{Percent Yield} \approx 78.53\% \][/tex]
Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container of volume 0.500 liters.At this temperature, the root mean square speed of methanol molecules is 605 m/sec.What is the average pressure in the container due to these 75 molecules?(The molar mass of methanol is 32.0 g/mol.)
Answer:
The average pressure in the container due to these 75 gas molecules is [tex]P=9.72 \times 10^{-16} Pa[/tex]
Explanation:
Here Pressure in a container is given as
[tex]P=\frac{1}{3} \rho <u^2>[/tex]
Here
P is the pressure which is to be calculatedρ is the density of the gas which is to be calculated as below[tex]\rho =\frac{mass}{Volume of container}[/tex]
Here
mass is to be calculated for 75 gas phase molecules as
[tex]m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times 10^{-21} g[/tex]
Volume of container is 0.5 lts
So density is given as
[tex]\rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3[/tex]
[tex]<u^2>[/tex] is the mean squared velocity which is given as[tex]<u^2>=RMS^2[/tex]
Here RMS is the Root Mean Square speed given as 605 m/s so
[tex]<u^2>=RMS^2\\<u^2>=(605)^2\\<u^2>=366025[/tex]
Substituting the values in the equation and solving
[tex]P=\frac{1}{3} \rho <u^2>\\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa[/tex]
So the average pressure in the container due to these 75 gas molecules is [tex]P=9.72 \times 10^{-16} Pa[/tex]
As an athlete exercises, sweat is produced and evaporated to help maintain a proper body temperature. On average, an athlete loses approximately 443 g of sweat during an hour of exercise. How much energy is needed to evaporate the sweat that is produced? The heat of vaporization for water is 2257 J/g.
Answer: 999851 Joules of energy is needed to evaporate the sweat that is produced
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 gram of liquid water to gaseous form at its boiling point.
Given: The heat of vaporization for water is 2257 J/g.
Thus if for 1 g , the heat required is = 2257 J
For 443 g , heat required is =[tex]\frac{2257}{1}\times 443=999851J[/tex]
Thus 999851 Joules of energy is needed to evaporate the sweat that is produced
To calculate the energy required to evaporate the sweat produced during exercise, we multiply the mass of the sweat (443 g) by the heat of vaporization for water (2257 J/g). The result is approximately 999,791 Joules.
Explanation:To calculate the amount of energy required to evaporate the sweat that an athlete produces during exercise, we can use the formula energy = mass × heat of vaporization. Given that the heat of vaporization for water is 2257 J/g and the average sweat loss is approximately 443 g, we can substitute these values into the formula: energy = 443 g × 2257 J/g = 999791 J. So, on average, an athlete needs about 999,791 Joules of energy to evaporate the sweat produced in an hour of exercise.
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You measure 3.03 mL of a 50% NaOH solution by weight (its density is 1.53 g mL-1) and dilute it to 500 mL total volume. What is the concentration of this NaOH solution? Write your answer to four decimal places (X.XXXX).
To calculate the concentration of the NaOH solution, convert the volume to grams and calculate the moles of NaOH. Then, use the moles and total volume to find the concentration.
Explanation:To find the concentration of the NaOH solution, we need to calculate the moles of NaOH present in the 3.03 mL volume. First, we convert the volume to grams using the density of the solution: 3.03 mL × 1.53 g/mL = 4.6419 g
Next, we calculate the moles of NaOH:
Moles of NaOH = \dfrac{4.6419 g}{40.00 g/mol} = 0.1160 mol
Finally, we use the moles of NaOH and the total volume of the solution (500 mL) to find the concentration:
Concentration = \dfrac{0.1160 mol}{0.5000 L} = 0.2320 M
Final answer:
The concentration of the diluted NaOH solution after measuring 3.03 mL of a 50% NaOH solution, with a density of 1.53 g/mL, and diluting to 500 mL total volume is 0.1159 M.
Explanation:
To calculate the concentration of the diluted NaOH solution, we need to determine the amount of NaOH in the initial 3.03 mL of 50% solution and then find the molarity after dilution to 500 mL. The density of the solution is given as 1.53 g/mL, so the mass of the 3.03 mL solution is:
mass = volume × density = 3.03 mL × 1.53 g/mL = 4.6359 g
Since the solution is 50% by weight, the mass of NaOH is:
mass of NaOH = 0.50 × 4.6359 g = 2.3180 g
The molar mass of NaOH is approximately 40.00 g/mol, so the number of moles of NaOH in the original solution is:
moles of NaOH = mass / molar mass = 2.3180 g / 40.00 g/mol = 0.05795 mol
After dilution to 500 mL, the molarity (M) is calculated as follows:
Molarity = moles / volume (L) = 0.05795 mol / 0.50 L = 0.1159 M
Therefore, the concentration of the diluted NaOH solution is 0.1159 M.
How many inner, outer, and valence electrons are present in an atom of each of the following elements?
(a) O (b) Sn (c) Ca (d) Fe (e) Se
Oxygen has 2 inner electrons, 6 outer electrons, and 6 valence electrons. Tin has 40 inner electrons, 10 outer electrons, and 10 valence electrons. Calcium has 18 inner electrons, 2 outer electrons, and 2 valence electrons. Iron has 18 inner electrons, 8 outer electrons, and 8 valence electrons. Selenium has 26 inner electrons, 6 outer electrons, and 6 valence electrons.
Explanation:Let's determine the number of inner, outer, and valence electrons for each element:
Oxygen (O): Atomic number 8 means it has 8 electrons. The distribution is 2 electrons in the first energy level, and the remaining 6 in the second energy level. Outer electrons: 6, Inner electrons: 2, Valence electrons: 6.Tin (Sn): Atomic number 50 means it has 50 electrons. The distribution is 2 electrons in the first energy level, 8 electrons in the second energy level, 18 electrons in the third energy level, and the remaining 22 in the fourth energy level. Outer electrons: 10, Inner electrons: 40, Valence electrons: 10.Calcium (Ca): Atomic number 20 means it has 20 electrons. The distribution is 2 electrons in the first energy level, 8 electrons in the second energy level, and the remaining 10 in the third energy level. Outer electrons: 2, Inner electrons: 18, Valence electrons: 2.Iron (Fe): Atomic number 26 means it has 26 electrons. The distribution is 2 electrons in the first energy level, 8 electrons in the second energy level, 14 electrons in the third energy level, and the remaining 2 in the fourth energy level. Outer electrons: 8, Inner electrons: 18, Valence electrons: 8.Selenium (Se): Atomic number 34 means it has 34 electrons. The distribution is 2 electrons in the first energy level, 8 electrons in the second energy level, 16 electrons in the third energy level, and the remaining 8 in the fourth energy level. Outer electrons: 6, Inner electrons: 26, Valence electrons: 6.Learn more about Electron distribution here:https://brainly.com/question/37566080
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The number of electrons in an atom can be determined by the electron configuration. For example, Oxygen has 2 inner, 6 outer, and 6 valence electrons while Tin has 50 inner, 14 outer, and 4 valence electrons.
Explanation:The electron configuration can be used to determine the number of inner, outer, and valence electrons in an atom.
O (Oxygen) has 2 inner electrons (in the 1s orbital), 6 outer electrons (in the 2s and 2p orbitals), and 6 valence electrons (also in the 2s and 2p orbitals). Sn (Tin) has 50 inner electrons, 14 outer electrons, and 4 valence electrons. Ca (Calcium) has 18 inner electrons, 2 outer electrons, and 2 valence electrons. Fe (Iron) has 18 inner electrons, 8 outer electrons, and 2 valence electrons. Se (Selenium) has 28 inner electrons, 6 outer electrons, and 6 valence electrons. Learn more about Electron Configuration here:
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If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of TES is 7.55.)
Answer:
The new pH after adding HCl is 7.07
Explanation:
The formula for calculating pH of a buffer is
pH = pKa + log([Conjugate base]/[Acid])
Before adding HCl,
7.55 = 7.55 + log([Conjugate base]/[Acid])
⇔ log([Conjugate base]/[Acid]) = 0
⇔ [Conjugate base] = [Acid] = 1/2 x 0.100 = 0.05 M
⇒ Mole of Conjugate base = Mole of Acid = 0.05 M x 0.12 mL = 0.006 mol
After adding HCl (3.00 mL, 1.00 M)
⇒ Mole of HCl = 0.003 x 1 = 0.003 mol)
New volume solution is 120 m L+ 3 mL = 123 mL
HCl is a strong acid, it will convert the conjugate base to acid form, or we can express
Mole of new Conjugate base = 0.006 - 0.003 = 0.003 mol
⇒ Concentration = 0.003/0.123 M
Mole of new Acid form = 0.006 + 0.003 = 0.009 mol
⇒ Concentration = 0.009/0.123 M
Use the formula
pH = pKa + log([Conjugate base]/[Acid])
= 7.55 + log(0.003 / 0.009) = 7.07
To determine the new pH after adding HCl to the TES buffer, we need to consider the Henderson-Hasselbalch equation. Initially, the concentration of TES is 0.100 M and the pH is 7.55. After adding 3.00 mL of 1.00 M HCl, we need to calculate the new concentration of TES and its conjugate base. Using the Henderson-Hasselbalch equation, we can find the new pH by substituting the new concentration of TES, its conjugate base, and the pKa of TES into the equation.
Explanation:To determine the new pH after adding HCl to the TES buffer, we need to consider the Henderson-Hasselbalch equation. The equation relates the pH of a buffer solution to the pKa of the acid and the ratio of the concentrations of the acid and its conjugate base. In this case, TES acts as the acid and its conjugate base is the salt.
The Henderson-Hasselbalch equation can be written as pH = pKa + log([A-]/[HA]), where [A-]/[HA] is the ratio of the salt to the acid. Initially, the concentration of TES is 0.100 M and the pH is 7.55. After adding 3.00 mL of 1.00 M HCl, we need to calculate the new concentration of TES and its conjugate base.
Using the equation c1V1 = c2V2, where c1 and V1 are the initial concentration and volume of HCl, and c2 and V2 are the final concentration and volume, we can find the new concentration of TES. The initial volume of TES is 120 mL. The final volume is the sum of the initial volume and the volume of HCl added. Calculate the new volume of TES from this information, and then substitute the values into the equation to find the new concentration of TES.
Once we have the new concentration of TES, we can use the Henderson-Hasselbalch equation to find the new pH. Substitute the new concentration of TES and the concentration of its conjugate base into the equation, along with the pKa of TES. Solve for the new pH to determine the answer.
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The nonvolatile, nonelectrolyte sucrose, C12H22O11 (342.3 g/mol), is soluble in water H2O.
Calculate the osmotic pressure (in atm) generated when 12.8 grams of sucrose are dissolved in 278 mL of a water solution at 298 K.
Answer: The osmotic pressure of the solution is 3.29 atm
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
or,
[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = ?
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of sucrose = 12.8 grams
Molar mass of sucrose = 342.3 g/mol
Volume of solution = 278 mL
R = Gas constant = [tex]0.082\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = 298 K
Putting values in above equation, we get:
[tex]\pi=1\times \frac{12.8\times 1000}{342.3\times 278}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\pi=3.29atm[/tex]
Hence, the osmotic pressure of the solution is 3.29 atm
What is the molarity of the resulting solution when 23.640 g of Mn(ClO4)2 · 6 H2O are added to 200.0 mL of water? WebAssign will check your answer for the correct number of significant figures. .326 Incorrect: Your answer is incorrect.
The molarity of the solution when 23.640 g of Mn(ClO4)2 · 6 H2O is added to 200.0 mL of water is 0.3014 M, calculated by dividing the moles of solute by the volume of solution in liters.
Explanation:To calculate the molarity of a solution when a certain amount of solute is added to a known volume of water, you must first determine the number of moles of solute. For the molarity of the resulting solution when 23.640 g of Mn(ClO4)2 · 6 H2O (molar mass approximately 392.13 g/mol) is dissolved in 200.0 mL of water, the calculation is as follows:
Calculate the moles of Mn(ClO4)2 · 6 H2O using its molar mass.Divide the moles by the volume of the solution in liters to find the molarity.To calculate the moles of Mn(ClO4)2 · 6 H2O: 23.640 g / 392.13 g/mol ≈ 0.06028 mol.
To find molarity (M), which is moles of solute per liter of solution: 0.06028 mol / 0.200 L = 0.3014 M.
A chemist dissolves of pure barium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (The temperature of the solution is .) Round your answer to significant decimal places.
Answer:
12.45
Explanation:
There is some info missing. I think this is the original question.
A chemist dissolves 169 mg of pure barium hydroxide in enough water to make up 70 mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Round your answer to 2 significant decimal places.
First, we will calculate the molarity of barium hydroxide.
M = mass / molar mass × liters of solution
M = 0.169 g / 171.34 g/mol × 0.070 L
M = 0.014 M
Barium hydroxide is a strong base that dissociates according to the following equation.
Ba(OH)₂ → Ba²⁺ + 2 OH⁻
The molar ratio of Ba(OH)₂ to OH⁻ is 1:2. The concentration of OH⁻ is 2 × 0.014 M = 0.028 M
The pOH is:
pOH = -log [OH⁻] = - log 0.028 = 1.55
The pH is:
pH = 14 - pOH = 14 - 1.55 = 12.45
For each of the following, give the sublevel designation, the allowable ml values, and the number of orbitals:
(a) n = 4, l = 2
(b) n = 5, l = 1
(c) n = 6, l = 3
Answer :
(a) n = 4, l = 2
At l = 2, [tex]m_l=+2,+1,0,-1,-2[/tex]
Number of orbitals = 5
(b) n = 5, l = 1
At l = 1, [tex]m_l=+1,0,-1[/tex]
Number of orbitals = 3
(c) n = 6, l = 3
At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
Number of orbitals = 7
Explanation:
There are 4 quantum numbers :
Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of [tex]m_l[/tex] will be -2, -1, 0, +1, +2.
Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.
(a) n = 4, l = 2
At l = 2, [tex]m_l=+2,+1,0,-1,-2[/tex]
Number of orbitals = 5
(b) n = 5, l = 1
At l = 1, [tex]m_l=+1,0,-1[/tex]
Number of orbitals = 3
(c) n = 6, l = 3
At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
Number of orbitals = 7
A first order reaction has a rate constant of 0.816 at 25 oC. Given that the activation energy is 22.7 kJ/mol, calculate the rate constant at 34 oC. (enter answer to 3 decimal places)
Answer:
1.067
Explanation:
Using the Arrhenius equation relates rate constants to the temperature is:
ln(k2/k1) = −Ea/R * [1/T2−1/T1]
where,
Ea = activation energy in kJ/mol ,
R = universal gas constant, and
T = temperature in K .
T1 = 25 + 273.15
= 298.15 K
T2 = 34 + 273.15
= 307.15 K
k1 = 0.816
Therefore,
k2/k1 = exp(-22.7/0.008314472) * [1/307.15 −1/298.15 ])
= 0.816 * 1.308
k2 = 1.067.
Consider the following multistep reaction:
C+D⇌CD(fast)
CD+D→CD2(slow)
CD2+D→CD3
Based on this mechanism, determine the rate law for the overall reaction.
The question is incomplete, here is the complete question:
Consider the following multistep reaction:
C+D⇌CD (fast)
CD+D→CD₂ (slow)
CD₂+D→CD₃ (fast)
C+3D→CD₃
Based on this mechanism, determine the rate law for the overall reaction.
Answer: The rate law for the reaction is [tex]\text{Rate}=k'[C][D]^2[/tex]
Explanation:
Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.
In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.
For the given chemical reaction:
[tex]C+3D\rightarrow CD_3[/tex]
The intermediate reaction of the mechanism follows:
Step 1: [tex]C+D\rightleftharpoons CD;\text{ (fast)}[/tex]
Step 2: [tex]CD+D\rightarrow CD_2;\text{(slow)}[/tex]
Step 3: [tex]CD_2+D\rightarrow CD_3;\text{(fast)}[/tex]
As, step 2 is the slow step. It is the rate determining step
Rate law for the reaction follows:
[tex]\text{Rate}=k[CD][D][/tex] ......(1)
As, [CD] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.
Applying steady state approximation for CD from step 1, we get:
[tex]K=\frac{[CD]}{[C][D]}[/tex]
[tex][CD]=K[C][D][/tex]
Putting the value of [CD] in equation 1, we get:
[tex]\text{Rate}=k.K[C][D]^2\\\\\text{Rate}=k'[C][D]^2[/tex]
Hence, the rate law for the reaction is [tex]\text{Rate}=k'[C][D]^2[/tex]
The following questions deal with safety issues in the Solutions and Spectroscopy experiment. (Select all that apply.) (a) Which hazards are associated with copper(II) sulfate? carcinogenic irritant toxic corrosive flammable (b) What actions should you take if you spill copper(II) sulfate solution on yourself? Dry the area with paper towels, but avoid getting it wet. Neutralize it with a solution of sodium bicarbonate. Neutralize it with a solution of sodium bisulfite. Flush the affected area with water. (c) What clothing is appropriate for the laboratory? shorts tank tops without sleeves shirt and pants that overlap 6" sandals pants that extend to the ankles shoes that cover heels and toes a shirt that covers the shoulders (has sleeves)
Answer:
(a) Which hazards are associated with copper(II) sulfate?
irritant toxic(b) What actions should you take if you spill copper(II) sulfate solution on yourself?
Flush the affected area with water.(c) What clothing is appropriate for the laboratory?
shirt and pants that overlap 6" pants that extend to the ankles shoes that cover heels and toes a shirt that covers the shoulders (has sleeves)Explanation:
Copper sulfate (ii) is a toxic compound by ingestion. Irritant in prolonged skin contact.
In case of skin contact, contaminated clothing and shoes should be removed, wash the affected areas with soap and large amounts of water. Seek medical assistance.
The shoe is recommended that this be completely closed, since if any substance falls do not burn the feet, and comfortable to perform the practice in the best way.
The clothes to be used in a laboratory should cover as much skin as possible, in addition it is necessary to use a cotton lab coat, long sleeve and that reaches the knees.
Copper II sulfate is a toxic irritant, if ever your skin comes in contact with it flush the affected area with water.
How to avoid accidents in the laboratory?In labs, there are many chemicals that can harm the student while haphazardly handling them.
For example - Copper II sulfate is a toxic irritant, if ever your skin comes in contact with it flush the affected area with water.
The students should wear full sleeve shirts, overlapping with pants and shoes that cover the ankle and toe.
Therefore, Copper II sulfate is a toxic irritant, if ever your skin comes in contact with it flush the affected area with water.
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The angle at which light strikes a surface is the same as the angle at which it is reflected. true or false
Answer: True
Explanation: the angle of incident Ray equals the angle of the reflected Ray.
As space exploration increases, means of communication with humans and probes on other planets are being developed. (a) How much time (in s) does it take for a radio wave of frequency 8.93 x 10⁷ s⁻¹ to reach Mars, which is 8.1 x 10⁷ km from Earth? (b) If it takes this radiation 1.2 s to reach the Moon, how far (in m) is the Moon from Earth?
Answer:
Explanation:
a ) Speed of radio waves in space = speed of light in space
= 3 x 10⁸ m /s
Time taken = distance / speed
= 8.1 x 10⁷ x 10³ / 3 x 10⁸ s
= 2.7 x 10² s
= 270 s
b )
Distance = speed x time
= 3 x 10⁸ x 1.2 m
= 3.6 x 10⁸ m
= 3.6 x 10⁵ km
Write the ground-state electron configurations of the following transition metal ions. (a) Sc3 (b) Cr3 (c) Cu (d) Au
The ground-state electron configurations for Sc3+, Cr3+, Cu, and Au are [Ar], [Ar]18 3d3, [Ar]18 3d10 4s1, and [Xe]6s1 4f14 5d10 respectively. Electrons are removed from the 4s orbital first and then from 3d in transition metals.
Explanation:The ground-state electron configurations of the transition metal ions are expressed as follows:
Sc3+: [Ar] (Scandium loses three electrons from 4s and 3d orbitals.)Cr3+: [Ar]18 3d3 (Chromium ion loses one electron from 4s and two from 3d orbital.)Cu: [Ar]18 3d10 4s1 (Copper has one electron in 4s and 10 in 3d in its ground state.) Au: [Xe]6s1 4f14 5d10 (Gold has one electron in 6s, ten in 5d, and 14 in 4f in its ground state.)It's essential to remember that for transition metals, electrons are removed from the 4s orbital first and then from the 3d orbital.
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The ground-state electron configurations for Sc3+, Cr3+, Cu, and Au ions are [Ar], [Ar] 3d3, [Ar] 3d10, and [Xe] 4f14 5d9, respectively, with the higher energy electrons being removed first to form the ions.
To write the ground-state electron configurations of the given transition metal ions, we must understand how electrons are removed when forming an ion. For transition metals, electrons are removed from the s orbital before the d orbital. Now let's determine the electron configurations:
Sc3+: The atomic number of Scandium (Sc) is 21. The electron configuration for Sc is [Ar] 3d1 4s2. When forming Sc3+, we remove three electrons, two from the 4s orbital and one from the 3d orbital, resulting in [Ar] as the electron configuration for Sc3+.Cr3+: Chromium (Cr) has an atomic number of 24. Its electron configuration is [Ar] 3d5 4s1, but upon forming Cr3+, three electrons are removed (one from the 4s orbital and two from the 3d orbital), leading to [Ar] 3d3 as the electron configuration for Cr3+.Cu: Copper (Cu) has an atomic number of 29. The electron configuration of Cu is [Ar] 3d10 4s1. In its monatomic ion form (Cu+), only one electron (from the 4s orbital) is removed, resulting in [Ar] 3d10 as the electron configuration for Cu+.Au: Gold (Au) has an atomic number of 79. Its electron configuration is [Xe] 4f14 5d10 6s1. The most common monatomic ion of gold is Au3+, which means three electrons are removed (two from the 6s orbital and one from the 5d orbital), giving [Xe] 4f14 5d9 as the electron configuration for Au3+.The atomic masses of two isotopes of silver are 105 and 111. The chemical symbol for silver is Ag. (a) What are the number of protons and neutrons in each of the two isotopes?(b) What is the number of orbiting electrons in each of the two isotopes when each is electrically neutral?
Answer:
A. 47 protons and 58 neutrons, 47 protons and 64 neutrons
B. 47 electrons
Explanation:
While it is possible for two isotopes of a particular element to have different atomic masses, it is. It is impossible for the number of protons or the atomic number of the isotopes to be different.
To answer this question properly, we would be needing the proton number or the atomic number of silver. The atomic number of silver is 47. Let’s say the isotopes are A and B respectively.
The number of protons in the same is equal which is 47. The number of neutrons is different and can be obtained by subtracting the number of protons from the mass number.
For A: neutrons = 105 - 47 = 58
For B: neutrons = 111 - 47 = 64
Since Both are electrically neutral, the number of orbiting electrons is also equals the number of protons which is equal to 47.
Silver (Ag) has 47 protons. Its isotopes Ag-105 and Ag-111 have 58 and 64 neutrons respectively. In neutral conditions both isotopes have 47 electrons.
Explanation:The atomic number for silver (Ag) is 47, which means it has 47 protons. Isotopes of a particular element have the same number of protons but different number of neutrons. Isotope Ag-105 will have 47 protons and 105 - 47 = 58 neutrons, and Isotope Ag-111 will have 47 protons and 111 - 47 = 64 neutrons. In a state of electrical neutrality (no charge), an atom will have the same number of protons and electrons. Thus, both Isotopes Ag-105 and Ag-111 will each have 47 electrons.
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How many moles of sebacoyl chloride do you have if you measure out 5 mL of a 9% volume/volume solution in cyclohexane?
Answer:
moles sebacoyl chloride = 2.1 x 10-3 mol
Explanation:
Concentration sebacoyl chloride = 9% (v/v) = 9 mL sebacoyl chloride / 100 mL solution Volume of solution measured = 5 mL volume of sebacoyl chloride = (Volume of solution measured) * (Concentration sebacoyl chloride) volume of sebacoyl chloride = (5 mL) * (9 mL sebacoyl chloride / 100 mL solution) volume of sebacoyl chloride = (5 * 9) / 100 mL volume of sebacoyl chloride = 0.45 mL mass sebacoyl chloride = (volume of sebacoyl chloride) * (density of sebacoyl chloride) mass sebacoyl chloride = (0.45 mL) * (1.121 g/mL) mass sebacoyl chloride = 0.50445 g moles sebacoyl chloride = (mass sebacoyl chloride) / (molar mass sebacoyl chloride) moles sebacoyl chloride = (0.50445 g) / (239.14 g/mol) moles sebacoyl chloride = 2.1 x 10-3 molCobalt-60 is a radioactive isotope used to treat cancers. A gamma ray emitted by this isotope has an energy of 1.33 MeV (million electron volts; 1 eV = 1.602 x 10¹⁹ J). What is the frequency (in Hz) and the wavelength (in m) of this gamma ray?
Answer:
E = 1.33 MeV = 2.13 x [tex]10^{-13}[/tex] J
v = wavelength = E / h = 2.13 x [tex]10^{-13}[/tex] / 6.626 x [tex]10^{-34}[/tex] = 3.2 x [tex]10^{20}[/tex] m
f = frequency = c / 3.2 x [tex]10^{20}[/tex] m = 3 x [tex]10^{8}[/tex] / 3.2 x [tex]10^{20}[/tex] = 9.375 x [tex]10^{-13}[/tex] Hz
Many calculators use photocells to provide their energy. Find the maximum wavelength needed to remove an electron from silver (Φ = 7.59 x 10⁻¹⁹ J). Is silver a good choice for a photocell that uses visible light?
Answer:
[tex]\lambda=2.61\times 10^{-9}\ m[/tex] = 261 nm
Silver is not a good choice.
Explanation:
[tex]E=\frac {h\times c}{\lambda}[/tex]
Where,
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
[tex]\lambda[/tex] is the wavelength of the light
Given that:- Energy = [tex]7.59\times 10^{-19}\ J[/tex]
[tex]7.59\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]
[tex]7.59\times \:10^{26}\times \lambda=1.99\times 10^{20}[/tex]
[tex]\lambda=2.61\times 10^{-9}\ m[/tex] = 261 nm
Visible range has a spectrum of 380 to 740 nm
So, Silver is not a good choice.
The maximum wavelength needed to remove an electron from silver is approximately 262 nm. Silver is not a good choice for a photocell that uses visible light.
Explanation:To find the maximum wavelength needed to remove an electron from silver, we can use the work function of silver, which is Φ = 4.73 eV. The threshold wavelength for observing the photoelectric effect in silver can be calculated using Equation 6.16, which is λ = hc/Φ. Substituting the given values, we have λ = (1240 eV⋅nm) / (4.73 eV), which gives us a threshold wavelength of approximately 262 nm. Since visible light ranges from 400 to 700 nm, silver is not a good choice for a photocell that uses visible light.
Learn more about photoelectric effect in silver here:https://brainly.com/question/35875610
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In a game of "Clue," Ms. White is killed in the conservatory. You have a device in each room to help you find the murderer— a spectrometer that emits the entire visible spectrum to indicate who is in that room. For example, if someone wearing yellow is in a room, light at 580 nm is reflected. The suspects are Col. Mustard, Prof. Plum, Mr. Green, Ms. Peacock (blue), and Ms. Scarlet. At the time of the murder, the spectrometer in the dining room recorded a reflection at 520 nm, those in the lounge and study recorded reflections of lower frequencies, and the one in the library recorded a reflection of the shortest possible wavelength. Who killed Ms. White? Explain.
Answer: Ms. Scarlet is the murderer.
Explanation: Each color has a wavelength band, some visible to the human eye (visible spectrum), some for the other animals. In this game of "Clue", the color are yellow (Col. Mustard), violet (Prof. Plum), green (Mr. Green), blue (Ms. Peacock) and red (Ms. Scarlet). Using the device and knowing each band, it's determined that at the time of murder, Mr. Green was in the dining room, because green light has the wavelength between 495 to 570nm; Prof. Plum was in the library, due to violet's light has the shortest wavelength; Col. Mustard and Ms. Peacock were either in the lounge or in the study, because of their color's light wavelength measurement. So, the person responsible for the murderer was Ms. Scarlet, as the other devices in the other places didn't register a higher wavelength, which is compatible to the red light.
According to the spectrometer readings, Ms. Peacock is the murderer of Ms. White in the conservatory, as the reflected colors in the other rooms correspond to the other characters' colors and leave only Ms. Peacock unaccounted for at the crime scene.
To determine who killed Ms. White in the conservatory in a game of "Clue" using a spectrometer, we need to match the color associated with the wavelength of light reflected in each room with the corresponding suspect's color. The dining room spectrometer recorded a reflection at 520 nm, which corresponds to green.
Since Mr. Green reflects green and that is his character color, he is in the dining room. The lounge and study recorded reflections of lower frequencies i.e., longer wavelengths than green, which suggest colors towards the red end of the spectrum, likely placing Ms. Scarlet (red) and Prof. Plum (purple) in those rooms.
The library recorded a reflection of the shortest possible wavelength, meaning violet, which is typically associated with Prof. Plum; however, since he is presumed to be in a room with a longer wavelength, that leaves Col. Mustard (who wears yellow) as the suspect in the library because yellow requires the absorption of wavelengths at the violet end of the spectrum.
Therefore, Ms. Peacock, whose color is blue and wasn't indicated in the other rooms, must have been in the conservatory and is determined to be the murderer according to the spectrometer readings.