Answer:
The correct answer is option C) "equilibrium price of peanut butter definitely falls".
Explanation:
The equilibrium price of a product is the market price established at a point where the quantity of products is equal to the quantity demanded by the consumers. In this case, the market of peanut butter is facing the decrease in the price of two substitutes of peanut butter: deli turkey slices and peanut brittle. As a result, the equilibrium price of peanut butter definitely falls since the quantity demanded by the consumers will certainly fall.
Final answer:
A decrease in the price of deli turkey slices and peanut brittle affects the demand and supply for peanut butter differently, making the final impact on the market for peanut butter ambiguous without further information.
Explanation:
The question considers the effects of changes in the prices of substitute goods in consumption (deli turkey slices) and production (peanut brittle) on the market for peanut butter. A decrease in the price of deli turkey slices, a substitute in consumption, would lead to consumers substituting away from peanut butter towards deli turkey slices, thereby decreasing the demand for peanut butter. Conversely, a decrease in the price of peanut brittle, a substitute in production, might signal that manufacturers could switch to producing more peanut brittle as it becomes more profitable, potentially decreasing the supply of peanut butter if resources are diverted. Consequently, the equilibrium quantity of peanut butter might increase or decrease, depending on the relative magnitudes of the shifts in supply and demand, while the equilibrium price might rise or fall for similar reasons. Therefore, without additional information on the extents of these changes, the final impact on the equilibrium price and quantity of peanut butter cannot be definitively determined, making option E the correct answer.
A 1.00-kg block of copper at 100°C is placed in an in- sulated calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0°C. Find the entropy change of (a) the cop- per block, (b) the water, and (c) the universe.
Answer:
the entropy change of the copper block = - 117.29 J/K
the entropy change of the water = 138.01 J/K
the entropy change of the universe = 20.72 J/K
Explanation:
For Copper block:
the mass of copper block [tex](m_c)[/tex] = 1.00 kg
Temperature of block of copper [tex](T_c)[/tex] = 100°C
= (100+273)K
= 373K
Standard Heat capacity for copper [tex](C_c)[/tex] = 386 J/kg.K
For water:
We know our volume of liquid water to be = 4.00 L
At 0.0°C Density of liquid water = 999.9 kg/m³
As such; we can determine the mass since : [tex]density = \frac{mass}{volume}[/tex]
∴ the mass of 4.00 L of liquid water at 0.0°C will be its density × volume.
= 999.9 kg/m³ × [tex]\frac{4}{1000}m^3[/tex]
= 3.9996 kg
so, mass of liquid water [tex](m_w)[/tex] = 3.9996 kg
Temperature of liquid water [tex](T_w)[/tex] at 0.0°C = 273 K
Standard Heat Capacity of liquid water [tex](C_w)[/tex] = 4185.5 J/kg.K
Let's determine the equilibruium temperature between the copper and the liquid water. In order to do that; we have:
[tex]m_cC_c \delta T_c =m_wC_w \delta T_w[/tex]
[tex]1.00*386*(373-T_\theta)=3.996*4185.5*(T _\theta-273)[/tex]
[tex]386(373-T_\theta)=16725.26(T_\theta-273)[/tex]
[tex](373-T_\theta)=\frac{16725.26}{386} (T_\theta-273)[/tex]
[tex](373-T_\theta)=43.33 (T_\theta-273)[/tex]
[tex](373-T_\theta)=43.33 T_\theta-11829.09[/tex]
[tex]373+11829.09=43.33 T_\theta+T_\theta[/tex]
[tex]12202.09 =43.33T_\theta[/tex]
[tex]T_\theta= 275.26 K[/tex]
∴ the equilibrium temperature = 275.26 K
NOW, to determine the Entropy change of the copper block; we have:
[tex](\delta S)_{copper}=m_cC_cIn(\frac{T_\theta}{T_c} )[/tex]
[tex](\delta S)_{copper}=1.0*386In(\frac{275.26}{373} )[/tex]
[tex](\delta S)_{copper}=-117.29 J/K[/tex]
The entropy change of the water can also be calculated as:
[tex](\delta S)_{water}=m_wC_wIn(\frac{T_\theta}{T_w} )[/tex]
[tex](\delta S)_{water}=3.9996*4185.5In(\frac{275.26}{373} )[/tex]
[tex](\delta S)_{water}=138.01J/K[/tex]
The entropy change of the universe is the combination of both the entropy change of copper and water.
[tex](\delta S)_{universe}=(\delta S)_{copper}+(\delta S)_{water}[/tex]
[tex](\delta S)_{universe}=(-117.29+138.01)J/K[/tex]
[tex](\delta S)_{universe}=20.72J/K[/tex]
A liquid is added to a buret, and the initial measurement is found to be 0.75 mL. After liquid has been added to the flask from the buret, the buret reads 15.20 mL. How much liquid (in mL) was dispensed from the buret, with the correct number of significant figures?
Answer:
14.45 mL
Explanation:
The rule apply for the addition and subtraction for significant digits is :
The least precise number present after the decimal point determines the number of significant figures in the answer.
Initial Burette reading = 0.75 mL ( 2 significant digits)
Final Burette reading = 15.20 mL ( 4 significant digits)
Liquid dispensed = 15.20 mL - 0.75 mL = 14.45 mL ( Answer to two decimal places )
To find the amount of liquid dispensed from the buret, subtract the initial volume (0.75 mL) from the final volume (15.20 mL), resulting in 14.45 mL of liquid dispensed. Ensure that the answer is reported with the same level of precision as the measurements from the buret.
Explanation:To calculate the amount of liquid dispensed from the buret, you subtract the initial measurement from the final measurement. In this case, the initial measurement is 0.75 mL and the final measurement after the liquid is added to the flask is 15.20 mL. Therefore, the amount of liquid dispensed is 15.20 mL - 0.75 mL = 14.45 mL.
Burets are commonly used in titration analyses and allow volume measurements to the nearest 0.01 mL, which is why our final answer should be reported with the same level of precision as the buret readings. According to the proper significant figures rules, the result should be reported with the same number of decimal places as the measurement with the fewest decimal places. Since the readings from the buret are given to two decimal places (0.75 mL and 15.20 mL), our calculated volume of dispensed liquid also should be reported to two decimal places as 14.45 mL.
Write the full ground-state electron configuration for each:
(a) S (b) Kr (c) Cs
Answer:
S: [Ne] 3s² 3p⁴
[Ar] 3d¹⁰ 4s² 4p⁶
[Xe] 6s1
Explanation:
The electronic configuration in ground state
In what sense is the wave motion of a guitar string analogous to the motion of an electron in an atom?
Explanation:
A guitar string vibrates when we strikes it. It starts vibrating in several modes simultaneously. It stretches between the saddle and the nut. This distance represents one-half wavelength.
Now if we consider that the string forms a circle, then we have an interpretation of an electron which vibrates in the orbit surrounding the nucleus. We are aware that electrons have wavelength. If the circumference of the orbit happens to be the integer multiple of wavelength , then the orbit is "allowed" since "the electron will retraces its own path."
This explains the line spectrum and not a continuous spectrum.
A line spectrum refers an electron that jumps between the specific energy levels, thus producing only specific colors.
An archer pulls her bowstring back 0.370 m by exerting a force that increases uniformly from zero to 264 N. (a) What is the equivalent spring constant of the bow?
Answer:
713.51 N/m
Explanation:
Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.
From hook's law,
F = ke ...........................Equation 1
Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.
Make k the subject of the equation,
k = F/e ............................ Equation 2
Given: F = 264 N, e = 0.37 m.
Substitute into equation 2
k = 264/0.37
k = 713.51 N/m
Hence the spring constant of the bow = 713.51 N/m
Analysis of a sample of a compound, containing only carbon, nitrogen, hydrogen, and oxygen, determined that it contained 20.0% C, 6.7% H, 46.6% N and the balance O. What is the empirical formula of the compound?
Answer:
CH₄N₂O
Explanation:
Let's assume 100 g of the compound, thus, the mass of each substance is it percent multiplied by 100:
C = 20.0 g (20% = 0.20, and 0.20*100 = 20.0)
H = 6.7 g
N = 46.6 g
O = 100 - (20+6.7+46.6) = 26.7 g
The number of moles of each compound is the mass divided by the molar mass. The molar masses are C = 12.011 g/mol, H = 1.00794 g/mol, N = 14.01 g/mol, and O = 15.999 g/mol, so:
nC = 20/12.011 = 1.67 mol
nH = 6.7/1.00794 = 6.65 mol
nN = 46.6/14.01 = 3.33 mol
nO = 26.7/15.999 = 1.67 mol
The empirical formula is the formula with the minimum possible number of the moles of each compound, which must be proportional to the percent of each one. So, we must divide each number of moles for the smallest, 1.67:
C = 1.67/1.67 = 1
H = 6.65/1.67 = 4
N = 3.33/1.67 = 2
O = 1.67/1.67 = 1
So, the empirical formula is CH₄N₂O.
Determine the point groups for a. Naphthalene b. 1,8-Dichloronaphthalene c. 1,5-Dichloronaphthalene d. 1,2-Dichloronaphthalene
Answer:
a). Nepthalene is a D2h molecule because it has 3 C2 axes which are perpendicular and it has a mirror plane which is horizontal. Moreover C2 is taken as principal axis.
b). 1,8-dichloronaphthalene is a C2v molecule beccause it has 1 C2 axis, two rings are joined by the C-C bond and it also has two mirro planes.
c). 1,5-dichloronaphthalene is a C2h molecule because it has only 1 C2 axis which is pependicular to the plane, it has an inversion center and also a mirror plane which is horizontal in position.
d). 1,2-dichloronaphthalene is a Cs molecule because it has only a mirror plane.
g Nitrogen in the atmosphere consists of two nitrogen atoms covalently bonded together (N2). Knowing that nitrogen is atomic number 7, what type of covalent bond holds those two atoms together?
Explanation:
Since, the atomic number of nitrogen is 7 and its electronic distribution is 2, 5. So, in order to attain stability it needs to gain 3 electrons.
Hence, when it chemically combines another nitrogen atom then as both the atoms are non-metals. So, sharing of electrons will take place.
Also, there is no difference in electronegativity of two nitrogen atoms. Hence, compound formed [tex]N_{2}[/tex] is non-polar covalent in nature.
Answer:
non polar bond
Explanation:
Write the full electron configuration of the Period 3 element with the following successive IEs (in kJ/mol):
IE₁ = 738
IE₂ = 1450
IE₃ = 7732
IE₄ = 10,539
IE₅ = 13,628
Answer:
Magnesium (Mg)
Electronic Configuration of Magnesium (Mg): [tex]1s^22s^22p^63s^2[/tex]
Explanation:
Ionization Energy:
It is the amount of energy required to remove valance shell electron of an atom in gaseous phase. The more the electron is closer to the nucleus the more energy is required to remove the electron.
Trend:
Increases from left to right in period.
Decreases generally from top to bottom in group.
In our case the data is:
IE₁ = 738
IE₂ = 1450
IE₃ = 7732
IE₄ = 10,539
IE₅ = 13,628
This data matches the Ionization energies of Magnesium (Mg). Magnesium has 12 electrons with 2 electrons in valance shell. It has 12 Ionization energies.
Electronic Configuration of Magnesium (Mg): [tex]1s^22s^22p^63s^2[/tex]
The electron configuration of the Period 3 element with the given successive IEs is;
1s² 2s² 2p⁶ 3s²The ionization energy data given is in synchrony with that of Magnesium, Mg.
In essence, the element in question is Magnesium, Mg.
The electronic configuration of Magnesium is therefore;
1s² 2s² 2p⁶ 3s².Read more;
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A single penny is 1.52 mm thick. The distance to the next nearest star other than our own (Alpha Centauri) is 4.22 light-years. If it were possible to stack one mole of pennies, how many times would the stack go between the earth and Alpha Centauri? Use the unit factoring method to determine the answer and show your work. You will need to find or look up the appropriate conversion factors to solve the problem. Your answer should be in scientific notation and have the correct number of significant figures in order to get full credit. (Please note that the text editing functions/buttons below for this essay question allows you to show exponents by using the button show as "x2"in the controls. To use it type the number followed by the exponent such as 104, highlight the 4 and hit the x2 button and you will end up with 104 as the result)
Answer:
2.29 × 10⁴ times
Explanation:
A single penny is 1.52 mm thick. The distance covered by 1 mole of pennies (6.02 × 10²³ pennies) is:
6.02 × 10²³ p × (1.52 mm/1 p) = 9.15 × 10²³ mm = 9.15 × 10²³ × 10⁻³ m = 9.15 × 10²⁰ m
The distance to the next nearest star other than our own (Alpha Centauri) is 4.22 light-years. Considering 1 ly = 9.46 × 10¹⁵ m, this distance in meters is:
4.22 ly × (9.46 × 10¹⁵ m/1 ly) = 3.99 × 10¹⁶ m
The times that the stack would go between the earth and Alpha Centauri are:
9.15 × 10²⁰ m / 3.99 × 10¹⁶ m = 2.29 × 10⁴
(6)The first-order rate constant for the decomposition of N2O5, N2O5(g) 2NO2(g) + O2(g)At 70C is 6.810-3s-1. Suppose we start with 0.0250 mol of N2O5(g) in a volume of 1.0 L. (a)How many moles of N2O5 will remain after 2.5 min
Answer:
0.009 moles
Explanation:
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given that:
The rate constant, k = [tex]6.8\times 10^{-3}[/tex] s⁻¹
Initial concentration [tex][A_0][/tex] = 0.0250 mol
Final concentration [tex][A_t][/tex] = ?
Time = 2.5 min = 2.5 x 60 seconds = 150 sec
Applying in the above equation, we get that:-
[tex][A_t]=0.0250e^{-6.8\times 10^{-3}\times 150}\ moles=0.025\times \frac{1}{e^{\frac{51}{50}}}\ moles=\frac{0.025}{2.77319}\ moles=0.009\ moles[/tex]
Express your answer as a balanced chemical equation. Identify all of the phases in your answer.
1. Li(s)+N2(g)→Li3N(s)
2. TiCl4(l)+H2O(l)→TiO2(s)+HCl(aq)
3. NH4NO3(s)→N2(g)+O2(g)+H2O(g)
4. Ca3P2(s)+H2O(l)→Ca(OH)2(aq)+PH3(g)
5. Al(OH)3(s)+H2SO4(aq)→Al2(SO4)3(aq)+H2O(l)
6.AgNO3(aq)+Na2SO4(aq)→Ag2SO4(s)+NaNO3(aq)
7. C2H5NH2(g)+O2(g)→CO2(g)+H2O(g)+N2(g)
Explanation:
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
1.[tex]Li(s)+N_2(g)\rightarrow Li_3N(s)[/tex]
The balanced equation is:
[tex]6Li(s)+N_2(g)\rightarrow 2Li_3N(s)[/tex]
2. [tex]TiCl_4(l)+H_2O(l)\rightarrow TiO_2(s)+HCl(aq)[/tex]
The balanced equation is:
[tex]TiCl_4(l)+2H_2O(l)\rightarrow TiO_2(s)+4HCl(aq)[/tex]
3. [tex]NH_4NO_3(s)\rightarrow N_2(g)+O_2(g)+H_2O(g)[/tex]
The balanced equation is:
[tex]2NH_4NO_3(s)\rightarrow 2N_2(g)+O_2(g)+4H_2O(g)[/tex]
4.[tex] Ca3P2(s)+H2O(l)\rightarrow Ca(OH)2(aq)+PH3(g)[/tex]
The balanced equation is:
[tex] Ca_3P_2(s)+6H_2O(l)\rightarrow 3Ca(OH)_2(aq)+2PH_3(g)[/tex]
5. [tex]Al(OH)_3(s)+H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+H_2O(l)[/tex]
The balanced equation is:
[tex]2Al(OH)_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+6H_2O(l)[/tex]
6.[tex] AgNO_3(aq)+Na_2SO_4(aq)\rightarrow Ag_2SO_4(s)+NaNO_3(aq)[/tex]
The balanced equation is:
[tex] 2AgNO_3(aq)+Na_2SO_4(aq)\rightarrow Ag_2SO_4(s)+2NaNO_3(aq)[/tex]
7. [tex]C_2H_5NH_2(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)+N_2(g)[/tex]
The balanced equation is:
[tex]4C_2H_5NH_2(g)+15O_2(g)\rightarrow 8CO_2(g)+14H_2O(g)+2N_2(g)[/tex]
A balanced chemical equation is a representation of a chemical reaction using chemical formulas and symbols, where the number of atoms of each element on the left side (reactants) is equal to the number of atoms of the same element on the right side (products). In other words, it obeys the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction
1. Li(s)+N2(g)→Li3N(s)
The balanced equation is :
[tex](Li(s) + N_2(g) \rightarrow Li_3N(s)\)[/tex]
2. TiCl4(l)+H2O(l)→TiO2(s)+HCl(aq)
The balanced equation is :
[tex]\(TiCl_4(l) + 4H_2O(l) \rightarrow TiO_2(s) + 4HCl(aq)\)[/tex]
3. NH4NO3(s)→N2(g)+O2(g)+H2O(g)
The balanced equation is :
[tex]\(NH_4NO_3(s) \rightarrow N_2(g) + 2O_2(g) + 2H_2O(g)\)[/tex]
4. Ca3P2(s)+H2O(l)→Ca(OH)2(aq)+PH3(g)
The balanced equation is :
[tex]\(Ca_3P_2(s) + 6H_2O(l) \rightarrow 3Ca(OH)_2(aq) + 2PH_3(g)\)[/tex]
5. Al(OH)3(s)+H2SO4(aq)→Al2(SO4)3(aq)+H2O(l)
The balanced equation is :
[tex]\(2Al(OH)_3(s) + 3H_2SO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 6H_2O(l)\)[/tex]
6.AgNO3(aq)+Na2SO4(aq)→Ag2SO4(s)+NaNO3(aq)
The balanced equation is :
[tex]\(2AgNO_3(aq) + Na_2SO_4(aq) \rightarrow Ag_2SO_4(s) + 2NaNO_3(aq)\)[/tex]
7. C2H5NH2(g)+O2(g)→CO2(g)+H2O(g)+N2(g)
The balanced equation is :
[tex]\(2C_2H_5NH_2(g) + 9O_2(g) \rightarrow 4CO_2(g) + 10H_2O(g) + 2N_2(g)\)[/tex]
These equations show the balanced chemical reactions along with the respective phases of the substances involved.
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A 20.3 mass % aqueous solution of iron(III) chloride has a density of 1.280 g/mL. Calculate the molality of the solution. Give your answer to 2 decimal places.
Answer:
Molality for the solution is 1.57 m
Explanation:
Molality is mol of solute in 1kg of solvent.
20.3 % by mass means that 20.3 g of solute (FeCl₃) are contained in 100 g of solution..
Let's determine the mass of solvent.
Mass of solution = Mass of solvent + Mass of solute
100 g = Mass of solvent + 20.3 g
100 g - 20.3 g = Mass of solvent → 79.7 g
Let's convert the mass in g to kg
79.7 g . 1kg / 1000 g = 0.0797 kg
Let's determine the moles of solute (mass / molar mass)
20.3 g / 162.2 g/mol = 0.125 mol
Molality = 0.125 mol / 0.0797 kg → 1.57 m
Final answer:
The molality of the 20.3% aqueous solution of iron(III) chloride is 1.60 m.
Explanation:
This question demands basic understanding of molatility.
To calculate the molality of the solution, we need to determine the moles of solute (iron(III) chloride) and the mass of the solvent (water).
First, convert the mass percent to grams of solute:
Mass of solute = (20.3%)(1.280 g/mL)(1000 mL) = 260.48 g
Next, calculate the moles of solute:
Moles of solute = (260.48 g)/(162.2 g/mol) = 1.603 mol
Finally, calculate the molality:
Molality = (1.603 mol)/(1 kg) = 1.60 m
Therefore, The molality of the 20.3% aqueous solution of iron(III) chloride is 1.60 m.
Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces. 32P → 32S + e− 15 16 Rate = k [32P], where k = 4.85 ✕ 10−2 day−1 What is the instantaneous rate (in mol/L/d) of production of electrons in a sample with a phosphorus concentration of 0.0031 M?
Answer:
Rate = 1.50x10⁻⁴ mol*day⁻¹*L⁻¹
Explanation:
The equation of the study of biochemical reaction mechanisms is:
³²P → ³²S + e⁻
The rate of that reaction is:
Rate = k [³²P]
where k: is the rate constant and [³²P] is the concentration of ³²P.
Hence, having that: k = 4.85x10⁻² day⁻¹ and, [³²P] = 0.0031 M, the instantaneous rate of production of electrons is:
Rate = k [³²P] = (4.85x10⁻² day⁻¹)(0.0031 mol*L⁻¹) = 1.50x10⁻⁴ mol*day⁻¹*L⁻¹
Therefore, the instantaneous rate of production of electrons is 1.50x10⁻⁴ mol*day⁻¹*L⁻¹.
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A 4.00 gram sample of a solution of sodium chloride in water was heated until all the water had evaporated. The sodium chloride that remained weighed 1.22 grams. Calculate the percentage of water in the original 4.00 grams of solution.
The weight of the water in the original solution is 2.78 grams. The percentage of water in the original 4.00 grams solution of sodium chloride, once the water was evaporated, is therefore approximately 69.5%.
Explanation:The weight of the water in the original solution can be found by subtracting the weight of the remaining sodium chloride from the total weight of the solution. So, water = total weight - weight of sodium chloride = 4.00 grams - 1.22 grams = 2.78 grams.
The percentage of water in the solution can be found by the following formula:
Percentage = (Weight of water / Total weight) * 100 = (2.78 grams / 4.00 grams) * 100 = 69.5%
So, the percentage of water in the 4.00 gram solution of sodium chloride after the water was evaporated, is approximately 69.5%.
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A student ran the following reaction in the laboratory at 600 K: COCl2(g) CO(g) + Cl2(g) When he introduced COCl2(g) at a pressure of 0.822 atm into a 1.00 L evacuated container, he found the equilibrium partial pressure of COCl2(g) to be 0.351 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction.
The equilibrium constant, Kp, for the reaction COCl2(g) → CO(g) + Cl2(g) at 600 K is calculated to be 0.634.
Explanation:To calculate the equilibrium constant, Kp, for the reaction COCl2(g) → CO(g) + Cl2(g) at 600 K, we must first use the given equilibrium partial pressure of COCl2(g) to determine the change in pressure for CO and Cl2 since the reaction started with only COCl2 present. The initial pressure of COCl2 was 0.822 atm, and the equilibrium partial pressure was 0.351 atm, which implies the pressure change (∆P) for CO and Cl2 is 0.822 atm - 0.351 atm = 0.471 atm.
Since the reaction shows that 1 mole of COCl2 produces 1 mole of CO and Cl2 each, the equilibrium partial pressures of CO and Cl2 are also 0.471 atm each. Now, we can write the expression for Kp, which is Kp = (PCO)(PCl2) / (PCOCl2). Plugging in the values, we get Kp = (0.471 atm)(0.471 atm) / (0.351 atm), and thus Kp is calculated to be 0.634 at 600 K.
Name the element described in each of the following:
(a) Smallest atomic radius in Group 6A(16)
(b) Largest atomic radius in Period 6
(c) Smallest metal in Period 3
(d) Highest IE₁ in Group 4A(14)
(e) Lowest IE₁ in Period 5
(f) Most metallic in Group 5A(15)
(g) Group 3A(13) element that forms the most basic oxide
(h) Period 4 element with filled outer level
(i) Condensed ground-state electron configuration of [Ne] 3s²3p²
(j) Condensed ground-state electron configuration of [Kr] 5s²4d⁶
(k) Forms 2+ ion with electron configuration [Ar] 3d³
(l) Period 5 element that forms 3+ ion with pseudo–noble gas configuration
(m) Period 4 transition element that forms 3+ diamagnetic ion
(n) Period 4 transition element that forms 2+ ion with a halffilled d sublevel
(o) Heaviest lanthanide
(p) Period 3 element whose 2- ion is isoelectronic with Ar
(q) Alkaline earth metal whose cation is isoelectronic with Kr
(r) Group 5A(15) metalloid with the most acidic oxide
Answer:
a. Smallest atomic radius in 6A – Oxygen (O)
b. Largest atomic radius in Period 6 – Cesium (Cs)
c. Smallest metal in period 3 – Aluminum (Al)
d. Highest IE1 in Group 4A –Carbon (C)
e. Lowest IE1 in period 5 – Rubidium (Rb)
f. Most metallic in Group 5A – Bismuth (Bi) or element 115
g. Group 3A element that forms the most basic oxide – Thallium
(Tl) or element 113
h. Period 4 element with filled outer level – Krypton (Kr)
i. Condensed gound state configuration is [Ne]3s23p2 –
Germanium (Ge)
j. Condensed ground state configuration is [Kr]5s24d6 –
Ruthenium (Ru)
k. Forms 2+ ion with electron configuration of [Ar]3d3 – Vanadium
(V)
l. Period 5 element that forms 3+ ion with pseudo‐noble gas
configuration – Indium (In)
m. Period 4 transition element that forms 3+ diamagnetic ion –
Scandium (Sc)
n. Period 4 transition element that forms 2+ ion with half‐filled d
sublevel – Manganese (Mn)
o. Heaviest Lanthanide – Lutetium (Lu)
p. Period 3 element whose 2‐ ion is isoelectronic with Ar – Sulfur
(S)
q. Alkali earth metal whose cation is isoelectronic with Kr –
Strontium (Sr)
r. Group 5 A metalloid with the most acidic oxide – Arsenic (As) or
Antimony (Sb)
Elements in a group are chemically similar to each other.
The periodic table is an arrangement of elements in groups and periods. The elements in the same group share a lot of chemical similarity with each other. The elements that are in the same period only have the same number of valence shells.
The elements described by each statement is;
(a) Smallest atomic radius in Group 6A(16) - oxygen
(b) Largest atomic radius in Period 6 - cesium
(c) Smallest metal in Period 3 - Aluminium
(d) Highest IE₁ in Group 4A(14) - carbon
(e) Lowest IE₁ in Period 5 - Bismuth
(g) Group 3A(13) element that forms the most basic oxide - Thallium
(h) Period 4 element with filled outer level - krypton
(i) Condensed ground-state electron configuration of [Ne] 3s²3p² - silicon
(j) Condensed ground-state electron configuration of [Kr] 5s²4d⁶- xenon
(k) Forms 2+ ion with electron configuration [Ar] 3d³ - vanadium
(l) Period 5 element that forms 3+ ion with pseudo–noble gas configuration - Indium
(m) Period 4 transition element that forms 3+ diamagnetic ion - Scandium
(n) Period 4 transition element that forms 2+ ion with a halffilled d sublevel - Manganese
(o) Heaviest lanthanide - Lutetium
(p) Period 3 element whose 2- ion is isoelectronic with Ar - sulfur
(q) Alkaline earth metal whose cation is isoelectronic with Kr - Strontium
(r) Group 5A(15) metalloid with the most acidic oxide - nitrogen
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Draw a structure containing only carbon and hydrogen that is a stable alkyne of five carbons containing a ring.
Answer:
Ethynylcyclopropane is the stable isomer for given alkyne.
Explanation:
In order to solve this problem we will first calculate the number of Hydrogen atoms. The general formula for alkynes is as,
CₙH₂ₙ₋₂
Putting value on n = 5,
C₅H₂.₅₋₂
C₅H₈
Also, the statement states that the compound contains one ring therefore, we will subtract 2 hydrogen atoms from the above formula i.e.
C₅H₈ ------------(-2 H) ----------> C₅H₆
Hence, the molecular formula for given compound is C₅H₆
Below, 4 different isomers with molecular formula C₅H₆ are attached.
The first compound i.e. ethynylcyclopropane is stable. As we know that alkynes are sp hybridized. The angle between C-C-H in alkynes is 180°. Hence, in this structure it can be seen that the alkyne part is linear and also the cyclopropane part is a well known moiety.
Compounds 3-ethylcycloprop-1-yne, cyclopentyne and 3-methylcyclobut-1-yne are highly unstable. The main reason for the instability is the presence of triple bond in three, five and four membered ring. As the alkynes are linear but the C-C-H bond in these compound is less than 180° which will make them highly unstable.
Potassium (K) has a work function value of 2.29 eV. What is the wavelength of light required to begin to generate a current in an apparatus like the one Hertz used with potassium metal?
Answer: The wavelength of light required is [tex]5.43\times 10^{-7}m[/tex]
Explanation:
To calculate the threshold wavelength for a given work function, we use the equation:
[tex]\phi =h\nu_o[/tex]
where,
[tex]\phi[/tex] = work function of the potassium metal = [tex]2.29eV=3.66\times 10^{-19}J[/tex] (Conversion factor: [tex]1eV=1.6\times 10^{-19}[/tex] )
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
[tex]\nu_o=\frac{c}{\lambda _o}[/tex]
c = speed of light = [tex]3\times 10^9m/s[/tex]
[tex]\lambda_o[/tex] = wavelength of light
Putting values in above equation:
[tex]3.66\times 10^{-19}J=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{\lambda_o}\\\\\lambda_o=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.66\times 10^{-19}J}=5.43\times 10^{-7}m[/tex]
Hence, the wavelength of light required is [tex]5.43\times 10^{-7}m[/tex]
27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams of the excess reagent remain after the reaction? Hint: Write a balanced chemical equation for the reaction first. Enter your answer to 1 decimal place.
Answer: The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For aluminium nitrite:Given mass of aluminium nitrite = 27.4 g
Molar mass of aluminium nitrite = 41 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol[/tex]
For ammonium chloride:Given mass of ammonium chloride = 169.9 g
Molar mass of ammonium chloride = 53.5 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol[/tex]
The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:
[tex]Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride
So, 0.668 moles of aluminium nitrite will react with = [tex]\frac{3}{1}\times 0.668=2.004mol[/tex] of ammonium chloride.
As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.
Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.
Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles
Calculating the mass of ammonium chloride by using equation 1, we get:
Excess moles of ammonium chloride = 1.172 moles
Molar mass of ammonium chloride = 53.5 g/mol
Putting values in equation 1, we get:
[tex]1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g[/tex]
Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams
Write the rate law for the reaction 2A + B → C if the reaction
(1) is second order in B and overall third order, –rA = ______
(2) is zero order in A and first order in B, –rA = ______
(3) is zero order in both A and B, –rA = ______
(4) is first order in A and overall zero order. –rA = ______
Answer:
1. [tex]R=k[A]^1[B]^2[/tex]
2. [tex]R=k[B]^1[/tex]
3. [tex]R=k[A]^0[B]^0=k[/tex]
4. [tex]R=k[A]^1[B]^{-1}[/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
(1) is second order in B and overall third order.
2A + B → C
Order of the reaction = sum of stoichiometric coefficient
= x + 2 = 3
x = 1
Rate of the reaction =R
[tex]R=k[A]^1[B]^2[/tex]
(2) is zero order in A and first order in B.
2A + B → C
Rate of the reaction =R
[tex]R=k[A]^0[B]^1=k[B]^1[/tex]
Order of the reaction = sum of stoichiometric coefficient
= 0 + 1 = 1
(3) is zero order in both A and B .
2A + B → C
Order of the reaction = sum of stoichiometric coefficient
= 0 + 0 = 0
Rate of the reaction =R
[tex]R=k[A]^0[B]^0=k[/tex]
(4) is first order in A and overall zero order.
2A + B → C
Order of the reaction = sum of stoichiometric coefficient
= 1 + x = 0
x = -1
Rate of the reaction = R
[tex]R=k[A]^1[B]^{-1}[/tex]
Which of the following types of molecules always has a dipole moment? Linear molecules with two identical bonds. Trigonal pyramid molecules (three identical bonds). Trigonal planar molecules (three identical bonds equally spaced). Tetrahedral molecules (four identical bonds equally spaced). None has a dipole moment.
Answer:
Trigonal pyramid molecules (three identical bonds)
Explanation:
In trigonal pyramidal molecule like molecule of ammonia , the vector some of intra- molecular dipole moment is not zero because the bonds are not symmetrically oriented . In other molecules , bonds are symmetrically oriented in space so the vector sum of all the internal dipole moment vectors cancel each other to make total dipole moment zero.
The type of molecules that always have a dipole moment is: B. Trigonal pyramid molecules (three identical bonds).
A molecule can be defined as a group of two (2) or more atoms that are chemically bonded together by means of shared electrons.
Also, a molecule form the smallest, fundamental ((basic) unit of a chemical compound or substance and they are capable of taking part in a chemical reaction.
In Chemistry, a dipole moment occurs whenever there is a separation of charge between molecules.
Trigonal pyramidal molecules have three (3) atoms at their trigonal base corners and one (1) atom at the top.
Additionally, a trigonal pyramidal molecule with three (3) identical bonds always have a dipole moment because bond the can't cancel one another.
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In the laboratory you are asked to make a 0.303 m cobalt(II) sulfate solution using 275 grams of water. How many grams of cobalt(II) sulfate should you add?
Answer: 12.92g of CoSO4
Explanation:
Molar Mass of CoSO4 = 59 + 32 + (16x4) = 59 + 32 +64 = 155g/mol
Molarity of CoSO4 = 0.303mol/L
Mass conc. In g/L = Molarity x molar Mass
= 0.303x155 = 46.965g/L
275 grams of water = 0.275L of water
46.965g of CoSO4 dissolves in 1L
Therefore Xg of CoSO4 will dissolve in 0.275L i.e
Xg of CoSO4 = 46.965x0.275 = 12.92g
Therefore 12.92g of CoSO4 is needed
From the choices provided below, list the reagent(s) in order that will react with 3-pentanone to form 3-phenyl-2-pentene. (List your answer as a letter (single-step transformation), or series of letters (multi-step transformation) in the order the reagents are used, with no commas separating them. No more than four steps are required for this synthesis.)a. H2 / Pt e. 1. PhMgBr 2. H3O+b. H2SO4, Δ f. H3O+c. OsO4, H2O2 g. RCO3Hd. 1. ch3mgbr2. h30+ h. KOH
Final answer:
To form 3-phenyl-2-pentene from 3-pentanone, phenylmagnesium bromide (PhMgBr) is used to introduce the phenyl group, then acid (H3O+) protonates the alkoxide to form an alcohol, followed by acidic dehydration (H2SO4, Δ) to eliminate water and form the double bond. The answer is e followed by b.
Explanation:
To convert 3-pentanone to 3-phenyl-2-pentene, we'll need to form a new carbon-carbon bond and then eliminate a carbonyl group while introducing a double bond. The first step is the Grignard reaction to introduce the phenyl group. We use phenylmagnesium bromide (PhMgBr) as the Grignard reagent to attack the carbonyl carbon of 3-pentanone and then protonate the resulting alkoxide with acid (H3O+) to give 3-phenyl-3-pentanol. The next step is to eliminate water from the alcohol to form the desired double bond. This is done by using acidic dehydration (H2SO4, Δ) to yield 3-phenyl-2-pentene. Therefore, the correct order of reagents is e followed by b.
The correct answer is f, e, d. The correct sequence of reagents to synthesize 3-phenyl-2-pentene from 3-pentanone is f (PhMgBr), followed by e (H3O+), and then d (H3O+ for hydrolysis and elimination).
To synthesize 3-phenyl-2-pentene from 3-pentanone, we can follow a multi-step transformation process involving the given reagents. Here's the step-by-step
1. Phenylmagnesium bromide (PhMgBr) - Reagent e:
The first step involves the formation of a Grignard reagent by reacting 3-pentanone with phenylmagnesium bromide. This reaction will result in the addition of a phenyl group to the carbonyl carbon of 3-pentanone, forming a tertiary alcohol with a phenyl group attached to the carbon adjacent to the hydroxyl group.
The reaction can be represented as follows:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{COR} + \text{PhMgBr} \rightarrow \text{CH}_3\text{CH}_2\text{CH}(\text{OH})\text{CH}(\text{Ph})\text{R} \][/tex]
Note that R represents the rest of the molecule.
2. Water (H3O+) - Reagent d (Step 2):
The second step is the hydrolysis of the Grignard reaction product. The tertiary alcohol formed in the first step is protonated by water, which is acidic in nature (H3O+), to form a good leaving group (water). This sets up the next step, which is an elimination reaction.
The reaction can be represented as follows:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}(\text{OH})\text{CH}(\text{Ph})\text{R} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{C}=\text{CHCH}(\text{Ph})\text{R} + \text{H}_2\text{O} \][/tex]
This step results in the formation of an alkene, specifically 3-phenyl-2-pentene, through an E1 or E2 elimination mechanism.
The other reagents listed in the question are not suitable for the synthesis of 3-phenyl-2-pentene from 3-pentanone:
- H2 / Pt (Reagent a): This would reduce the ketone to an alcohol, which is not the desired transformation.
- H2SO4, (Reagent b): This would lead to dehydration of the ketone, but not in a controlled manner to give the desired alkene.
- OsO4, H2O2 (Reagent c): This would oxidize the ketone to a diol, which is not on the pathway to the desired product.
- RCO3H (Reagent g): This is a reagent for the oxidation of primary alcohols to carboxylic acids and would not be useful in this synthesis.
- CH3MgBr, H3O+ (Reagent d): Methylmagnesium bromide would add a methyl group to the ketone, which is not what we want for the synthesis of 3-phenyl-2-pentene.
- KOH (Reagent h): While KOH could be used to deprotonate the alpha-hydrogen to form an enolate, which could then react with an electrophile, it is not the reagent of choice for the synthesis of 3-phenyl-2-pentene from 3-pentanone.
Therefore, the correct sequence of reagents to synthesize 3-phenyl-2-pentene from 3-pentanone is f (PhMgBr), followed by e (H3O+), and then d (H3O+ for hydrolysis and elimination).
What mass of Na2CrO4 is required to precipitate all of the silver ions from 73.6 mL of a 0.150 M solution of AgNO3?
Answer:
We need 0.894 grams of Na2CrO4
Explanation:
Step 1: Data given
Volume of a 0.150 M AgNO3 = 73.6 mL = 0.0736 L
Step 2: Calculate moles of Ag+
Moles Ag+ = moles AgNO3
Moles Ag+ = volume * molarity
moles Ag+ = 0.0736 L x 0.150 M = 0.01104 moles
Step 3: Calculate moles Na2CrO4
2AgNO3 + Na2CrO4 → Ag2CrO4 (s) + 2NaNO3
For 2 moles AgNO3 we need 1 mol Na2CRO4
For 0.01104 moles AgNO3 we need 0.01104/2 = 0.00552 moles Na2CrO4
Step 4: Calculate mass of Na2CrO4
Mass Na2CrO4 = moles * molar mass
Mass Na2CrO4 = 0.00552 moles * 161.97 g/mol
Mass Na2CrO4 = 0.894 grams
We need 0.894 grams of Na2CrO4
To precipitate all the silver ions, 0.893 g of Na2CrO4 is required.
Explanation:To calculate the mass of Na2CrO4 required to precipitate all of the silver ions from the solution, we need to use the stoichiometry of the reaction. The balanced equation for the reaction is:
2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3
From the equation, we can see that 2 moles of AgNO3 react with 1 mole of Na2CrO4 to form 1 mole of Ag2CrO4. The molarity of AgNO3 is given as 0.150 M and the volume is 73.6 mL. First, we need to convert the volume to liters:
73.6 mL × (1 L / 1000 mL) = 0.0736 L
Next, we can use the molarity and volume to calculate the number of moles of AgNO3:
moles of AgNO3 = molarity × volume = 0.150 M × 0.0736 L = 0.01104 moles
Since the reaction ratio is 2 moles of AgNO3 to 1 mole of Na2CrO4, we can calculate the number of moles of Na2CrO4:
moles of Na2CrO4 = (0.01104 moles of AgNO3) / 2 = 0.00552 moles
Finally, we can use the molar mass of Na2CrO4 to calculate the mass:
mass of Na2CrO4 = moles of Na2CrO4 × molar mass of Na2CrO4
From the periodic table, the molar mass of Na2CrO4 is:
molar mass of Na2CrO4 = 22.99 g/mol + (2 × 35.45 g/mol) + 4 × 16.00 g/mol = 161.97 g/mol
Therefore, the mass of Na2CrO4 required to precipitate all of the silver ions is:
mass of Na2CrO4 = 0.00552 moles × 161.97 g/mol = 0.893 g
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What is the density of atoms/nm^2 on the (110) plane of a zinc blende lattice with lattice spacing 0.546 nm. Three significant digits, fixed point notation.
Answer:
The density is 10.25 [tex]\frac{atom}{nm^{2} }[/tex]
Explanation:
From the question we are given that the lattice spacing nm
V = [tex](0.546)^{3} }[/tex] × [tex]nm^{3}[/tex]
= 0.1628 [tex]nm^{3}[/tex]
zinc blende lattice has 4 atom per unit cell.
This means that 4 atoms is contained in 0.15056 [tex]nm^{3}[/tex]
Density of the cubic unit in terms of atom is given as =
=26.565 [tex]\frac{atoms}{nm^{3} }[/tex]
For plane (110) the spacing between the cubic unit in the crystal denoted by [tex]d_{hkl}[/tex] is given as = [tex]\frac{a}{\sqrt{1^{2} + 1^{2} +0^{2} } }[/tex] Where a is the lattic spacing = 0.546
= [tex]\frac{0.546}{\sqrt{2} }[/tex] = 0.376 nm
The density of the lattice in (110) plane =26.565 [tex]\frac{atoms}{nm^{3} }[/tex] × 0.386 nm
= 10.25 [tex]nm^{2}[/tex]
Consider the second-order reaction: 2HI(g)→H2(g)+I2(g). Use the simulation to find the initial concentration [HI]0and the rate constant k for the reaction. What will be the concentration of HI after t = 1.01×10^10 s ([HI]t) for a reaction starting under the condition in the simulation?
Explanation:
The given reaction equation is as follows.
[tex]2HI(g) \rightarrow H_{2}(g) + I_{2}(g)[/tex]
[tex]\frac{-d[HI]}{dt} = k[HI]^{2}[/tex]
[tex]-\int_{[HI]_{o}}^{[HI]_{t}} \frac{d[HI]}{[HI]^{2}} = k \int_{o}^{t} dt[/tex]
[tex]-[\frac{-1}{[HI]}]^{[HI]_{t}}_{[HI]_{o}} = kt[/tex]
[tex]\frac{1}{[HI]_{t}} - \frac{1}{[HI]_{o}} = kt[/tex] .......... (1)
where, [tex][HI_{o}][/tex] = Initial concentration
[tex][HI]_{t}[/tex] = concentration at time t
k = rate constant
t = time
Now, we will calculate the initial concentration of HI as follows.
Initial rate = [tex]1.6 \times 10^{-7} mol/sec[/tex]
k = [tex]6.4 \times 10^{-9}[/tex]
R = [tex]k[HI]^{2}_{o}[/tex]
[tex][HI]^{2}_{o} = \frac{R}{k}[/tex]
= [tex]\frac{1.6 \times 10^{-7}}{6.4 \times 10^{-9}}[/tex]
[tex][HI]_{o}[/tex] = 5 M
Now, we will calculate the concentration of [tex][HI]_{t}[/tex] at t = [tex]1.01 \times 10^{10}[/tex] sec as follows.
Using equation (1) as follows.
k = [tex]6.4 \times 10^{-9}[/tex]
[tex]\frac{1}{[HI]_{t}} - \frac{1}{5}[/tex] = [tex](6.4 \times 10^{-9}) \times 1.01 \times 10^{10}[/tex]
[tex]\frac{1}{[HI]_{t}}[/tex] = 64.44
[tex][HI]_{t}[/tex] = 0.0155 M
Thus, we can conclude that concentration of HI at t = [tex]1.01 \times 10^{10}[/tex] sec is 0.0155 M.
A reception subservient on the second and first-order reactants is called a second-order reaction.
The correct answer is:
The concentration of HI at t = 1.01 × 10¹⁰ sec is 0.0155 M.
The equation according to the question is:
2 HI (g) ⇒ H₂ (g) + I₂ (g)
[tex]\dfrac{\text{-d}\left[\begin{array}{ccc}\text{HI}\end{array}\right] }{\text{dt}} & = \text{k} \left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2}[/tex]
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] _{\text{t}} } - \dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] _{\text{o}} } &= \text{kt}[/tex] .......equation (1)
Where, the initial concentration can be represented as: [tex]\left[\begin{array}{ccc}\text{HI}_{o} \end{array}\right][/tex]
The concentration at time t = [tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}[/tex]
Rate constant will be = k
The time will be = t
The initial concentration of HI can be calculated as:
The initial rate = 1.6 × 10⁻⁷ mol/sec
k = 6.4 × 10⁻⁹
[tex]\text{R} & = \text{k} \left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2} _{0}[/tex]
[tex]\dfrac{\text{R}}{\text{k}} & =\left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2} _{0}[/tex]
= [tex]\dfrac{1.6 \times 10^{-7} }{6.4 \times 10^{-9} }[/tex]
[tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right]\end{array}\right] _{0} &= 5 \;\text{M}[/tex]
To calculate the concentration [tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}[/tex] at time (t) = 1.01 × 10 ¹⁰ sec.
Now, using the above equation: (1)
k = 6.4 × 10⁻⁹
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} - \dfrac{1}{5}[/tex]
= (6.4 × 10⁻⁹) × 1.01 × 10¹⁰
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} = 64.44[/tex]
[tex]{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} = 0.0155 \;\text{M}[/tex]
Therefore, concentration of HI at t = 1.01 × 10¹⁰ sec is 0.0155 M.
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Are compounds of these ground-state ions paramagnetic?
(a) Ti²⁺ (b) Zn²⁺ (c) Ca²⁺ (d) Sn²⁺
Answer:
No, (a) Ti²⁺ is only paramagnetic
Explanation:
Paramagnetic are those which has unpaired electrons and diamagnetic are those in which all electrons are paired.
(a) Ti²⁺
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{2}[/tex]
The electrons in 3d orbital = 2 (Unpaired)
Thus, the ion is paramagnetic as the electrons are unpaired.
(b) Zn²⁺
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}[/tex]
The electrons in 3d orbital = 10 (paired)
Thus, the ion is diamagnetic as the electrons are paired.
(c) Ca²⁺
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^6[/tex]
The electrons in 3p orbital = 6 (paired)
Thus, the ion is diamagnetic as the electrons are paired.
(d) Sn²⁺
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^2[/tex]
The electrons in 5s orbital = 2 (paired)
Thus, the ion is diamagnetic as the electrons are paired.
The compounds of these ground-state ions (a) Ti²⁺, (b) Zn²⁺, (c) Ca²⁺, and (d) Sn²⁺ are not paramagnetic.
Explanation:The compounds of these ground-state ions are not paramagnetic.
Paramagnetic substances have unpaired electrons in their atoms or ions, which cause them to be attracted to a magnetic field. To determine whether an ion is paramagnetic, we need to consider its electron configuration.
(a) Ti²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d². The 3d² sublevel has two unpaired electrons, so Ti²⁺ is paramagnetic.
(b) Zn²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰. All the sublevels in its electron configuration are filled, so Zn²⁺ is not paramagnetic.
(c) Ca²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s². All the sublevels are filled, so Ca²⁺ is not paramagnetic.
(d) Sn²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰. All the sublevels are filled, so Sn²⁺ is not paramagnetic.
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Which contains the greatest mass of oxygen: 0.75 moles of ethanol ( C2H5OH ), 0.60 mole of formic acid ( HCO2H ), or 1.0 mole of water ( H2O )? Why?
Answer:
0.60 mole of formic acid contains the greatest mass of oxygen
(19.2 grams)
Explanation:
Step 1: Data given
Moles of ethanol = 0.75 moles
Molar mass ethanol = 46.07 g/mol
Moles of formic acid = 0.60 moles
Molar mass of formic acid = 46.03 g/mol
Moles of H2O = 1.0
Molar mass of H2O = 18.02 g/mol
Step 2: Calculate moles of oxygen in each
Ethanol: For 1 mol ethanol we have 1 mol oxygen
For 0.75 moles ethanol we have 0.75 moles O
Mass O = 0.75 moles * 16.0 g/mol = 12.0 grams O
Formic acid: For 1 mol formic acid, we have 2 moles O
For 0.60 moles formic acid, we have 2*0.60 = 1.20 moles O
Mass O = 1.20 moles * 16.0 g/mol = 19.2 grams O
H2O: for 1.0 mol H2O we have 1 mol O
Mass O = 18.02 g/mol * 1.0 mol = 18.02 grams
0.60 mole of formic acid contains the greatest mass of oxygen
(19.2 grams)
Formic acid (HCO₂H) contains the greatest mass of oxygen among the three compounds with 19.2 grams of oxygen.
Step 1: Molar Mass of Oxygen
The molar mass of oxygen (O) is approximately 16 grams per mole (g/mol).
Step 2: Identify the Number of Oxygen Atoms in Each Compound
Ethanol (C₂H₅OH) has 1 oxygen atom.Formic acid (HCO₂H) also has 2 oxygen atoms.Water (H₂O) has 1 oxygen atom.Step 3: Calculate the Mass of Oxygen in Each Compound
Ethanol (C₂H₅OH):
Moles of oxygen = 0.75 moles of C₂H₅OH × 1 O atom/mole = 0.75 moles of O. Mass of oxygen = 0.75 moles × 16 g/mol = 12 g.Formic Acid (HCO₂H):
Moles of oxygen = 0.60 moles of HCO₂H × 2 O atoms/mole = 1.20 moles of O. Mass of oxygen = 1.20 moles × 16 g/mol = 19.2 g.Water (H₂O):
Moles of oxygen = 1.0 mole of H₂O × 1 O atom/mole = 1.0 mole of O. Mass of oxygen = 1.0 mole × 16 g/mol = 16 g.Step 4: Compare the Masses of Oxygen
Ethanol: 12 g Formic Acid: 19.2 g Water: 16 gWhat feature of an orbital is related to each of the following quantum numbers?
(a) Principal quantum number (n)
(b) Angular momentum quantum number (l)
(c) Magnetic quantum number (ml)
Each of the following quantum numbers is related to an orbital characteristic in the following ways: n, the primary quantum number
What is quantum number?The values of conserved quantities in the dynamics of a quantum system are explained by quantum numbers in quantum physics and chemistry.Quantum numbers can be used to describe the electrons in atomic orbitals. An atom's or ion's electron has four quantum numbers that characterize its state. Consider the electrons as important variables in an equation that describes the three-dimensional position of the electrons in a certain atom. The primary quantum number (n), the orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number are the four quantum numbers that make up an atom (ms).To learn more about quantum number refer to:
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Final answer:
The principal quantum number determines the energy range and distance of an electron from the nucleus. The angular momentum quantum number determines the shape or type of the orbital. The magnetic quantum number determines the orientation of the orbital in space.
Explanation:
The feature of an orbital related to the principal quantum number (n) is the general range for the value of energy and the probable distances that the electron can be from the nucleus.
The feature of an orbital related to the angular momentum quantum number (l) is the shape or type of the orbital. The values of 1 range from 0 to (n-1), and orbitals with the same principal quantum number and the same l value belong to the same subshell.
The feature of an orbital related to the magnetic quantum number (m₁) is the orientation of the orbital in space. The values of m₁ range from -l to +l, and orbitals with the same l value have different orientations.