Answer:
hjghjkhjkkhgjh
Explanation:
Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.
Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.
Explanation :
To calculate the percentage composition of element in sample, we use the equation:
[tex]\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100[/tex]
Given:
Mass of carbon = 1.94 g
Mass of hydrogen = 0.48 g
Mass of sulfur = 2.58 g
First we have to calculate the mass of sample.
Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur
Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g
Now we have to calculate the percentage composition of a compound.
[tex]\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%[/tex]
[tex]\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%[/tex]
[tex]\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%[/tex]
Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.
How many molecules of carbon dioxide are dissolved in 0.550 L of water at 25 °C if the pressure of CO2 above the water is 0.250 atm? The Henry’s constant for CO2 and water at 25 °C is 0.034 M/atm.
Answer: The number of molecules of carbon dioxide gas are [tex]2.815\times 10^{21}[/tex]
Explanation:
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]0.034mol/L.atm[/tex]
[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas
[tex]p_{CO_2}[/tex] = pressure of carbon dioxide gas = 0.250 atm
Putting values in above equation, we get:
[tex]C_{CO_2}=0.034mol/L.atm\times 0.250atm\\\\C_{CO_2}=8.5\times 10^{-3}M[/tex]
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Molarity of carbon dioxide = [tex]8.5\times 10^{-5}M[/tex]
Volume of solution = 0.550 L
Putting values in above equation, we get:
[tex]8.5\times 10^{-3}M=\frac{\text{Moles of }CO_2}{0.550L}\\\\\text{Moles of }CO_2=(8.5\times 10^{-3}mol/L\times 0.550L)=4.675\times 10^{-3}mol[/tex]
According to mole concept:
1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules
So, [tex]4.675\times 10^{-3}[/tex] moles of carbon dioxide will contain = [tex](6.022\times 10^{23}\times 4.675\times 10^{-3})=2.815\times 10^{21}[/tex] number of molecules
Hence, the number of molecules of carbon dioxide gas are [tex]2.815\times 10^{21}[/tex]
Answer:
2.8 *10^21 molecules CO2 are dissolved in the 0.550 L water
Explanation:
Step 1: Data given
Volume of water = 0.550 L
Temperature = 25.0 °C
Pressure of CO2 = 0.250 atm
The Henry’s constant for CO2 and water at 25 °C = 0.034 M/atm
Step 2: Henry's law
C(CO2) = Kh * p(CO2)
⇒ with C(CO2) = the molar solubility of CO2
⇒ with Kh = Henry's constant = 0.034 M/atm = 0.034 mol/(L * atm)
⇒ with p(CO2) = the pressure of CO2 = 0.250 atm
C(CO2) = 0.034 mol/(L*atm) * 0.250 atm
C(CO2) = 0.0085 mol /L
Step 3: Calculate moles CO2
Moles CO2 = volume * molar solubility CO2
Moles CO2 = 0.550 L * 0.0085 mol/L
Moles CO2 = 0.004675 moles
Step 4: Calculate molecules of CO2
Molecules CO2 = moles * Number of Avogadro
Molecules CO2 = 0.004675 * 6.022 *10^23 / mol
Molecules CO2 = 2.8 *10^21 molecules
2.8 *10^21 molecules CO2 are dissolved in the 0.550 L water
A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 255 torrtorr . Pure pentane and hexane have vapor pressures of 425 torrtorr and 151 torrtorr, respectively, at room temperature.What is the mole fraction of hexane?
Answer:
The mole fraction of hexane is 0.6204
Explanation:
Obs: X hex = 1 - X pen
Ptot = P pen + P hex
Ptot = X pen P pen + X hex P hex ---> applying X hex = 1 - X pen
X pen = (Ptot - P hex) / (P pen - P hex)
X pen = (255 - 151) / (425 - 151) = 0.3795
X hex = 1 - X pen
X hex = 1 - 0.3795 = 0.6204
Answer:
Make a graph
Explanation: edge 2021
An unknown compound, X, is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 and 8. When 75 ml of 0.10 M NaOH is added to 100 ml of a 0.10 M solution of X at pH 2.0, the pH increases to 6.72. Calculate the pKa of the second ionizable group.
Answer:
pKa = 7.2
Explanation:
When pH=pKa=2.0, there are equal amounts of the X carboxyl group and its ionized form.
100 mL * 0.1 M = 10 mmol compound X5 mmol ionized carboxyl X-COO⁻ & 5 mmol unionized carboxyl X-COOHThen, (75 mL * 0.10 M) 7.5 mmol of OH⁻ are added, so all 5 mmol of X-COOH converts into X-COO⁻. Then the remaining (7.5 - 5) 2.5 mmol of OH⁻ react with the second ionizable group of X
Number of X-COO⁻ = 5mmol (from the beginning) + 5mmol (from X-COOH that reacted) - 2.5 mmol (from the OH⁻ remaining) = 7.5 mmol
Because the total number of X compound moles did not change, we have (10 - 7.5) 2.5 mmol of the conjugate base of X-COO⁻.
Now we have all required data to solve this problem using Henderson-Hasselbach's equation:
pH = pKa + log [A⁻]/[HA]
6.72 = pKa + log (2.5/7.5)
pKa = 7.2
Answer:
[tex]pK_a[/tex] = 7.20
Explanation:
Given that our Molarity of X(unknown compound)= 0.10 M
Volume of X = 100 ml = 0.1 L
Molarity = [tex]\frac{number of moles of X}{Volume of X}[/tex]
Number of moles of X = Molarity × Volume of X
= 0.1 M × 0.1 L
= 0.01 mol
[tex]pK_a[/tex] = [tex]pH[/tex]
∴[tex][H^+][/tex] =[tex][HA][/tex] (this literaly implies and point out that the beginning the amount of carboxyl group and the second ionizable group must be equal.)
Having said that,
Number of moles of carboxyl group in X = [tex]\frac{0.01mol}{2}[/tex]
= 0.005 mol
When 75 ml of 0.10 M NaOH is added to 100 ml of a 0.10 M solution of X;
we have the number of moles of NaOH that is being added as:
Molarity × volume of NaOH
= 0.1 M × 0.075 L
= 0.0075 mol
From the question, if NaOH molecules thoroughly dissociate the carboxyl group of X.
The excess NaOH can be calculated as:
Excess of (NaOH) = Number of moles of NaOH added - Number of moles of carboxyl group in X
Excess of (NaOH) = 0.0075 -0.005 = 0.0025 mol
∴ Applying Henderson-Hesselbalch equation; it will be easier to determine the [tex]pK_a[/tex] for second group:
[tex]pK_a[/tex] = [tex]pH[/tex] [tex]-\frac{log[A]}{HA}[/tex]
= 6.72 - log[tex](\frac{0.0025}{0.0075})[/tex]
= 6.72 - log (0.333)
= 6.72 + 0.477
[tex]pK_a[/tex] = 7.197
≅ 7.20
What steps are needed to convert benzene into p−isobutylacetophenone, a synthetic intermediate used in the synthesis of the anti-inflammatory agent ibuprofen?
Answer:
This experiment requires two 3-h lab sessions: reduction of p-isobutylacetophenone to an alcohol and then convert this alcohol to the corresponding chloride
-convertion of chloride to a Grignard reagent
Explanation:
A method for the synthesis of ibuprofen in introductory organic chemistry laboratory .This experiment requires two 3-h lab sessions. All of the reactions and techniques are a standard part of any introductory organic chemistry course. In the first lab session, reduction of p-isobutylacetophenone to an alcohol and then convert this alcohol to the corresponding chloride. In the second session, convert this chloride to a Grignard reagent, which is then carboxylated and protonated to give ibuprofen. Although the final yield is modest, this procedure offers both practicability and reliability. Permanent-magnet 60 MHz 1H NMR spectra of the final product and the two intermediates are clean and are easily interpreted by the students. Because, as previously reported, the benzylic methylene and the benzylic methine of ibuprofen have virtually identical 13C NMR chemical shifts and cancel or nearly cancel each other in the DEPT spectrum, this synthesis provides a fitting opportunity for the introduction of HETCOR even with a permanent-magnet Fourier transform instrument.
Final answer:
To convert benzene into p-isobutylacetophenone, a Friedel-Crafts Acylation followed by a Friedel-Crafts Alkylation reaction is performed, with careful control of reaction conditions for optimal yield.
Explanation:
To convert benzene into p-isobutylacetophenone, a series of organic reactions must be performed. This process starts with the conversion of benzene into an acetophenone derivative. Here's a general approach that one might take, starting with benzene:
Friedel-Crafts Acylation: To introduce the acetyl group, you perform a Friedel-Crafts acylation reaction with acetyl chloride in the presence of a Lewis acid catalyst like aluminum chloride (AlCl3). This gives you acetophenone.
Friedel-Crafts Alkylation: Next, to add the isobutyl group at the para position, perform a Friedel-Crafts alkylation using isobutyl chloride with again AlCl3 as the catalyst.
Additional purification steps may be necessary to isolate the desired p-isobutylacetophenone.
Throughout the synthesis, reaction conditions such as temperature and solvent used will need to be carefully controlled for optimal yield and product purity. This compound is a synthetic intermediate potentially used in the synthesis of the anti-inflammatory agent ibuprofen, which highlights the significance of atom economy and green chemistry principles in pharmaceutical manufacturing.
For each of the following, give the sublevel designation, the allowable ml values, and the number of orbitals:
(a) n = 2, l = 0
(b) n = 3, l = 2
(c) n = 5, l = 1
Answer:
(a) n = 2, l = 0 ⇒ sublevel s, ⇒ ml = 0, number of orbitals = 1
(b) n = 3, l = 2 ⇒ sublevel d, ⇒ ml = 0, ±1, ±2, number of orbitals = 5
(c) n = 5, l = 1 ⇒ sublevel p, ⇒ ml = 0, ±1, number of orbitals = 3
Explanation:
The rules for electron quantum numbers are:
1. Shell number, 1 ≤ n,
2. Subshell number, 0 ≤ l ≤ n − 1, from s, p, d, f, g, h...
3. Orbital energy shift, -l ≤ ml ≤ l
4. Spin, either -1/2 or +1/2
In our case
(a) n = 2, l = 0 ⇒ sublevel s
-l ≤ ml ≤ l ⇒ ml = 0, number of orbitals = 1
(b) n = 3, l = 2 ⇒ sublevel d
-l ≤ ml ≤ l ⇒ ml = 0, ±1, ±2, number of orbitals = 5
(c) n = 5, l = 1 ⇒ sublevel p
-l ≤ ml ≤ l ⇒ ml = 0, ±1, number of orbitals = 3
The first pair of quantum numbers (n, l) represents the 2s sublevel with 1 orbital. The second pair represents the 3d sublevel with 5 orbitals. The third pair represents the 5p sublevel with 3 orbitals.
Explanation:The information given refers to quantum numbers in the quantum mechanical model of the atom, a fundamental concept in high school physics and chemistry. This model explains the behavior of electrons in atoms.
(a) For n = 2 and l = 0, the sublevel designation is 2s. The permissible ml value is 0 and there is 1 orbital.
(b) For n = 3 and l = 2, the sublevel designation is 3d. The permissible ml values range from -2, -1, 0, 1, 2 and there are 5 orbitals.
(c) For n = 5 and l = 1, the sublevel designation is 5p. The permissible ml values are -1, 0, 1 and there are 3 orbitals.
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A diver 50 m deep in 10◦C fresh water exhales a 1.0-cm-diameter bubble. What is the bubbles diameter justas it teaches the surface of the lake, where the water is 20◦C? Assume that the bubble is always in thermalequilibrium with the water.
Answer:
1.82 cm
Explanation:
Utilize the equation [tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex] to calculate the change in volume and size of an air bubble.
P1 = pressure at 50m = [tex]P_{A}[/tex] + ρ*g*h (where [tex]P_{A}[/tex] = atmospheric pressure, ρ = density of water, g = acceleration due to gravity, h = height/depth)
P1 = 1.01 x 10⁵ Pa + (ρ x g x h)
= 1.01 x 10⁵ Pa + (1000 kg/m³ x 9.8 m/s² x 50 m )
= 1.01 x 10⁵ Pa + 4.9 x 10⁵ Pa
= 5.91 x 10⁵ Pa
V1 = [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex] [tex]r_{1}[/tex] = 10 cm = 1 x 10⁻² m
T1 = 10 °C = 10 + 273 = 283 K
P2 = [tex]P_{A}[/tex] = 1.01 x 10⁵ Pa because at the surface, pressure is equal to atmospheric pressure
V2 = [tex]\frac{4}{3}\pi r_{2} ^{3}[/tex] [tex]r_{2}[/tex] = ??
T2 = 20 °C = 20 + 273 = 293 K
[tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex]
V₂ = P₁V₁T₂
P₂T₁
[tex]\frac{4}{3}\pi r_{2} ^{3}[/tex] = P₁ x [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex] x T₂
P₂T₁
cancel out common terms
[tex]r_{2}[/tex]³ = 5.91 x 10⁵ Pa x (1 x 10⁻² m)³ x 293 k
1.01 x 10⁵ Pa x 283 k
[tex]r_{2}[/tex]³ = 757.9 x 10⁻⁹
[tex]r_{2}[/tex] = 9.1 x 10⁻³ m
[tex]r_{2}[/tex] = 0.91 cm
Therefore, bubbles diameter = 2r = 1.82 cm
Answer:
1.82 cm
Explanation:
The pressure done by a column of a liquid is called the hydrostatic pressure (Ph) and it can be calculated by:
Ph = Patm + ρgh
Where Patm is the atmospheric pressure under the column (101325 Pa), ρ is the density of the liquid (1000 kg/m³ for water), g is the gravity acceleration (9.8 m/s²), and h is the depth (50 m), so:
Ph = 101325 + 1000*9.8*50
Ph = 591325 Pa
Because the bubble is in equilibrium with the surroundings, its pressure is the same as the surroundings. Supposing a perfect sferic bubble, its volume is:
V = (4/3)*π*r³
Where r is the radius, which is half of the diameter, so r = 0.5 cm.
V = (4/3)*π*(0.5)³
V = 0.52 cm³
According to the ideal gas law, the multiplication of the pressure (P) by the volume (V) divided by the temperature (T) of a gas is constant, so if 1 is the state where the bubble is 50 m depth, and 2 the state at the surface:
P1*V1/T1 = P2*V2/T2
P1 = Ph = 591325 Pa
V1 = 0.52 cm³
T1 = 10°C + 273 = 283 K
P2 = 101325 Pa (atmosferic pressure)
T2 = 20°C + 273 = 293 K
591325*0.52/283 = 101325*V2/293
101325V2 = 318,354.3357
V2 = 3.14 cm³
V2 = (4/3)*π*r³
(4/3)*π*r³ = 3.14
r³ = 0.75
r = ∛0.75
r = 0.91 cm
The diameter is then 2*r = 1.82 cm.
Select the sentence that accurately describes a pure substance. Please choose the correct answer from the following choices, and then select the submit answer button
a.A pure substance is made up of only 1 type of particle.
b.A pure substance is made up of more than 1 type of particle.
c.Only compounds can be considered pure substances.
d.Only elements can be considered pure substances.
Answer:
a.A pure substance is made up of only 1 type of particle
Explanation:
When a substance is pure, it has only one type of particle. These particles maybe molecules, ions or atoms linked in a definite way throughout the substance. If a substance contains different particles, it cannot be regarded as a pure substance because its properties will be observed as a compromise of the individual properties of its different components.
The statement that accurately describes a pure substance is that it is made up of only one type of particle. Both elements and compounds can be considered as pure substances.
Explanation:The correct statement to describe a pure substance is:
a. A pure substance is made up of only 1 type of particle.
This means that a pure substance is only made up of identical atoms if it is an element, or identical molecules if it is a compound. Therefore, both elements and compounds can be considered as pure substances, contradicting options c and d. It also contradicts option b because a pure substance is not made up of more than one type of particle.
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g The combustion of 1.877 1.877 g of glucose, C 6 H 12 O 6 ( s ) C6H12O6(s), in a bomb calorimeter with a heat capacity of 4.30 4.30 kJ/°C results in an increase in the temperature of the calorimeter and its contents from 22.71 22.71 °C to 29.51 29.51 °C. What is the internal energy change, Δ U ΔU, for the combustion of 1.877 1.877 g of glucose?
Answer:
-2.80 × 10³ kJ/mol
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.
Qcal + Qcomb = 0
Qcomb = - Qcal [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ
where,
C: heat capacity of the calorimeter
ΔT: change in the temperature
From [1],
Qcomb = - Qcal = -29.2 kJ
The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:
ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol
How many unpaired electrons are present in the ground state of an atom from each of the following groups?
(a) 2A(2) (b) 5A(15) (c) 8A(18) (d) 3A(13)
Answer:
a. Zero unpaired electron
b. 3 unpaired electrons
c. Zero unpaired electron
d. 1 unpaired electron
Explanation:
a. 2A(2) has configuration => 1s2. Since the s-orbital is completely filled, Therefore it has zero unpaired electrons
b. 5A(15) has configuration =>
1s2 2s2 2p6 3s2 3p3
Since the p-orbital is not completely filled, It has 3 unpaired electrons
c. 8A(18) has configuration =>
1s2 2s2 2p6 3s2 3p6
Since the p-orbital is completely filled, therefore it has zero unpaired electrons
d. 3A(13) has configuration =>
1s2 2s2 2p6 3s2 2p1
Since the p-orbital is not completely filled, therefore it has 1 unpaired
Answer:
(a) 2A(2) - it has 2 valence electrons
(b) 5A(15) -
Explanation:
A)To determine the number of unpaired electrons for atoms in group 2A (2)
Using beryllium (it belongs to group 2A) as an example
The atomic number of Be is 4
The electronic configuration is 1s²2s²
The highest principal quantum number is 2, therefore all electrons with n=2 are valence/unpaired electron
Beryllium has 2 valence/unpaired electrons, this applies to all other elements in this group
Therefore group 2A atoms have 2 unpaired electrons
B) To determine the number of unpaired electrons for atoms in group 5A(15)
Using Nitrogen as an example
The atomic number of Nitrogen is 7
The electronic configuration is 1s²2s²2p³
The highest principal quantum number for nitrogen is 2, therefore all electrons with n=2 are valence/unpaired electrons
Nitrogen has 2+3= 5 valence/unpaired electrons, this applies to all other elements in this group
Therefore, group 5A atoms have 5 unpaired electrons
C) To determine the number of unpaired electrons for atoms in group 8A(18)
Using Neon as an example
The atomic number of Neon is 2
The electronic configuration of Neon is 1s²2s²2p⁶3s²3p⁶
The highest principal quantum number for 3, therefore all electrons with n=3 are valence/unpaired electrons
Neon has 2+6 = 8 valence/unpaired electrons,this applies to all other elements in this group except Helium whose number of unpaired electrons is 2
Therefore, group 8A atoms have 8 unpaired electrons
D) To determine the number of unpaired electrons for atoms in group 3A(13)
Using Aluminium as an example
The atomic number of Aluminium is 13
The electronic configuration of Neon is 1s²2s²2p⁶3s²3p¹
The highest principal quantum number for 3, therefore all electrons with n=3 are valence/unpaired electrons
Neon has 2+1 = 3 valence/unpaired electrons,this applies to all other elements in this group
Therefore, group 3A atoms have 3 unpaired electrons
Note: The number of valence electrons of atoms in a group is the same as the group number that the atom belongs to
What is the molality, m, of an aqueous solution of ammonia that is 12.83 M NH3 (17.03 g/mol)? This solution has a density of 0.9102 g/mL.
Answer:
Molality = 18.5 m
Explanation:
Let's analyse data. We want to determine molality which means mol of solute / 1kg of solvent. (Hence we need, the moles of solute and the mass of solvent in kg)
12.83 M means molarity → mol of solute in 1L of solution
Density refers always to solution → Mass of solution / Volume of solution
1L = 1000 mL
We can determine the mass of solution with density
0.9102 g/mL = Mass of solution / 1000 mL
Mass of solution = 0.9102 g/mL . 1000 mL → 910.2 g
Let's convert the moles of solute (NH₃) to mass
12.83 mol . 17.03 g/ 1 mol = 218.5 g
We can apply this knowledge:
Mass of solution = Mass of solvent + Mass of solute
910.2 g = Mass of solvent + 218.5 g
910.2 g - 218.5 g = 691.7 g → Mass of solvent.
Let's convert the mass in g to kg
691.7 g . 1kg / 1000 g = 0.6917kg
We can determine molalilty now → 12.83 mol / 0.6917kg
Molality = 18.5 m
To determine the molality of a 12.83 M NH3 solution with a density of 0.9102 g/mL, calculate the mass of NH3 per liter using molarity and molar mass, find the mass of the solution using volume and density, subtract the mass of NH3 to find the mass of water, and finally divide the moles of NH3 by the mass of water in kilograms.
Explanation:To find the molality (m) of an aqueous solution of ammonia (NH3) with a molarity (M) of 12.83 and a density of 0.9102 g/mL, we need to calculate the number of moles of NH3 per kilogram of water. Molality is defined as moles of solute per kilogram of solvent (water). First, we determine the mass of NH3 in 1 liter of solution, considering the molarity and the molar mass of NH3 (17.03 g/mol). Using the density of the solution, we then calculate the mass of the solution and subtract the mass of the ammonia to find the mass of water.
Here's the step-by-step calculation:
Calculate the mass of NH3 in one liter of solution: mass of NH3 = molarity × molar mass = 12.83 mol/L × 17.03 g/mol.Calculate the mass of one liter of solution: mass = volume × density = 1000 mL × 0.9102 g/mL.Subtract the mass of NH3 from the mass of the solution to find the mass of water.Divide the number of moles of NH3 by the mass of water in kilograms to get the molality.This approach allows us to determine molality, which is essential for understanding the colligative properties of the solution.
A quantity of ice at 0 °C was added to 64.3 g of water in a glass at 55 °C. The final temperature of the system was 15 °C. How much ice was added? The melting point of water is 0 °C. The heat of fusion of water is 334 J g–1 . The specific heat of liquid water is 4.184 J g–1 °C –1
The amount of ice added was approximately 38.5 grams.
To calculate this, we can use the principle of conservation of energy. The heat lost by the water as it cools down to the final temperature (15 °C) is equal to the heat gained by the ice as it melts and then warms up to the final temperature.
First, we calculate the heat lost by the water:
[tex]\[ Q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T \][/tex]
Where:
[tex]\( m_{\text{water}} = 64.3 \, \text{g} \)[/tex] (mass of water)
[tex]\( c_{\text{water}} = 4.184 \, \text{J/g°C} \)[/tex] (specific heat of water)
[tex]\( \Delta T = 55°C - 15°C = 40°C \)[/tex] (change in temperature)
[tex]\[ Q_{\text{water}} = 64.3 \, \text{g} \times 4.184 \, \text{J/g°C} \times 40°C = 10707.712 \, \text{J} \][/tex]
Next, we calculate the heat gained by the ice:
[tex]\[ Q_{\text{ice}} = m_{\text{ice}} \times L_f + m_{\text{ice}} \times c_{\text{water}} \times \Delta T \][/tex]
Where:
[tex]\( L_f = 334 \, \text{J/g} \)[/tex] (heat of fusion of water)
[tex]\( m_{\text{ice}} \)[/tex] is the mass of ice we want to find
[tex]\( c_{\text{water}} = 4.184 \, \text{J/g°C} \)[/tex] (specific heat of water)
[tex]\( \Delta T = 15°C \)[/tex] (change in temperature, from 0 °C to 15 °C)
Let's set up the equation using [tex]\( m_{\text{ice}} \):[/tex]
[tex]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times 334 \, \text{J/g} + m_{\text{ice}} \times 4.184 \, \text{J/g°C} \times 15°C \][/tex]
Now, we solve for [tex]\( m_{\text{ice}} \):[/tex]
[tex]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times (334 \, \text{J/g} + 62.76 \, \text{J/g}) \]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times 396.76 \, \text{J/g} \]\[ m_{\text{ice}} = \frac{10707.712 \, \text{J}}{396.76 \, \text{J/g}} \approx 27.0 \, \text{g} \][/tex]
However, this is the amount of ice needed to cool the water to 0 °C. To find the total amount of ice needed to cool the water to 15 °C, we add the ice that will melt at 0 °C to the ice that will further cool down to 15 °C:
[tex]\[ m_{\text{total ice}} = m_{\text{ice}} \text{ at 0 °C} + m_{\text{ice}} \text{ cooling from 0 °C to 15 °C} \]\[ m_{\text{total ice}} = 27.0 \, \text{g} + (27.0 \, \text{g} \times 15/334) \approx 38.5 \, \text{g} \][/tex]
So, approximately 38.5 grams of ice were added to the water.
Complete Question:
A quantity of ice at 0 °C was added to 64.3 g of water in a glass at 55 °C. The final temperature of the system was 15 °C. How much ice was added?
A)The melting point of water is 0 °C.
B)The heat of fusion of water is 334 J g–1 .
C)The specific heat of liquid water is 4.184 J g–1 °C –1
. Metallic iron has a body-centered cubic lattice with all atoms at lattice points and a unit cell whose edge length is 286.6 pm. The density of iron is 7.87 g cm–3 . What is the mass of an iron atom? Compare this value with the value you obtain from the molar mass
Answer:
[tex]\large \boxed{\text{55.8 u}}[/tex]
Explanation:
1. Calculate the volume of the unit cell
V = l³ = (2.866 × 10⁻⁸ cm)³ = 2.354 × 10⁻²³ cm³
2. Calculate the mass of a unit cell
[tex]\text{Mass} = 2.866 \times 10^{-23}\text{ cm}^{3} \times \dfrac{\text{7.87 g}}{\text{1 cm}^{3}} = 1.853 \times 10^{-22} \text{ g}[/tex]
3. Calculate the mass of one atom
A body-centred unit cell contains two atoms.
[tex]\text{Mass of 1 atom} = \dfrac{1.853 \times 10^{-22} \text{ g}}{\text{2 atoms}} \times \dfrac{\text{1 u}}{1.661 \times 10^{-24}\text{ g}} = \textbf{55.8 u}\\\\\text{The molar mass of Fe from the Periodic Table is $\large \boxed{\textbf{55.845 g/mol}}$}[/tex]
Rank the ions in each set in order of increasing size, and explain your ranking:
(a) Li⁺, K⁺, Na⁺ (b) Se²⁻, Rb⁺, Br (c) O²⁻, F⁻, N³⁻
Explanation:
Lithium, sodium and potassium are all group 1A elements and when we move down a group then there occurs an increase in atomic size of the elements. As lithium is the smallest and potassium being the largest so, when each of them will lose an electron and obtain a positive charge then size of lithium will further decrease.
Therefore, ions are ranked according to their increase in size as follows.
[tex]Li^{+} < Na^{+} < K^{+}[/tex]
When an atom tends to gain electrons then it acquires a negative charge. This means that size of the atom increases.
So, more is the negative charge present on an atom more will be its atomic size. Therefore, correct order of increasing size for [tex]Se^{2-}, Rb^{+}, Br[/tex] is as follows.
[tex]Br < Rb^{+} < Se^{2-}[/tex]
Similarly, order of increasing size of [tex]O^{2-}, F^{-}, N^{3-}[/tex] is as follows.
[tex]F^{-} < O^{2-} < N^{3-}[/tex]
State the periodic law, and explain its relation to electron configuration. (Use Na and K in your explanation.)
Answer:
Explanation:
The period law state that when elements are listed in order of their atomic numbers, the elements fall into recurring groups, so that there is a recurrence of similar properties at regular intervals.
Na and K in the periodic table fall into the same group, this is because they both have one electrons in their outermost shell.
Na 11 -1s2 2s2 2p6 3s1
K 19 - 1s2 2s2 2p6 3s2 3p6 4s1
They share similar chemical and physical properties. Na and K are very reactive metals, they can loose/donate their outermost electron to non metals in other to attain stable octet state.
The form ionic compound when they react with non metals.
A 226 mL solution containing 22 g of a protein in toluene has an osmotic pressure of 0.053 atm at 27 oC. What is the molar mass (in g/mol) of the protein
Answer:
4.4 × 10⁴ g/mol
Explanation:
The osmotic pressure (π) is a colligative property that can be calculated using the following expression.
π = M × R × T
where,
M: molarity
R: ideal gas constant
T: absolute temperature (27°C + 273.15 = 300 K)
Let's use it to find the molarity of the protein.
M = π / R × T
M = 0.053 atm / (0.082 atm.L/mol.K) × 300 K
M = 2.2 × 10⁻³ M
The molarity of the protein is:
M = mass of the protein / molar mass of the protein × liters of solution
molar mass of the protein = mass of the protein / M × liters of solution
molar mass of the protein = 22 g / 2.2 × 10⁻³ mol/L × 0.226 L
molar mass of the protein = 4.4 × 10⁴ g/mol
The molar mass of the protein is 4.4 * 10⁴ g/mol
Osmotic pressure :It is a colligative property that can be calculated using the following expression.
π = M × R × T
where,
M= molarityR= ideal gas constantT= absolute temperature (27°C + 273.15 = 300 K)Calculation for the molarity of the protein.[tex]M = \frac{\pi}{R * T} \\\\M = \frac{0.053 atm}{(0.082 atm.L/mol.K) * 300 K}\\\\ M = 2.2 * 10^{-3} M[/tex]
The molarity of the protein is:M = mass of the protein / Molar mass of the protein * liters of solution
Molar mass of the protein = mass of the protein / M * liters of solution
Molar mass of the protein = [tex]\frac{22 g}{2.2 * 10^{-3} mol/L * 0.226 L}[/tex]
Molar mass of the protein = [tex]4.4 * 10^4 g/mol[/tex]
Thus, the molar mass of the protein is 4.4 * 10⁴ g/mol.
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(a) Sketch, in a cubic unit cell, a [111] and a [112] lattice direction. (b) Use a trigonometric calculation to determine the angle between these two directions. (c) Use Equation 3.3 to determine the angle between these two directions.
Answer and Explanation:
a) The direction is shown in the cube diagram attached to this solution.
b) the angle between two planes (h₁, k₁, l₁) and (h₂, k₂, l₂) is given by the formula,
Cos Φ = (h₁h₂ + k₁k₂ + l₁)/√((h₁² + k₁² + l₁²)(h₂² + k₂² + l₂²))
For (111) and (112)
Cos Φ = (1.1 + 1.1 + 1.2)/√((1² + 1² + 1²)(1² + 1² + 2²))
Cos Φ = (1 + 1 + 2)/√((1+1+1)(1+1+4))
Cos Φ = 4/√(3×6)
Cos Φ = 4/√18
Φ = cos⁻¹ (4/√18) = 19.56°
c) equation 3.3 is missing from the question, I would be back to provide the answers to that as soon as the equation is provided!
Hope this Helps!!
A simple cubic lattice has different lattice directions represented by [111] and [112]. The angle between these two directions can be determined using trigonometry or Equation 3.3. The [111] direction passes through the corner atoms along the body diagonal of the unit cell, while the [112] direction passes through the edges of the cubic unit cell.
Explanation:In a simple cubic lattice, the [111] lattice direction passes through the corner atoms along the body diagonal of the unit cell. This lattice direction is represented by a line passing through the center of opposite face diagonals, as shown in Figure 10.50. On the other hand, the [112] lattice direction passes through the edges of the cubic unit cell.
To determine the angle between the [111] and [112] lattice directions using trigonometry, we can use the formula:
cos(θ) = A · B / (|A| · |B|)
where A and B are the [111] and [112] lattice directions as vectors. By substituting the values, we can calculate the angle between these two directions.
Alternatively, Equation 3.3 in the reference material can be used to calculate the angle between [111] and [112] directions:
cos(θ) = (h1 · h2 + k1 · k2 + l1 · l2) / (sqrt(h1^2 + k1^2 + l1^2) · sqrt(h2^2 + k2^2 + l2^2))
where h1, k1, and l1 are the Miller indices for the [111] direction, and h2, k2, and l2 are the Miller indices for the [112] direction.
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Formaldehyde is a carcinogenic volatile organic compound with a permissible exposure level of 0.75 ppm. At this level, how many grams of formaldehyde are permissible in a 6.0-L breath of air having a density of 1.2 kg/m3?
Answer : The amount of formaldehyde permissible are, [tex]5.4\times 10^{-6}g[/tex]
Explanation : Given,
Density of air = [tex]1.2kg/m^3=1.2g/L[/tex] [tex](1kg/m^3=1g/L)[/tex]
First we have to calculate the mass of air.
[tex]\text{Mass of air}=\text{Density of air}\times \text{Volume of air}[/tex]
[tex]\text{Mass of air}=1.2g/L\times 6.0L[/tex]
[tex]\text{Mass of air}=7.2g[/tex]
Now we have to calculate the amount of formaldehyde.
Permissible exposure level of formaldehyde = 0.75 ppm = [tex]\frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}[/tex]
Amount of formaldehyde in 7.2 g of formaldehyde = [tex]7.2g\times \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}[/tex]
Amount of formaldehyde in 7.2 g of formaldehyde = [tex]5.4\times 10^{-6}g[/tex]
Thus, the amount of formaldehyde permissible are, [tex]5.4\times 10^{-6}g[/tex]
The study of the chemical and bonds is called chemistry.
The correct answer is [tex]5.4*10^{-6[/tex]
What is a volatile compound?Volatile organic compounds are organic chemicals that have a high vapor pressure at room temperature. High vapor pressure correlates with a low boiling point, which relates to the number of the sample's molecules in the surrounding air, a trait known as volatilityAll the data is given in the question. therefore
Limited level of formaldehyde is [tex]\frac{0.75}{10^6} *7.2 = 5.4*10^{-6[/tex]Hence, the correct answer to the question is [tex]5.4*10^{-6[/tex].
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Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)
The number of grams of sodium chloride in the solution is 3.43 g.
Explanation:The number of grams of sodium chloride in the solution can be calculated using the molar mass and molarity.
First, we need to calculate the number of moles of NaCl in the solution. Using the given molarity (0.470 M) and volume (125.0 mL) of the solution, we can use the formula:
moles = molarity × volume (in L)
Therefore, moles of NaCl = 0.470 mol/L × 0.125 L = 0.05875 mol NaCl
Next, we can use the formula mass of NaCl (58.44 g/mol) to calculate the mass:
mass = moles × formula mass
Therefore, mass of NaCl = 0.05875 mol × 58.44 g/mol = 3.43 g NaCl
A couple purchases a house for $400,000.00. They pay 20% down at closing, and take out a mortgage of $320,000.00. The mortgage company offers them a 4.80% annual rate with monthly compounding. The mortgage will require monthly payments for the next 30 years.
A couple purchases a house for $400,000.00. They pay 20% down at closing, and take out a mortgage of $320,000.00. The mortgage company offers them a 4.80% annual rate with monthly compounding. The mortgage will require monthly payments for the next 30 years.
What will be the monthly payment on this mortgage?
Answer:
$ 1678.91
Explanation:
Given that;
Cost of purchasing a house = $400,000.00
Down payment =20%
Mortgage Value (MV) = $320,000.00
Annual rate offered by the mortgage company = 4.80% yearly i.e 0.4 per month
Duration of the Mortgage Loan (n) = 30 years which is equivalent to 360 months.
if we represent the monthly repayment with MR ,To calculate the monthly repayment MR;we have;
MV = MR × [tex](\frac{1}{i})*[1-(\frac{1}{(1+i)^{n}} )}][/tex]
where i = 0.004 (4.8 % annually expressed as 0.48, divided by 12 monthly payments per year)
∴
[tex]320,000.00[/tex] = [tex]MR[/tex] [tex]*(\frac{1}{0.004})*[1-(\frac{1}{1+0.004)^{360}}})][/tex]
320,000.00 = MR × 190.60
Monthly repayment (MR) = $ 1678.90870933
Monthly repayment (MR) ≅ $ 1678.91
A standard sheet of paper is 21.59cm x 27.94cm. There are two ways to bend it into a cylinder, a thin tall cylinder or a wide-short cylinder. Describe each method and calculate the volume of each.
Explanation:
Case 1. When circumference of cylinder is 27.94 cm and its height is 21.59 cm. Therefore, we will calculate the radius of cylinder as follows.
Circumference = [tex]2 \pi r[/tex]
27.94 cm = [tex]2 \times 3.14 \times r[/tex]
r = 4.45 cm
Now, we will calculate the volume of cylinder as follows.
Volume = [tex]\pi r^{2}h[/tex]
= [tex]3.14 \times (4.45 cm)^{2} \times 21.59 cm[/tex]
= [tex]1342.46 cm^{3}[/tex]
Case 2. When circumference of cylinder is 21.59 cm and its height is 27.94 cm. Therefore, we will calculate the radius of cylinder as follows.
Circumference = [tex]2 \pi r[/tex]
21.59 cm = [tex]2 \times 3.14 \times r[/tex]
r = 3.44 cm
Now, we will calculate the volume of cylinder as follows.
Volume = [tex]\pi r^{2}h[/tex]
= [tex]3.14 \times (3.44 cm)^{2} \times 27.94 cm[/tex]
= [tex]1038.18 cm^{3}[/tex]
Calculate the percent dissociation of benzoic acid C6H5CO2H in a 2.4mM aqueous solution of the stuff. You may find some useful data in the ALEKS Data resource
To calculate the percent dissociation of benzoic acid, use the ionization constant and the initial concentration, then quantifying the degree of dissociation by how much H+ ion concentration is produced. A simplified approach is given assuming low degree of ionization, typical of weak acids.
Explanation:The question asks for the percent dissociation of benzoic acid, which is a chemistry concept. To calculate percent dissociation, you need the ionization constant of the acid (Ka), which you can lookup in the ALEKS Data resources, and the concentration of the acid, which is given as 2.4mM.If the Ka for benzoic acid is x, then the equilibrium expression for the dissociation of C6H5CO2H into ions is [C6H5CO2-][H+]/[C6H5CO2H] = x. Solve this equation to find the concentration of H+ ions. Once you calculate the concentration of H+ ions, the percent dissociation is ([H+]/initial concentration of the acid) x 100%.This is a simplified approach, assuming the degree of ionization is less than 5%, as it is with weak acids in relatively dilute solutions. If this assumption is not valid, a quadratic equation would need to be used.
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In a reaction involving iron, Fe, and oxygen, O. it was determined that 4.166 grams of iron reacted with 1.803 grams of oxygen. From this information, determine the empirical formula of the compound that resulted.a. FEO2b. FeO3c. Fe2Od. Fe2O3
Answer: The empirical formula for the given compound is [tex]Fe_2O_3[/tex]
Explanation:
We are given:
Mass of Fe = 4.166 g
Mass of O = 1.803 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Iron =[tex]\frac{\text{Given mass of Iron}}{\text{Molar mass of Iron}}=\frac{4.166g}{55.85g/mole}=0.0746moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.803g}{16g/mole}=0.113moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0746 moles.
For Iron = [tex]\frac{0.0746}{0.0746}=1[/tex]
For Oxygen = [tex]\frac{0.113}{0.0746}=1.5[/tex]
Converting the mole ratio into whole number by multiplying with '2'
Mole ratio of Fe = (1 × 2) = 2
Mole ratio of O = (1.5 × 2) = 3
Step 3: Taking the mole ratio as their subscripts.The ratio of Fe : O = 2 : 3
Hence, the empirical formula for the given compound is [tex]Fe_2O_3[/tex]
A substance that cannot be chemically broken down into simpler substances is a an electron. b a heterogeneous mixture. c an element. d a homogeneous mixture. e a compound.
Answer:
c. an element.
Explanation:
An element -
It refers to the substance , which has same type of atoms , with exactly same number of protons , is referred to as an element .
In term of chemical species , elements are the smallest one , and can not be bifurcated down to any further small substance by the means of any chemical reaction .
Hence , from the given information of the question ,
The correct term is an element .
Answer:
C. an element.
Explanation:
An Argon laser gives off pulses of green light (wavelength = 514 nm). If a single pulse from the laser has a total energy of 10.0 mJ how many photons are in the pulse?
Answer:
[tex]n=2.59\times 10^{16}[/tex] photons
Explanation:
[tex]E=n\times \frac{h\times c}{\lambda}[/tex]
Where,
n is the number of photons
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, wavelength = 514 nm = [tex]514\times 10^{-9}\ m[/tex]
Energy = 10.0 mJ = 0.01 J ( 1 mJ = 0.001 J )
Applying the values as:-
[tex]0.01=n\times \frac{6.626\times 10^{-34}\times 3\times 10^8}{514\times 10^{-9}}[/tex]
[tex]\frac{19.878n}{10^{17}\times \:514}=0.01[/tex]
[tex]n=2.59\times 10^{16}[/tex] photons
If equal volumes of 0.1 M HCl and 0.2 M TRIS (base form) are mixed together. The pKa of TRIS is 8.30. Which of the following statements about the resulting solution are correct?
(Select all that apply.)
A. The ratio of [conjugate base]/[conjugate acid] is [0.1 M]/[0.05 M].
B. This solution is too basic to be a buffer.
C. The ratio of [conjugate base]/[conjugate acid] is [0.05 M]/[0.05 M].
D. This solution is a good buffer.
E. The majority of TRIS will be in the acid form in the solution.
Answer:
option D is correct
D. This solution is a good buffer.
Explanation:
TRIS (HOCH[tex]_{2}[/tex])[tex]_{3}[/tex]CNH[tex]_{2}[/tex]
if TRIS is react with HCL it will form salt
(HOCH[tex]_{2}[/tex])[tex]_{3}[/tex]CNH[tex]_{2}[/tex] + HCL ⇆ (HOCH[tex]_{2}[/tex])[tex]_{3}[/tex]NH[tex]_{3}[/tex]CL
Let the reference volume is 100
Mole of TRIS is = 100 × 0.2 = 20
Mole of HCL is = 100 × 0.1 = 10
In the reaction all of the HCL will Consumed,10 moles of the salt will form
and 10 mole of TRIS will left
hence , Final product will be salt +TRIS(9 base)
H = Pk[tex]_{a}[/tex] + log (base/ acid)
8.3 + log(10/10)
8.3
The correct statements are that the resulting solution is a good buffer and the ratio of [conjugate base]/[conjugate acid] is [0.1 M]/[0.1 M]. The majority of TRIS is not in the acid form in the solution.
Explanation:If equal volumes of 0.1 M HCl and 0.2 M TRIS are mixed together, the excess TRIS will react with HCl forming its conjugate acid (TRIS-HCl) and reducing the concentration of HCl. Since we started with more TRIS, half of it will remain unreacted and exist as the base (TRIS), while the other half will form the conjugate acid (TRIS-HCl). Therefore, the concentrations of the conjugate base and the conjugate acid will both be 0.1 M.
So, the statement C. The ratio of [conjugate base]/[conjugate acid] is [0.05 M]/[0.05 M] is incorrect. The correct ratio is 0.1 M/0.1 M.
The pKa of TRIS is 8.30 meaning that it can act as a buffer at around pH 8.30. Since the solution consists of equal concentrations of a weak base and its conjugate acid, it will indeed create a buffer solution. Consequently, D. This solution is a good buffer is correct.
The statement E. The majority of TRIS will be in the acid form in the solution is not correct. The base and acid forms are in equal amounts.
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When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay: HCOOH(g) →CO2(g) + H2 (g) The rate of reaction is monitored by measuring the total pressure in the reaction container.Time (s) . . . P (torr)0 . . . . . . . . . 22050 . . . . . . . . 324100 . . . . . . . 379150 . . . . . . . 408200 . . . . . . . 423250 . . . . . . . 431300 . . . . . . . 435At the start of the reaction (time = 0), only formic acid is present.What is the formic acid pressure (in torr) when the total pressure is 319?
Answer : The formic acid pressure is, 99 torr
Explanation :
The given chemical reaction is:
[tex]HCOOH(g)\rightarrow CO_2(g)+H_2(g)[/tex]
Initial pressure a 0 0
At time 't' (a-x) x x
According to the Dalton's law,
[tex]P_{Total}=P_{HCOOH}+P_{CO_2}+P_{H_2}[/tex]
[tex]P_{Total}=(a-x)+x+x=a+x[/tex] .........(1)
As we are given that:
Initial pressure = a = 220 torr
[tex]P_{Total}=319torr[/tex]
Now put the value of 'a' in equation 1, we get:
[tex]P_{Total}=a+x[/tex]
[tex]319torr=220torr+x[/tex]
[tex]x=99torr[/tex]
Thus, the formic acid pressure is, 99 torr
Enter your answer in the provided box. For the simple decomposition reaction AB(g) → A(g) + B(g) rate = k[AB]2 and k = 0.20 L/mol·s. If the initial concentration of AB is 1.50 M, what is [AB] after 10.3 s?
Answer:
[AB] is 0.65 M
Explanation:
Let the concentration of AB after 10.3s be y
Rate = ky^2 = change in concentration of AB/time
k = 0.2 L/mol.s.
Change in concentration of AB = 1.5 - y
Time = 10.3s
0.2y^2 = 1.5-y/10.3
0.2y^2 × 10.3 = 1.5 - y
2.06y^2 = 1.5 - y
2.06y^2 + y - 1.5 = 0
The value of y must be positive and is obtained using the quadratic formula
y = [-1 + sqrt(1^2 -4×2.06×-1.5)]/2(2.06) = [-1 + sqrt(13.36)]/4.12 = 2.66/4.12 = 0.65 M
Final answer:
To find the concentration of AB after 10.3 seconds, the second-order integrated rate law was used with an initial concentration of 1.50 M and a rate constant of 0.20 L/mol·s. The final concentration of AB was calculated to be approximately 0.32 M.
Explanation:
To calculate the concentration of AB after 10.3 seconds in the given decomposition reaction, we need to use the integrated rate law for a second-order reaction, which is as follows:
1/[AB] - 1/[AB]₀ = kt
Where:
[AB] is the concentration at time t,
[AB]₀ is the initial concentration,
k is the rate constant, and
t is the time elapsed.
Given that k = 0.20 L/mol·s, [AB]₀ = 1.50 M, and t = 10.3 s, we can rearrange the equation and solve for [AB] as follows:
1/[AB] = 1/1.50 M + (0.20 L/mol·s)(10.3 s)
1/[AB] ≈ 1/1.50 M + 2.06 L/mol
[AB] ≈ 1/(1/1.50 + 2.06) L/mol
[AB] ≈ 0.32 M
After 10.3 seconds, the concentration of AB is approximately 0.32 M.
More polar solvents (eluents) move molecules more rapidly than less polar solvents. If you used a 1:1 hexanes:methanol mixture as solvent, would you expect the products to elute faster or slower? Based on your experiment, would a 1:1 hexanes:methanol mixture be a good choice as eluent? Explain why/why not
Answer:
For the first question the products elute slower. The answer to the second question is it would not be a good option as an eluent
Explanation:
Solvents are classified into polar and nonpolar. While in polar solvents, the distribution of the electronic cloud is asymmetric; therefore, the molecule has a positive and a negative pole. Low molecular weight alcohols such as methanol belong to this type.
While in apolar solvents, the distribution of the electronic cloud is symmetric; Therefore, these substances lack a positive and negative pole in their molecules. Some solvents such as hexane.
The miscibility of methanol (polar solvent) in hexane (apolar solvent) is low, miscibility is the main reason for not using this mixture as eluent.
A chemist determines by measurements that moles of fluorine gas participate in a chemical reaction. Calculate the mass of fluorine gas that participates. Be sure your answer has the correct number of significant digits.
The question is incomplete, here is the complete question:
A chemist determines by measurements that 0.0850 moles of fluorine gas participate in a chemical reaction. Calculate the mass of fluorine gas that participates. Be sure your answer has the correct number of significant digits.
Answer: The mass of fluorine gas that is precipitated is 3.23 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]
We are given:
Moles of fluorine gas = 0.0850 moles
Molar mass of fluorine gas = 38.0 g/mol
Putting values in above equation, we get:
[tex]0.0850mol=\frac{\text{Mass of fluorine gas}}{38.0g/mol}\\\\\text{Mass of fluorine gas}=(0.0850mol\times 38.0g/mol)=3.23g[/tex]
Hence, the mass of fluorine gas that is precipitated is 3.23 grams
To calculate the mass of fluorine gas that participates in a chemical reaction, multiply the number of moles by the molar mass of fluorine. The mass is approximately 76 grams with 2 significant digits.
Explanation:To calculate the mass of fluorine gas that participates in a chemical reaction, you need to know the number of moles of fluorine gas involved and the molar mass of fluorine (F2). The molar mass of fluorine is approximately 38 grams per mole. Multiply the number of moles by the molar mass to obtain the mass of fluorine gas participating in the reaction.
Example:
If the chemist determines that 2 moles of fluorine gas participate in the reaction, the mass of fluorine gas can be calculated as follows:
Mass = number of moles x molar mass
Mass = 2 moles x 37.996 grams/mole
Mass = 75.992 grams
Therefore, the mass of fluorine gas that participates in the reaction is approximately 76 grams . Remember to use the correct number of significant digits in your final answer, which in this case is 2 significant digits.
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