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3 years ago

Please show all of your work! :)

Chemistry

2 answers:

Airida [17]3 years ago

6 0

Answer:

The problem says that carbon dioxide was collected at 740.4 mmHg, but this is a pressure that includes the pressure of the water. We want to only use the pressure of the carbon dioxide, so subtract the pressure of the water (21.0 mmHg) from 740.4 mmHg:

740.4 - 21.0 = 719.4 mmHg

Now, we can use the ideal gas law to find the number of moles of carbon dioxide we have: PV = nRT.

- the pressure P is 719.4 mmHg

- the volume V is 38.82 mL, but we need Litres, so divide 38.82 by 1000: 38.82 / 1000 = 0.03882 L

- the moles n is what we want to find

- the gas constant R is 62.36 L mmHg / (mol K)

- the temperature T is 23.0°C, but we need this in Kelvins, so add 273 to 23.0: 23.0 + 273 = 296 K

Plug all these in:

PV = nRT

(719.4 mmHg) * (0.03882 L) = n * (62.36) * (296 K)

n ≈ 0.00151 mol

Note that this is moles of carbon dioxide, but since we want CaCO3, we need to convert moles using the reaction. It's just a 1 to 1 ratio, so we still have 0.00151 moles of CaCO3. Now, convert moles to grams. The molar mass of CaCO3 is 40.08 + 12.01 + 3 * 16 = 100.09 g/mol.

To find the mass percent, divide 0.151 by the total mass of the tablet, which is 0.3211 g:

0.151 g / 0.3211 g = 0.471 = 47.1%

The Answer is D

Explanation:

Hope this help

allsm [11]3 years ago

5 0

Answer:

Explanation:

740.4 - 21.0 = 719.4 mmHg

Now, we can use the ideal gas law to find the number of moles of carbon dioxide we have: PV = nRT.

- the pressure P is 719.4 mmHg

- the volume V is 38.82 mL, but we need Litres, so divide 38.82 by 1000: 38.82 / 1000 = 0.03882 L

- the moles n is what we want to find

- the gas constant R is 62.36 L mmHg / (mol K)

- the temperature T is 23.0°C, but we need this in Kelvins, so add 273 to 23.0: 23.0 + 273 = 296 K

Plug all these in:

PV = nRT

(719.4 mmHg) * (0.03882 L) = n * (62.36) * (296 K)

n ≈ 0.00151 mol $CO_2$

$0.00151 molCaCO_3*\frac{100.09g}{1mol} =0.151gCaCO_3$

To find the mass percent, divide 0.151 by the total mass of the tablet, which is 0.3211 g:

0.151 g / 0.3211 g = 0.471 = 47.1%

The answer is D.

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B. -127.7 kJ/mol

C. -133.9 kJ/mol

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Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

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Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

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