MULTIPLE CHOICE QUESTION

On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and sulfur gives 2.0 L of CO2, 3.0 L of H2O vapor, and 1.0 L of

Murrr4er [49]

Answer:

The empirical formula of the organic compound is = $C_2H_6S_1$

Explanation:

At STP, 1 mole of gas occupies 22.4 L of volume.

Moles of $CO_2$ gas at STP occupying 2.0 L = n

$n\times 22.4L=2.0L$

$n=\frac{2.0 L}{22.4 L}=0.08929 mol$

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of $H_2O$ gas at STP occupying 3.0 L = n'

$n'\times 22.4L=3.0L$

$n'=\frac{3.0 L}{22.4 L}=0.1339 mol$

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of $SO_2$ gas at STP occupying 1.0 L = n''

$n''\times 22.4L=1.0L$

$n''=\frac{1.0 L}{22.4 L}=0.04464 mol$

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

Moles of carbon , hydrogen and sulfur constituent of that organic compound .

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

For empirical; formula divide the least number of moles from all the moles of elements.

carbon = $\frac{0.08920 mol}{0.04464 mol}=2$

Hydrogen = $\frac{0.2678 mol}{0.04464 mol}=6$

Sulfur = $\frac{0.04464 mol}{0.04464 mol}=1$

The empirical formula of the organic compound is = $C_2H_6S_1$

3 0

3 years ago

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.

timofeeve [1]

Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU + w

q = 475 J + 93 J

q = 568 J

Therefore, the magnitude of q for the process 568 J.

6 0

3 years ago

130cals.com

gregori [183]

Answer:x=3.76

Explanation:

6 0

3 years ago

A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio

Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

$nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol$

$nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol$

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

NaOH + HCl ⇒ NaCl + H₂O

Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0

Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³

Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³

The concentration of NaOH is:

$[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M$

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0

3 years ago

What is the formula to find the area of circle

Advocard [28]

Answer:

A= π * r *2

Explanation:

6 0

3 years ago