All categories

2 years ago

11. Propane (C3Hg) is a fuel commonly used in gas grills.

Chemistry

1 answer:

Mrac [35]2 years ago

5 0

Answer:

81.71%

Explanation:

One mole of propane contains 3 moles of carbon atoms and 8 moles of hydrogen atoms, as seen from the molecular formula of $C_3H_8$ . In order to calculate the percent of carbon in propane by mass, we need to remember that %w/w (or percent mass) formula states that:

$\omega=\frac{m_{component}}{m_{total}}\cdot100\%$

That is, we need to divide the mass of the component of interest by the total mass of the compound and multiply by 100 to obtain the percentage.

For simplicity, let's take 1 mole of propane and find the mass of 1 mole (hence, we'll be finding the molar mass of propane). To do that, we add the 3 molar masses of carbon and 8 molar masses of hydrogen to obtain a total of:

$M_{C_3H_8}=3M_C+8M_H=3\cdot12.011 \frac{g}{mol}+8\cdot1.00784\frac{g}{mol}=44.096 \frac{g}{mol}$

Now that we have the molar mass of propane, we also need to find the total mass of carbon in 1 mole of propane. We know that we have a total of 3 moles of carbon which corresponds to:

$M_C=3\cdot12.011 \frac{g}{mol}=36.033 \frac{g}{mol}$

Dividing the mass of carbon present by the total mass of the compound will yield the mass percentage as defined by the formula we introduced:

$\omega_C=\frac{36.033\frac{g}{mol}}{44.096 \frac{g}{mol}}\cdot 100\%=81.71\%$

You might be interested in

9-11 is best? for new york

valkas [14]

NEW YORK IS A STATE BECAUSE IT IS

5 0

3 years ago

Two gases X and Y are found in the atmosphere in only trace amounts because they decompose quickly. When exposed to ultraviolet

igomit [66]

Answer:

1. Yes

2. After 68.1 mins, pX < pY.

Explanation:

Assuming the total gas pressure is 1 atm, let the partial pressure of Y be y, partial pressure of X will be 0.25y + y = 1.25y

1.25y + y = 1atm

2.25y = 1 atm

y = 1 atm / 2.25 = 0.44 atm

Then partial pressure of X = 0.56 atm

The partial pressure of a gas in a mixture of gases is directly proportional to the mole fraction of the gas.

Therefore the mole fraction of X and Y is 0.56 and 0.44 respectively.

The partial pressures of X and Y becomes half of their original values at 1.25 h = 85 min and Y at 150 min respectively.

The partial pressure after some time can be calculated from the half-life equation :

m = m⁰ * 1/2ⁿ

Where m = the remaining mass, m⁰ = initial mass, and n is number of half-lives undergone.

Let partial pressures represent the mass, and n for X and Y be a and b respectively:

pX = 0.56/2ᵃ

pY = 0.44/2ᵇ

We then determine when Partial pressure of X, pX = Partial pressure of Y, pY

0.56/2ᵃ = 0.44/2ᵇ

2ᵃ/2ᵇ = 0.56/0.44

2ᵃ/2ᵇ = 1.27

2ᵃ⁻ᵇ = 2⁰°³⁴⁵

a - b = 0.345

Let this time be t, therefore,

For X; t = 85a and For Y: t = 150b

85a = 150b

then, a = 1.76b

1.76b - b = 0.345

0.760b = 0.345

b = 0.454 and,

a = 0.345 + 0.454 = 0.799

So, X goes through 0.779 half-lives while Y goes through 0.454 half-lives Then, the time for both X and Y to have the same amount is:

t = 150 * 0.454 = 68.1 min

After 68.1 mins, pX < pY.

5 0

3 years ago

The reactants side of an equation is shown. How many silver atoms should the products side of the equation have after a chemical

erastovalidia [21]

6 h(2) O?bhrjjxiddjxu

6 0

3 years ago

How many grams of NaOH are produced from 20.0 grams of Na2CO3?

natita [175]

Answer:

Hope this helps!

Explanation:

Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams

4 0

2 years ago

Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago