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FrozenT [24]
3 years ago
5

PLEASE HELP

Chemistry
1 answer:
Step2247 [10]3 years ago
4 0

Answer:

See explanation

Explanation:

The equation of the reaction is;

C3H8 + 5O2 ----> 3CO2 + 4H2O

Number of moles of C3H8 = 132.33g/44g/mol = 3 moles

1 mole of C3H8 yields 3 moles of CO2

3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2

Number of moles of oxygen = 384.00 g/32 g/mol = 12 moles

5 moles of oxygen yields 3 moles of CO2

12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2

Hence C3H8 is the limiting reactant.

Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2

1 moles of C3H8 yields 4 moles of water

3 moles of C3H8 yields 3 × 4/1 = 12 moles of water

Mass of water = 12 moles of water × 18 g/mol = 216 g of water

b) Actual yield = 269.34 g

Theoretical yield = 396 g

% yield = actual yield/theoretical yield × 100/1

% yield = 269.34 g /396 g × 100

% yield = 68%

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C. They are members of the same species.
5 0
3 years ago
A mixture of 14.2 g of H2 and 36.7 g of Ar is placed in a 100.0 L container at
TiliK225 [7]

a) The total pressure of the system is  1.79 atm

b) The mole fraction and partial pressure of hydrogen is  0.89 and 1.59 atm respectively

c) The mole fraction and the partial pressure  of argon is 0.11 and 0.19 atm.

<h3>What is the total pressure?</h3>

We know tat we can be able to obtain the total pressure in the system by the use of the ideal gas equation. We would have from the equation;

PV = nRT

P = pressure

V = volume

n = Number of moles

R = gas constant

T = temperature

Number of moles of hydrogen = 14.2 g/2g = 7.1 moles

Number of moles of Argon = 36.7 g/40 g/mol

= 0.92 moles

Total number of moles =  7.1 moles + 0.92 moles = 8.02 moles

Then;

P = nRT/V

P = 8.02 * 0.082 * 273/100

P = 1.79 atm

Mole fraction of hydrogen = 7.1/8.02 = 0.89

Partial pressure of hydrogen =  0.89 * 1.79 atm

= 1.59 atm

Mole fraction of argon = 0.92 / 8.02

= 0.11

Partial pressure of argon = 0.11  *  1.79 atm

= 0.19 atm

Learn more about partial pressure:brainly.com/question/13199169

#SPJ1

3 0
1 year ago
A chemist has one solu6on that is 40% sulfuric acid and one that is 10% sulfuric acid. How much of each should she use to make 2
kipiarov [429]

Answer:

12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.

Explanation:

For first solution of sulfuric acid :

C₁ = 40% , V₁ = ?

For second solution of sulfuric acid :

C₂ = 10% , V₂ = ?

For the resultant solution of sulfuric acid:

C₃ = 28% , V₃ = 20L

Also,

<u>V₁ + V₂ = V₃ = 20L</u> ......................................(1)

Using

<u>C₁V₁ + C₂V₂ = C₃V₃</u>

<u>40×V₁ + 10×V₂ = 28×20</u>

So,

40V₁ + 10V₂ = 560........................................(2)

Solving 1 and 2 as:

V₂ = 20 - V₁

Applying in 2

40V₁ + 10(20 - V₁)  = 560

40V₁ + 200 - 10V₁ = 560

30V₁ = 360

<u>V₁ = 12 L</u>

So,

<u>V₂ = 20 - V₁ = 8L</u>

<u><em>12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.</em></u>

4 0
4 years ago
If there are 50 grams of U-238 on day zero of radioactive decay, how much will there be after 4.5 billion years? A) 0.0 grams B)
Natalija [7]

Answer:

= 25 g

Explanation:

Using the formula;

A = A₀ (1/2)^(t/h)

where A is the final amount,

A₀ is the initial amount of the substance,

t is the time and

h is the half-life of the substance,

In this case; the half life of U-238 h is equal to 4.47 billion years.

A = A₀ (1/2)^(t/h)

A = 50 (1/2)^(4.5 / 4.47)

   = 24.88

  <u> = 25 g</u>

7 0
3 years ago
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lbvjy [14]

Answer:

Atoms gain or lose electrons

Explanation:

6 0
3 years ago
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