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Anna35 [415]
3 years ago
5

(0)A mixture of dihydrogen and dioxygen at one bar pressure contains 70% by weight of dioxygen. Calculate the partial pressure o

f dioxygen.
a (ii)Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure. R = 0.083 bar L K-1 mol-1


​
Chemistry
1 answer:
ser-zykov [4K]3 years ago
6 0

Answer:

a) 0.13 bar

b) 5.05 L

Explanation:

Let us take the mass of the mixture to be 100 g. Hence, 70% by weight of dioxygen corresponds to 70 g

Mass of dihydrogen = 100g - 70 g = 30g

Number of moles of dioxygen = 70g/32 g/mol = 2.2 moles of dioxygen

Number of moles of dihydrogen = 30g/2g/mol = 15 moles of dihydrogen

Total number of moles = 2.2 + 15 = 17.2 moles

Mole fraction of dioxygen = 2.2/17.2 = 0.13

Partial pressure = mole fraction * total pressure

Partial pressure of dioxygen = 0.13 * 1 = 0.13 bar

ii) number of moles in 8.8 g of CO2 = 8.8g/44g/mol = 0.2 moles

T = 31.1 + 273 = 304.1 K

P = 1 bar

V= ?

R = 0.083 bar L K-1 mol-1

From

PV=nRT

V = nRT/P

V= 0.2 * 0.083 * 304.1/1

V= 5.05 L

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Answer:

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Explanation:

Assuming the given percentages are by mass, we can solve this problem via imagining we have <em>100 g of the compound</em>, if that were the case we would have:

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Now we <u>convert those masses into moles</u>, using the<em> elements' respective molar masses</em>:

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As the number of C moles and O moles is roughly the same, the empirical formula for the compound is <em>CO</em>.

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The half-life of thorium-227 is 18.72 days. How many days are required for three-fourths of a given amount to decay?
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The dissociation of calcium carbonate has an equilibrium constant of Kp= 1.20 at 800°C. CaCO3(s) ⇋ CaO(s) + CO2(g)
Fed [463]

Explanation:

(a)   Formula that shows relation between K_{c} and K_{p} is as follows.

                 K_c = K_p \times (RT)^{-\Delta n}

Here, \Delta n = 1

Putting the given values into the above formula as follows.

        K_c = K_p \times (RT)^{-\Delta n}

                  = 1.20 \times (RT)^{-1}

                  = \frac{1.20}{0.0820 \times 1073}

                  = 0.01316

(b) As the given reaction equation is as follows.

               CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)

As there is only one gas so ,

                p[CO_{2}] = K_{p} = 1.20

Therefore, pressure of CO_{2} in the container is 1.20.

(c)   Now, expression for K_{c} for the given reaction equation is as follows.  

             K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}

                        = \frac{x \times x}{(a - x)}

                        = \frac{x^{2}}{(a - x)}[/tex]

where,    a = initial conc. of CaCO_{3}

                  = \frac{22.5}{100} \times 9.56

                  = 0.023 M

          0.0131 = \frac{x^{2}}{0.023 - x}

                  x = 0.017

Therefore, calculate the percentage of calcium carbonate remained as follows.

       % of CaCO_{3} remained = (\frac{0.017}{0.023}) \times 100

                                  = 75.46%

Thus, the percentage of calcium carbonate remained is 75.46%.

3 0
4 years ago
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