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Kisachek [45]
3 years ago
11

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm8 at 27 ∘C. (R =

0.083 bar dm8 K−1 mol−1​
Chemistry
1 answer:
maxonik [38]3 years ago
7 0

Answer:

Total pressure =  56.77 bar

Explanation:

Given data:

Mass of dioxygen = 8 g

Mass of dihydrogen = 4 g

Volume of vessel = 1 dm³

Temperature = 27°C (27+273 = 300 K)

R = 0.083 bar.dm³ / mol.K

Total pressure = ?

Solution:

Number of moles of dioxygen:

Number of moles = mass/molar mass

Number of moles = 8 g/ 32 g/mol

Number of moles = 0.25 mol

Pressure of dioxygen:

PV = nRT

P = nRT/V

P = 0.25 mol × 0.083 bar.dm³ / mol.K × 300 K / 1 dm³

P = 6.97 bar.dm³  /1 dm³

P = 6.97 bar

Number of moles of dihydrogen:

Number of moles = mass/molar mass

Number of moles = 4 g/ 2 g/mol

Number of moles = 2 mol

Pressure of dihydrogen:

PV = nRT

P = nRT/V

P = 2 mol × 0.083 bar.dm³ / mol.K × 300 K / 1 dm³

P = 49.8 bar.dm³  /1 dm³

P = 49.8 bar

Total pressure of mixture in a vessel:

Total pressure = P (O₂) + P(H₂)

Total pressure =  6.97 bar + 49.8 bar

Total pressure =  56.77 bar

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.
azamat

Answer:

(a) m_{gold}=7.322g

(b)

V_{gold}=0.379cm^3

V_{copper}=0.122cm^3

(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3}  \\\\V_{copper}=0.122cm^3

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

Best regards.

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