What pressure would a gas mixture in a 10.0 l tank exert if it were composed of 48.5 g he and 94.6 g co2 at 398 k?

Explanation:

1) Data:

a) V = 10.0 l

b) He, m₁ = 48.5 g

c) CO₂, m₂ = 94.6 g

d) T = 398 K

e) p = ?

2) Formulae:

a) ideal gas equation: pV = nRT

b) Number of moles, n: n = mass in grams / molar mass

3) Solution:

a) Number of moles of He:

b) Number of moles of CO₂:

c) Total number of moles of the gas mixture:

d) Pressure, p:

Pressure of gas mixture : P tot = 46,588 atm

Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases

Can be formulated:

The partial pressure is the pressure of each gas in a mixture

Since the gas mixture consists of He and CO₂, we determine the pressure of each gas then we add them together to get the total pressure

We assume the ideal gas, so we use the formula:

where

P = pressure, atm, N / m²

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K (P = atm, v = liter), or 8.314 J / mol K (P = Pa or N / m2, v = m³)

T = temperature, Kelvin

n = N / No

n = mole

No = Avogadro number (6.02.10²³)

n = m / m

m = mass

M = relative molecular mass

M = 4 g / mol

m = 48.5 g

n = m: M

n = 48.5: 4

n = 12,125 mole

PV = nRT

2. CO₂

M = 44 g / mol

m = 94.6 g

n = m: M

n = 94.6: 44

n = 2.15 mole

PV = nRT

P tot = P He + P CO₂

P tot = 39,571 + 7.0167

P tot = 46,588 atm

Which equation agrees with the ideal gas law  

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Which law relates to the ideal gas law  

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2b2t

explanation:

2b2t

a electronegativity decreases beacuse the size of nucleus is bigger.