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Kitty [74]
3 years ago
6

Calculate the molality of a 27.0% (by mass) aqueous solution of nitric acid.

Chemistry
1 answer:
sesenic [268]3 years ago
8 0

Answer is: molality is 5.87 m.

ω(HNO₃) = 27.0% = 0.27.

If we use mass of solution 100 grams:

m(solution) = 100 g.

m(HNO₃) = ω(HNO₃) · m(solution).

m(HNO₃) = 0.27 · 100 g.

m(HNO₃) = 27 g.

n(HNO₃) = m(HNO₃) ÷ M(HNO₃).

n(HNO₃) = 27 g ÷ 63 g/mol.

n(HNO₃) = 0.428 mol.

m(H₂O) = 100 g - 27 g.

m(H₂O) = 73 g ÷ 1000 g/kg.

m(H₂O) = 0.073 kg.

b(solution) = n(HNO₃) ÷ m(H₂O).

b(solution) = 0.428 mol ÷ 0.073 kg.

b(solution) = 5.87 mol/kg = 5.87 m.

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Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

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Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

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therefore

Vair × 0.0390/100 = 2038.73L

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Vair = 5.23 × 10⁶L

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a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

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