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Kryger [21]
3 years ago
9

Nitrogen was heated from 17°C to 37°C. The original volume of nitrogen was 8.5 L. Find the new volume in liters assuming P and n

remain the same.
Chemistry
1 answer:
Nata [24]3 years ago
7 0

Answer:

The new volume of the gas is 9.086 liters.

Explanation:

Let suppose that nitrogen has a behavior of ideal gas, the equation of state for ideal gases is:

P\cdot V = n \cdot R_{u}\cdot T (1)

Where:

P - Pressure, measured in atmospheres.

V - Volumen, measured in liters.

n - Molar amount, measured in moles.

T - Temperature, measured in Kelvin.

R_{u} - Ideal gas constant, measured in atmosphere-liters per mole-Kelvin.

If pressure and molar amount of the gas remain constant, then we construct the following relationship:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}} (2)

If we know that T_{1} = 290.15\,K, P_{1} = 8.5\,L and T_{2} = 310.15\,K, then the new volume of the gas is:

V_{2} = \left(\frac{T_{2}}{T_{1}}\right)\cdot V_{1}

V_{2} = \left(\frac{310.15\,K}{290.15\,K} \right)\cdot (8.5\,L)

V_{2} = 9.086\,L

The new volume of the gas is 9.086 liters.

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2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

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3 years ago
What volume of carbon dioxide (CO2) will be produced if 2.90 moles of it on (Fe) is produced?
dolphi86 [110]

Answer:

= 97.44 Liters at S.T.P

Explanation:

The reaction between Iron (iii) oxide and Carbon monoxide is given by the equation;

Fe2O3(s)+ 3CO(g) → 3CO2(g) + 2Fe(s)

From the reaction when the reactants react, 2 moles of Fe and 3 moles of CO2 are produced.

Therefore; Mole ratio of Iron : Carbon dioxide is 2:3

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                                                   = 4.35 moles

But; 1 mole of CO2 at s.t.p occupies 22.4 liters

Therefore;

Mass of CO2 = 22.4 × 4.35 Moles

                     = 97.44 L

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