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4 years ago

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A 24.1-g mixture of nitrogen and carbon dioxide is found to occupy a volume of 15.1 L when measured at 870.2 mmHg and 31.2oC. Wh

at is the mole fraction of nitrogen in this mixture?

1 answer:

zmey [24]4 years ago

3 0

Answer:

Mole fraction of nitrogen = 0.52

Explanation:

Given data:

Temperature = 31.2 °C

Pressure = 870.2 mmHg

Volume = 15.1 L

Mass of mixture = 24.1 g

Mole fraction of nitrogen = ?

Solution:

Pressure conversion:

870.2 /760 = 1.12 atm

Temperature conversion:

31.2 + 273 = 304.2 K

Total number of moles:

PV = nRT

n = PV/RT

n = 1.12 atm × 15.1 L / 0.0821 L.atm. mol⁻¹.K⁻¹ × 304.2 K

n = 16.9 L.atm. /25 L.atm. mol⁻¹

n = 0.676 mol

Number of moles of nitrogen are = x

Then the number of moles of CO₂ = 0.676 - x

Mass of nitrogen = x mol . 28 g/mol and for CO₂ Mass = 44 g/mol ( 0.676 - x)

24.1 = 28x + ( 29.7 -44x)

24.1 - 29.7 = 28x - 44x

-5.6 = -16 x

x = 0.35

Mole fraction of nitrogen:

Mole fraction of nitrogen = moles of nitrogen / total number of moles

Mole fraction of nitrogen = 0.35 mol / 0.676 mol

Mole fraction of nitrogen = 0.52

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Calculate the maximum numbers of moles and grams of iodic acid (HIO₃) that can form when 635 g of iodine trichloride reacts with

poizon [28]

What is Chemical Reaction?

A chemical reaction occurs when a certain group of molecules converts into another form without affecting their nuclei; only the transfer or sharing of electrons and the building and breaking of bonds occur.

Main Content

Known :

$\mathrm{ICl}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ICl}+\mathrm{HIO}_{3}+\mathrm{HCl}$

Mass of $ICI_3 = 635g$

Mass of $H_2O = 118.5g$

Calculations :

First, balance the given chemical equation by place 2,3 and 5 as the coefficients of $ICI_3, H_2O and HCl,\\$ Respectively

$2 \mathrm{ICl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ICl}+\mathrm{HIO}_{3}+5 \mathrm{HCl}$

We multiply the given mass of $ICl_3$ by the reciprocal of its molar mass to get the number of moles. The molar mass of $ICl_3 is 233.26g/mol$

$\text { Moles of } \mathrm{ICl}_{3}=635 \mathrm{~g} \mathrm{ICl}_{3} \times \frac{1 \mathrm{~mol} \mathrm{ICl}_{3}}{233.26 \mathrm{~g} \mathrm{ICl}_{3}}=2.7223 \mathrm{~mol} \mathrm{ICl}_{3}$

Them, we multiply the ratio between $ICl_3 and HIO_3$ . based on the chemical equation, the molar ratio 1 mol $HIO_3/2 mol ICl_3$ .

$\text { Moles of } \mathrm{HIO}_{3} \text { formed }=2.7223 \mathrm{~mol} \mathrm{ICl}_{3} \times \frac{1 \mathrm{~mol} \mathrm{HIO}_{3}}{2 \mathrm{~mol} \mathrm{ICl}_{3}}=1.3612 \mathrm{~mol} \mathrm{HIO}_{3}$

For water, we again multiply the given mass $H_2O$ by the reciprocal of its molar mass to get the number if moles. The molar mass of $H_2O$ is 18.02g/mol

$\text { Moles of } \mathrm{H}_{2} \mathrm{O}=118.5 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} \times \frac{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{18.02 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}}=6.5760 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}$

Then, we multiply the molar ratio between H2O and HIOs. Based on the chemical equation, the molar ratio is 1 mol $HIO_3/3 Mole H_2O$

$\text { Moles of } \mathrm{HIO}_{3} \text { formed }=6.5760 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O} \times \frac{1 \mathrm{~mol} \mathrm{HIO}_{3}}{3 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}=2.1920 \mathrm{~mol} \mathrm{HIO}_{3}$

We can see that the limiting reactant is $ICL_3$ since the given mass of $ICl_3$ forms lesser product than water does. Thus , the maximum number of moles of $HIO_3$ formed is 1.36 mol $HIO_3$ . We now multiply the molar mas of $HIO_3$ to the calculated number of moles. The molar mass of $HIO_3$ is 175.91 g/mol.

Mass of $HIO_3$ formed (Max) $=1.3612 \mathrm{~mol} \mathrm{HIO}_{3} \times \frac{175.91 \mathrm{~g} \mathrm{HIO}_{3}}{1 \mathrm{~mol} \mathrm{HIO}}=239 \mathrm{~g} \mathrm{HIO}_{3}$

We multiply the number of moles of $ICl_3$ by the molar ratio between $ICl_3$ and $H_2O$ which is $3 mol H_2O mol ICl_3$ we get the number of moles of $H_2O$ reacted. Then, we multiply the molar mass of water.

Mass of $H_2O$ reacted $=2.7223 \mathrm{~mol} \mathrm{ICl} 3 \times \frac{3 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{2 \mathrm{~mol} \mathrm{ICl}_{3}} \times \frac{18.02 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}$

Mass of $H_2O$ reacted = 73.6g $H_2O$

We subtract the mass of $H_2O$ reacted from the given mass of $H_2O$ .

Mass of $H_2O$ = 118.5 - 73.6g = $44.9g H_2O$

To learn more about Chemical Reaction

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2 years ago

The specific heat capacity of silver is 0.24 J/°C ·g.

kirza4 [7]

Answer:

A) 900 J

B) 27.96 J

C) 1,628 J ≅ 1.63 kJ

Explanation:

The heat absorbed by the metal (silver) - or energy required to heat it - is calculated as:

heat = mass x Cp x ΔT

Where Cp is the heat capacity (0.24 J/°C ·g) and ΔT is the change in temperature (final T - initial T).

A) Given:

mass = 150.0 g

final T = 298 K = 25°C

initial T = 273 K = 0°C

We calculate the energy in J to raise the temperature:

heat = mass x Cp x (final T - initial T)

= 150 .0 g x 0.24 J/°C ·g x (25°C - 0°C )

= 900 J

B) Given:

moles Ag= 1.0 mol

ΔT = 1.08°C

We first calculate the mass of silver (Ag) by multiplying the moles of Ag by the molar mass of Ag (MM = 107.9 g/mol)

mass = moles x MM = 1.0 mol Ag x 107.9 g/mol Ag = 107.9 g

Then, we calculate the heat required:

heat = mass x Cp x ΔT = 107.9 g x 0.24 J/°C ·g x 1.08°C = 27.96 J

C) Given:

heat = 1.25 kJ = 1,250 J

final T = 15.28°C

initial T = 12.08°C

We first calculate the change in temperature:

ΔT = final T - initial T = 15.28°C - 12.08°C = 3.2°C

Then, we calculate the mass of silver:

mass = heat/(Cp x ΔT) = 1,250 J/(0.24 J/°C ·g x 3.2°C) = 1,628 J ≅ 1.63 kJ

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3 years ago

Magnesium has three stable isotopes. The most commonly occurring isotope, 24 Mg, has an isotopic mass of 23.985 u and makes up 7

aleksandr82 [10.1K]

Answer:

Average atomic mass = 24.3051 amu

Explanation:

Average Atomic Mass = (Mass of Isotope 1 x Fractional Abundance of Isotope 1) + (Mass of Isotope 2 x Fractional Abundance of Isotope 2) + ......

For 24 Mg

Mass = 23.985 amu

Fractional Abundance = 0.7899

Mass * Fractional Abundance = 18.9458

For 25 Mg

Mass = 24.986 amu

Fractional Abundance = 0.10

Mass * Fractional Abundance = 2.4986

For 26 Mg

Mass = 25.983

Fractional Abundance = 0.1101

Mass * Fractional Abundance = 2.8607

Average atomic mass = 24 Mg + 25 Mg + 26 Mg

Average atomic mass = 18.9458 + 2.4986 + 2.8607

Average atomic mass = 24.3051 amu

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