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2 years ago

10

Given that the density of air is 14.4, then the vapour of sulphur IV oxide is A. 16 B. 32 C. 64 D. 128

1 answer:

Arisa [49]2 years ago

6 0

Answer:

32

Explanation:

The vapour density of a gas is the number of times a given volume of gas or vapour is as heavy as the same volume of hydrogen at a particular temperature and pressure.

Vapour density = 2 × relative molecular mass of the gas or vapour

Relative molecular mass of SO2 = 32 + 2(16) = 64

Hence;

Vapour density of SO2 = 64/2

Vapour density of SO2 = 32

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What kinds of elements form an ionic bond?

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3 years ago

A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch

ivanzaharov [21]

Answer:

The reactions free energy $\Delta G = -49.36 kJ$

Explanation:

From the question we are told that

The pressure of (NO) is $P_{NO} = 9.20 \ atm$

The pressure of (Cl) gas is $P_{Cl} = 9.15 \ atm$

The pressure of nitrosly chloride (NOCl) is $P_{(NOCl)} = 7.70 \ atm$

The reaction is

$2NO_{(g)} + Cl_2 (g)$ ⇆ $2 NOCl_{(g)}$

From the reaction we can mathematically evaluate the $\Delta G^o$ (Standard state free energy ) as

$\Delta G^o = 2 \Delta G^o _{NOCl} - \Delta G^o _{Cl_2} - 2 \Delta G^o _{NO}$

The Standard state free energy for NO is constant with a value

$\Delta G^o _{NO} = 86.55 kJ/mol$

The Standard state free energy for $Cl_2$ is constant with a value

$\Delta G^o _{Cl_2} = 0kJ/mol$

The Standard state free energy for $NOCl$ is constant with a value

$\Delta G^o _{NOCl} =66.1kJ/mol$

Now substituting this into the equation

$\Delta G^o = 2 * 66.1 - 0 - 2 * 87.6$

$= -43 kJ/mol$

The pressure constant is evaluated as

$Q = \frac{Pressure \ of \ product }{ Pressure \ of \ reactant }$

Substituting values

$Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}$

$= 0.0765$

The free energy for this reaction is evaluated as

$\Delta G = \Delta G^o + RT ln Q$

Where R is gas constant with a value of $R = 8.314 J / K \cdot mol$

T is temperature in K with a given value of $T = 25+273 = 298 K$

Substituting value

$\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765]$

$= -43-6.36$

$\Delta G = -49.36 kJ$

4 0

3 years ago

A closed system initially containing 1×10^-3 hydrogen 2×10^-3M iodine at 448 degree Celsius and is allowed to reach equilibrium.

GaryK [48]

Answer:

Kc = 50.5

Explanation:

We determine the reaction:

H₂ + I₂ ⇄ 2HI

Initially we have 0.001 molesof H₂

and 0.002 moles of I₂

If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.

H₂ + I₂ ⇄ 2HI

In: 0.001 0.002 -

R: x x 2x

Eq: 0.001-x 0.002-x 0.00187

x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted

So in the equilibrium we have:

0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂

0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂

Expression for Kc is = (HI)² / (H₂) . (I₂)

0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5

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3 years ago