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3 years ago

Write an equation that shows the formation of a chromium(ii) ion from a neutral chromium atom.

2 answers:

8 0

The equation that shows the formation of chromium (ii) ion from neutral chromium atom is as follow
Cr ---> cr^2+ + 2e-
Cr^2+ is the chromium ion with oxidation state of two which is one of the common ion of chromium. Other common ion of chromium include chromium of oxidation state 6 and 3

dangina [55]3 years ago

5 0

Answer : The balanced equation will be,

$Cr\rightarrow Cr^{2+}+2e^-$

Explanation :

Oxidation : It is the process of losing of one or more electrons.

Reduction : It is the process of gaining of one or more electrons.

The balanced equation will be,

$Cr\rightarrow Cr^{2+}+2e^-$

A neutral chromium atom is oxidized into chromium(II) ions by the loss of two electrons.

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2 years ago

When 229.0 J of energy is supplied as heat to 3.00 mol of Ar(g) at constant pressure the temperature of the sample increases by

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Answer:

The molar heat capacity at constant volume is 21.62 JK⁻¹mol⁻¹

The molar heat capacity at constant pressure is 29.93 JK⁻¹mol⁻¹

Explanation:

We can calculate the molar heat capacity at constant pressure from

$C_{p,m} = \frac{C_{p} }{n}$

Where $C_{p,m}$ is the molar heat capacity at constant pressure

${C_{p} }$ is the heat capacity at constant pressure

and $n$ is the number of moles

Also ${C_{p} }$ is given by

${C_{p} } = \frac{\Delta H}{\Delta T}$

Hence,

$C_{p,m} = \frac{C_{p} }{n}$ becomes

$C_{p,m} = \frac{\Delta H }{n \Delta T}$

From the question,

$\Delta H$ = 229.0 J

$n$ = 3.00 mol

$\Delta T$ = 2.55 K

Hence,

$C_{p,m} = \frac{\Delta H }{n \Delta T}$ becomes

$C_{p,m} = \frac{229.0}{(3.00) (2.55)}$

$C_{p,m} =$ 29.93 JK⁻¹mol⁻¹

This is the molar heat capacity at constant pressure

For, the molar heat capacity at constant volume,

From the formula

$C_{p,m} = C_{v,m} + R$

Where $C_{v,m}$ is the molar heat capacity at constant volume

and $R$ is the gas constant ( $R$ = 8.314 JK⁻¹mol⁻¹)

Then,

$C_{v,m} = C_{p,m} - R$

$C_{v,m} = 29.93 - 8.314$

$C_{v,m} =$ 21.62 JK⁻¹mol⁻¹

This is the molar heat capacity at constant volume

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