Chemistry: Write a balanced oxidation-reduction equation showing how zinc metal reduces the ferrous ion (fe2+) , to iron metal.

Write a balanced oxidation-reduction equation

melisa141b

14.03.2022

$Zn+Fe^{2+}\rightarrow Zn^{2+}+Fe$

Explanation:

Oxidation reaction : A reaction in which electrons are lost and there is an increase in oxidation state.

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

Reduction reaction : A reaction in which electrons are gained and there is a decrease in oxidation state number.

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

A redox reaction is one in which oxidation and reduction takes place in a single reaction.

$Zn+Fe^{2+}\rightarrow Zn^{2+}+Fe$

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alyssaflores

01.06.2022

The amount of air that could be held in lungs is 6.873g

Explanation:

To calculate the amount of air, we use the equation given by Ideal gas equation, which follows:

PV=nRT

where,

P = Pressure of the gas = 102 kPa

V = volume of the gas = 6 L

n = number of moles of gas = ? mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of the gas = 37°C = 310 K (Conversion factor: T(K)=T(^oC)+273)

Putting values in above equation, we get:

$102kPa\times 6L=n\times 8.31\text{L kPa }\times 310K\\\\n=0.237mol$

To calcultae the mass of the air, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We are given:

Moles of air entered = 0.237 mol

Molar mass of air = 29 g/mol

Putting values in above equation, we get:

$0.237mol=\frac{\text{Mass of air}}{29g/mol}\\\\\text{Mass of air}=6.873g$

Hence, the amount of air that could be held in lungs is 6.873g

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