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3 years ago

What is the term for the concentration expression that relates the moles of solute dissolved in each liter of solution?

Chemistry

1 answer:

OverLord2011 [107]3 years ago

3 0

Answer: -

Molarity is the term for the concentration expression that relates the moles of solute dissolved in each liter of solution.

Explanation: -

Molarity is defined as number of moles of solute dissolved per litre of the solution.

Molarity symbol is M.

1 M = 1 mol / L

Thus molarity is the term for the concentration expression that relates the moles of solute dissolved in each liter of solution

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Read 2 more answers

Automobile airbags contain solid sodium azide, NaN 3 , that reacts to produce nitrogen gas when heated, thus inflating the bag.

Yuri [45]

Answer: The value of work for the system is -935.23 J

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of sodium azide = 16.5 g

Molar mass of sodium azide = 65 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium azide}=\frac{16.5g}{65g/mol}=0.254mol$

The given chemical equation follows:

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

By Stoichiometry of the reaction:

2 moles of sodium azide produces 3 moles of nitrogen gas

So, 0.254 moles of sodium azide will produce = $\frac{3}{2}\times 0.254=0.381mol$ of nitrogen gas

To calculate volume of the gas given, we use the equation given by ideal gas, which follows:

$PV=nRT$

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = ?

T = Temperature of the gas = $22^oC=[22+273]K=295K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = 0.381 moles

Putting values in above equation, we get:

$1.00atm\times V=0.381mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 295K\\\\V=\frac{0.381\times 0.0821\times 295}{1.00}=9.23L$

To calculate the work done for expansion, we use the equation:

$W=-P\Delta V$

We are given:

P = pressure of the system = $1atm=1.01325\times 10^5Pa$ (Conversion factor: 1 atm = 101325 Pa)

$\Delta V$ = change in volume = $9.23L=9.23\times 10^{-3}m^3$ (Conversion factor: $1m^3=1000L$ )

Putting values in above equation, we get:

$W=-1.01325\times 10^5Pa\times 9.23\times 10^{-3}m^3\\\\W=-935.23J$

Hence, the value of work for the system is -935.23 J

7 0

4 years ago