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3 years ago

13

Where do the heaviest elements in our solar system come from?

1 answer:

Lady_Fox [76]3 years ago

5 0

The formation of elements heavier than iron and nickel requires the input of energy. Supernova explosions result when the cores of massive stars have exhausted their fuel supplies and burned everything into iron and nickel. The nuclei with mass heavier than nickel are thought to be formed during these explosions.

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

Yuri [45]

<u>Answer:</u> The value of work for the system is -935.23 J

<u>Explanation:</u>

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