What is the percent of aluminum found in Al2SO4

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

3 years ago

If 2.0 moles of A and 3.0 moles of B react according to the hypothetical reaction below, how many moles of the excess reactant w

valina [46]

.5 mol of A will be left over since 1.5 mol of A will be used for every 3 mol of B due to the 2:1 ratio established by the formula.

5 0

3 years ago

Read 2 more answers

Help please giving brainliest

Kitty [74]

Answer:

i think protons and neutrons

Explanation:

im not sure tho

8 0

3 years ago

Read 2 more answers

Which of the following describes the relationship between acid strength and ka value for weak acids?

Leona [35]

I think the correct answer would be when ka increases, the acid strength increases. Ka is the product of hydronium ions and the [A-] ions concentration over the acid concentration. Therefore, increasing the the Ka, the numerator in the definition would increase as well which correspondingly increase the strength of the acid.

8 0

4 years ago

A 0.3146 g sample of a mixture of NaCl ( s ) and KBr ( s ) was dissolved in water. The resulting solution required 45.70 mL of 0

Masteriza [31]

Answer:

The answer to the question is

The mass percentage of NaCl(s) in the mixture is 49.7%

Explanation:

The given variables are

mass of sample of mixture = 0.3146 g

Volume of AgNO₃ required to react comletely with the chloride ions = 45.70 mL

Concentration of the AgNO₃ added = 0.08765 M

The equations for the reactions oare

NaCl(aq) + AgNO₃ (aq) = AgCl(s) + NaNO₃(aq)

AgNO₃ (aq) + KBr (aq) → AgBr (s) + KNO₃

The equation for the reaction shows one mole of NaCl reacts with one mole of AgNO₃ to form one mole of AgCl

Thus 45.70 mL of 0.08765 M solution of AgNO₃ contains $\frac{45.7}{1000} (0.08765) = 0.004 moles$

Therefore the sum of the number of moles of Br⁻ and Cl⁻

precipitated out of the solution = 0.004 moles

Thus if the mass of NaCl in the sample = z then the mass of KBr = y

However the mass of the sample is given as 0.3146 g which means the molarity of the solution is 0.004 moles

given by

$\frac{z}{58.44} + \frac{y}{119} = 0.004 moles$ and z + y = 0.3146

Therefore z = 0.3146 - y which gives

$\frac{(0.3146-y)}{58.44} + \frac{y}{119} = 0.004 moles$

-8.7×10⁻³y +0.54×10⁻³ = 0.004

or 8.7×10⁻³y = 1.37769× 10⁻³

y = 0.158 g and z = 0.156 Thus the mass of NaCl = 0.156 g and the mass percentage = 0.156/0.3146×100 = 49.7% NaCl

The mass percentage of NaCl(s) in the mixture is 49.7%

8 0

3 years ago