2 H2 + O2 → 2 H2O What is the product of this chemical reaction

In a kitchen, during preparation of food or tea, salt and sugar are

Rzqust [24]

Answer:

A mixture of salt and sugar can be separated by the process of filtration.

5 0

2 years ago

Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N

Vitek1552 [10]

Answer : The value of work done for the system is 1144.69 J

Explanation :

First we have to calculate the moles of $NaN_3$

$\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}$

Molar mass of $NaN_3$ = 65.01 g/mole

$\text{Moles of }NaN_3=\frac{20.2g}{65.01g/mole}=0.311mole$

Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

As, 2 mole of $NaN_3$ react to give 3 mole of $N_2$

So, 0.311 moles of $NaN_3$ react to give $\frac{0.311}{2}\times 3=0.466$ moles of $N_2$

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = ?

n = number of moles $N_2$ = 0.466 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $22^oC=273+22=295K$

Putting values in above equation, we get:

$1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K$

$V=11.3L$

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

5 0

3 years ago

How many carbon atoms are in the smallest cyclic hydrocarbon?

SOVA2 [1]

There are 3 carbon atoms

3 0

3 years ago

Read 2 more answers

Consider a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For ammonia, pKb=4.75. Calculate the pH of 1.0 L of the so

vladimir2022 [97]

You have 0.50 mol of NH3 and 0.20 mol of NH4+ to start (NH4Cl dissolves completely), given the molarity and 1.0 L solution.

30.0 mL of 1.0 M HCl is 0.0300 mol of HCl. This will react with the NH3 to produced 0.030 mol of NH4+.

You now have 0.47 mol NH3 and 0.23 mol NH4+. Now use the Henderson-Hasselbach equation to calculate your pH. The equation says to use concentration of acid and base, but you can just use the moles of them because it doesn’t make a difference.

pH = pKa + log(base/acid)

pKa = 14 - pKb = 14 - 4.75 = 9.25

pH = 9.25 + log(0.47/0.23) = 9.56

5 0

3 years ago

The combustion reaction of c3h8o with o2 is represented by this balanced chemical equation. 2c3h8o + 9o2 6co2 + 8h2o when 3.00 g

MArishka [77]

<span>According to the question- 1 mol C3H8O = 60.096 g C3H8O 2 mol C3H8O = 9 mol O2 1 mol O2 = 31.998 g O2 [(3.00 g C3H8O)/1][(1 mol C3H8O)/(60.096)][(9 mol O2)/(2 mol C3H8O)][(32.998 g O2)/(1 mol O2)] = 7.1880435 g O2 Since 7.1880435 g of O2 is needed, and 7.38 g of O2 is available, 0.199565 g of O2 will be left over and oxygen is present in excess. Next, we need to convert 0.199565 g of O2 into moles of O2: [(0.199565 g O2)/1][(1 mol O2)/(31.998 g O2)] = 0.005999 mol O2, or 0.006 mol O2</span>

3 0

3 years ago