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spin [16.1K]
3 years ago
6

Calculate the number of molecules found in 35 g of Sodium Hydroxide?

Chemistry
2 answers:
Helen [10]3 years ago
7 0

Answer:

5.27*10^23 (rounded to 3 significant figures)

Explanation:

The amount of molecules in one mole of anything is equal to Avogadro's number: 6.022×10^23

To find the number of moles of NaOH in 35 grams of it, do 35 divided by the molar mass (39.997): 35/39.997=0.87506562 moles of NaOH

To find the number of molecules, multiply the moles of NaOH by Avogadro's number: 0.87506562×(6.022×10^23)=5.26964522*10^23

inn [45]3 years ago
4 0

Answer:

5.27x10²³ molecules

Explanation:

In order to solve this problem we first <u>convert 35 g of Sodium Hydroxide (NaOH) into moles</u>, using its <em>molar mass</em>:

  • 35 g ÷ 40 g/mol = 0.875 mol

Finally we<u> convert 0.875 moles into number of molecules</u>, using <em>Avogadro's number</em>:

  • 0.875 mol * 6.023x10²³ molecules/mol = 5.27x10²³ molecules
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Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N
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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

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