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3 years ago

One of the reactions used to inflate automobile air bags involves sodium azide (NaN₃).

Chemistry

1 answer:

AleksandrR [38]3 years ago

6 0

Answer:

161.54 g of N₂

Explanation:

The balanced equation for the reaction is given below:

2NaN₃ —> 2Na + 3N₂

Next, we shall determine the mass of NaN₃ that decomposed and the mass of N₂ produced from the balanced equation. This is illustrated below:

Molar mass of NaN₃ = 23 + (3×14)

= 23 + 42 = 65 g/mol

Mass of NaN₃ from the balanced equation = 2 × 65 = 130 g

Molar mass of N₂ = 2 × 14 = 28 g/mol

Mass of N₂ from the balanced equation = 3 × 28 = 84 g

SUMMARY:

From the balanced equation above,

130 g of NaN₃ decomposed to produce 84 g of N₂.

Finally, we shall determine the mass of N₂ produced by the decomposition of 250 g of NaN₃. This can be obtained as follow:

From the balanced equation above,

130 g of NaN₃ decomposed to produce 84 g of N₂.

Therefore, 250 g of NaN₃ will decompose to produce

= (250 × 84)/130 = 161.54 g of N₂

Thus, 161.54 g of N₂ were obtained from the reaction.

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Lakshmi has a sample of ammonium nitrate (NH4NO3) that has a mass of 40.10 g. She knows that the molar mass of NH4NO3 is 80.0432

avanturin [10]

The answer is 0.5010

Number of moles (n) is equal to the quotient of mass (m) and molar mass of a sample (Mr):
n = m/Mr

We have:
n = ?
m = 40.10 g
Mr = 80.0432 g/mol

n = 40.10 g : 80.0432 g/mol = 0.5010

40.10 has 4 significant digits,
80.0432 has 6 significant digits.
Since 4 is less than 6, we choose 4 significant digits

5 0

3 years ago

Sio2(s)+3c(s)→sic(s)+2co(g)

bonufazy [111]

Please explain in more detail

7 0

3 years ago

What is the mass of 48 ml of a liquid if its density is 1.25 g/ml?

soldi70 [24.7K]

Answer:

60 g

Explanation:

Mass = Density × Volume

Mass = 1.25 × 48

Mass = 60 g

I hope this helps!

4 0

3 years ago

Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.

o-na [289]

Answer:

pKa = 7.46

Explanation:

1) Data:

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

2) Strategy:

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

3) Solution:

a) pH

pH = - log [H₃O⁺]

4.64 = - log [H₃O⁺]

$[H_3O^+]= 10^{-4.64} = 2.29.10^{-5}$

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0

4 years ago

Would you expect the carbonyl carbon of benzaldehyde to be more or less electrophilic than that of acetaldehyde? Explain using r

Slav-nsk [51]

Answer:

Benzaldehyde is less electrophilic than acetaldehyde

Explanation:

Electrophilicity of carbonyl carbon depends on it's positive charge density.

In benzaldehyde, phenyl group can act as an electron donating group towards carbonyl group due to it's electron donating resonating effect.

In acetaldehyde, methyl group can act as an electron donating group towards carbonyl group due to it's electron donating inductive effect.

We know that resonating effect are more stronger than inductive effect.

Henece, it is expected that positive charge density on carbonyl carbon is much lower in benzaldehyde than acetaldehyde.

So, benzaldehyde is less electrophilic than acetaldehyde.

Resonating effect of phenyl group has been shown below.

4 0

3 years ago