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Mekhanik [1.2K]
3 years ago
10

Problem PageQuestion The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium

azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide () into solid sodium and gaseous dinitrogen. 2. Suppose of dinitrogen gas are produced by this reaction, at a temperature of and pressure of exactly . Calculate the mass of sodium azide that must have reacted. Round your answer to significant digits.
Chemistry
1 answer:
Shalnov [3]3 years ago
6 0

Answer:

1. 2NaN₃(s) → 2Na(s) + 3N₂(g)

2. 14.5 g NaN₃

Explanation:

The answer is incomplete, as it is missing the required values to solve the problem. An internet search shows me these values for this question. Keep in mind that if your values are different your result will be different as well, but the solving methodology won't change.

" The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen. 2. Suppose 71.0 L of dinitrogen gas are produced by this reaction, at a temperature of 16.0 °C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits. "

1. The <u>reaction that takes place is</u>:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

2. We use PV=nRT to <u>calculate the moles of N₂ that were produced</u>.

P = 1 atm

V = 71.0 L

n = ?

T = 16.0 °C ⇒ 16.0 + 273.16 = 289.16 K

  • 1 atm * 71.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 289.16 K
  • n = 0.334 mol

Now we <u>convert N₂ moles to NaN₃ moles</u>:

  • 0.334 mol N₂ * \frac{2molNaN_{3}}{3molN_2} = 0.223 mol NaN₃

Finally we <u>convert NaN₃ moles to grams</u>, using its molar mass:

  • 0.223 mol NaN₃ * 65 g/mol = 14.5 g NaN₃

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Reptile [31]

<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]

For the given chemical reaction:

C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]

We are given:

\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0
3 years ago
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Answer:

The change in the pH (ΔpH) is 2,17

Explanation:

The reaction:

CH₃NH₂(aq) + H₂O(aq) ⇌ CH₃NH₃⁺(aq) + OH⁻

kb = \frac{[OH^{-}][CH_{3}NH_{3}^+]}{[CH_{3}NH_{2}]} <em>(1)</em>

In equilibrium, a solution of CH₃NH₂ 4,7M produces:

[CH₃NH₂] = 4,7 - x

[CH₃NH₃⁺] = x

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Replacing in (1):

4,38x10^{-4} = \frac{x^2}{4,7-x}

x² + 4,38x10⁻⁴x - 2,0586x10⁻³ = 0

The solutions are:

x = -0,0456 No physical sense. There are not negative concentrations.

x = 0,04515 Real answer.

The concentration of [OH⁻] is 0,04515 M.

As pOH = -log [OH⁻] And pH+pOH = 14. The pH of this solution is:

<em>pH = 12,65</em>

The addition of 6,7M produce this changes in concentrations:

[CH₃NH₂] = 4,656 + x

[CH₃NH₃⁺] = 6,74515 - x

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Replacing in (1) you will obtain:

x² - 6,7907x + 0,3025 = 0

Solving for x:

x = 6,74586 No physical sense

x = 0,04484 Real answer.

Thus, [OH⁻] = 0,04515 - 0,044842 = 3,08x10⁻⁴M

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I hope it helps!

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Answer:

See Explanation

Explanation:

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1 mole of KHSO4 reacts with 1 mole of KOH

0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH

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Mass of excess KOH = 0.09 moles * 56.1056 g/mol  = 5 g of excess KOH

5 0
3 years ago
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