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ad-work [718]
3 years ago
6

A reaction requires 15.0 g of ammonium chloride. What volumeof a

Chemistry
1 answer:
Lubov Fominskaja [6]3 years ago
7 0

Answer:

103 cm³

Explanation:

A reaction requires 15.0 g of ammonium chloride. The percent by mass of ammonium chloride in the solution is 14%. The mass of the solution required is:

15.0 g ammonium chloride × (100 g solution/ 14 g ammonium chloride) = 107 g solution

The density of the solution is 1.04 g/cm³. The volume of solution corresponding to 107 g is:

107 g × (1 cm³/ 1.04 g) = 103 cm³

In 103 cm³ of a 14% solution, there will be 15.0 g of ammonium chloride.

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A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )
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Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

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The reaction rate of equation (1) is given by:

rate = k*[O_{3}][NO]     (2)

<u>We have:</u>

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s  

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

\frac{\Delta[NO_{2}]}{\Delta t} = rate

\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!                                

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