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dolphi86 [110]
2 years ago
5

Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). Match Term Definition Remo

ving ammonia A) No effect Removing hydrogen gas B) Shift to the right Adding a catalyst C) Shift to the left
Chemistry
1 answer:
pshichka [43]2 years ago
7 0

Answer:

The answers to the questions are given below.

Explanation:

According to Le Chatelier's principle, if an external constrain such as change in concentration, temperature or pressure is imposed on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.

A. Effective of removing ammonia, NH3.

N2(g) + 3H2(g) ⇌ 2NH3(g)

Removing NH3 from the reaction simply means we are left with more reactants and no product. Therefore, the reactant will react to produce the product. Hence, the equilibrium position will shift to the right.

2. Effect of removing H2

N2(g) + 3H2(g) ⇌ 2NH3(g)

Remoing H2 simply means we have more products and less reactant. Therefore, the product will be convert to reactant. Hence, the equilibrium position will shift to the left.

C. Effect of adding a catalyst.

N2(g) + 3H2(g) ⇌ 2NH3(g)

Catalyst does not affect the equilibrium position. It only creates an alternative path to arrive at the product within a short time. Hence, it has no effect.

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Cacodyl, which has an intolerable garlicky odor and is used in the manufacture of cacodylic acid, a cotton herbicide, has a mass
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Answer:

The molecular formula of cacodyl is C₄H₁₂As₂.

Explanation:

<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:

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Now we convert those masses into moles:

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1 year ago
Water at 4.4°C is to flow through a horizontal, commercial steel pipe having a length of 305 m at the rate of 150 gal/min. If a
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Answer:

d = 70.5 mm

Explanation:

given,

length of pipe = 305 m

discharge rate = 150 gal/min

pipe diameter = ?

1 gal/min = 6.30902 ×  10⁻⁵ m³/s

150 gal/min = 150 × 6.30902 ×  10⁻⁵ m³/s

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h = \dfrac{flv^2}{2gd}

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h = \dfrac{fl(\dfrac{Q}{A})^2}{2gd}

h = \dfrac{fl(\dfrac{Q}{\dfrac{\pi}{4}d^2})^2}{2gd}

h= \dfrac{8flQ^2}{\pi^2gd^5}

f = 0.048 from moody chart using P/D = 0.00015

\dfrac{1}{d^5}= \dfrac{6.1\times \pi^2\times 9.8}{8\times 0.048\times 305\times 0.00946^2}

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