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antoniya [11.8K]
2 years ago
15

What role does the oxidizing agent play in a redox reaction

Chemistry
2 answers:
kvv77 [185]2 years ago
3 0
The oxidizing agent receives electrons from the reducing agent. 2) The oxidation agent takes electrons from the reducing agent.
ycow [4]2 years ago
3 0

A for plato. The oxidizing agent receives electrons from the reducing agent.

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In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
What would be the formula of the precipitate that forms when pb(no3)2 (aq) and k2so4 (aq) are mixed?
maria [59]
The formula of the ppt. formed is PbSo4 , which is inslouble.
6 0
2 years ago
Write the balanced chemical equation between H2SO4H2SO4 and KOHKOH in aqueous solution. This is called a neutralization reaction
const2013 [10]

Answer:

0.185M sulfuric acid

Explanation:

Based on the reaction:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>

Initial moles of H₂SO₄ and KOH are:

H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>

KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>

The moles of sulfuric acis that react with KOH are:

0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.

Thus, moles that remain are:

0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>

As total volume is 0.700L + 0.750L = 1.450L, concentration is:

0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>

8 0
2 years ago
What are some synonyms for dissolved
Luba_88 [7]
*Go into solution
*Break down
*Liquefy
*Disintegrate
*Melt away
*Evaporate
*Dissapear

Hope it helps! Brainliest answer, PLZ? :)
4 0
3 years ago
Can someone help me
Veseljchak [2.6K]

Answer:

Is it D?

Explanation:

3 0
3 years ago
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