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Katyanochek1 [597]
3 years ago
9

Consider the following reaction:

Chemistry
1 answer:
lesantik [10]3 years ago
6 0

Answer:

the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Explanation:

Since

Br₂(g) + Cl₂(g) ⇌ 2BrCl(g) , Kp=1.112 at 150 K

denoting BC as BrCl , B as Br₂ , C as Cl₂, p as partial pressure , then

Kp = pBC²/[pB*pC]

solving for pBC

pBC = √(Kp*pB*pC)

replacing values

pBC = √(Kp*pB*pC) = √(1.112*751 torr*737 torr) = 784.52 torr

pBC = 784.52 torr

then the equilibrium partial pressure of BrCl is pBC = 784.52 torr

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5 0
3 years ago
Which example shows electrical energy being transformed into heat energy?
alekssr [168]

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B

Explanation:

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4 0
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The number of neutrons and protons in the nuclei of two atom A and B is as follow: [A] proton= 8 and neutron = 8, [B] proton = 8
sveticcg [70]
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4 0
3 years ago
The first-order decay of radon has a half-life of 3.823 days. How many grams of radon decomposes after 5.55 days if the sample i
Leni [432]

Answer:

The mass of radon that decompose = 63. 4 g

Explanation:

R.R = P.E/(2ᵇ/ⁿ)

Where R.R = radioactive remain, P.E = parent element, b = Time, n = half life.

Where P.E = 100 g , b = 5.55 days, n = 3.823 days.

∴ R.R = 100/2^{5.55/3.823}

  R.R = 100/2^{1.45}

  R.R = 100/2.73

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The mass of radon that decompose = Initial mass of radon - Remaining mass of radon after radioactivity.

Mass of radon that decompose = 100 - 36.63

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The mass of radon that decompose = 63. 4 g

8 0
3 years ago
An analytical chemist is titrating of a solution of hydrazoic acid with a solution of . The of hydrazoic acid is . Calculate the
serg [7]

Answer:

pH = 12.43

Explanation:

<em>...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it</em>.

To solve this question we need to know that hidrazoic acid reacts with KOH as follows:

HN3 + KOH → KN3 + H2O

<em>Moles KOH:</em>

0.5716L * (0.2900mol /L) =0.1658 moles of KOH

<em>Moles HN3:</em>

0.2127L * (0.6800mol/L) = 0.1446 moles HN3

As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:

0.1658 moles - 0.1446 moles =

0.0212 mol KOH

In 212.7mL + 571.6mL = 784.3mL = 0.7843L

The molarity of KOH = [OH-] is:

0.0212 mol KOH / 0.7843L = 0.027M = [OH-]

The pOH is defined as -log [OH-]

pOH = -log 0.027M

pOH = 1.57

pH = 14 - pOH

pH = 12.43

6 0
3 years ago
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