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SOVA2 [1]
3 years ago
9

The reaction 2PH3(g)+As2(g)⇌2AsH3(g)+P2(g) has Kp=2.9×10−5 at 873 K. At the same temperature, what is Kp for each of the followi

ng reactions?Parts A, B, and CPart A:2AsH3(g)+P2(g)⇌2PH3(g)+As2(g)Part B:6PH3(g)+3As2(g)⇌3P2(g)+6AsH3(g)Part C:2P2(g)+4AsH3(g)⇌2As2(g)+4PH3(g)
Chemistry
1 answer:
slavikrds [6]3 years ago
8 0

Answer:

Part A

Kp = 3.4 x 10⁴

Part B

Kp = 2.4 x 10⁻¹⁴

Part C

Kp = 1.2 x 10⁹

Explanation:

2PH₃(g) + As₂(g)  ⇌ 2 AsH₃(g) + P₂(g)  Kp = 2.9 x 10⁻⁵

Kp = [AsH₃]²[P₂]/[PH₃]²[As]  = 2.9 x 10⁻⁵

Part A

it is the inverse of the equilibrium given

Kp(A) = 1/ Kp = 1 / 2.9 x 10⁻⁵ = 3.4 x 10⁴

Part B

Is the equilibrium where the coefficients have been multiplied by 3,

Kp(B) = ( Kp )³ = ( 2.9 x 10⁻⁵ )³ = 2.4 x 10⁻¹⁴

Part C

This is the  reverse equilibrium multipled by 2.

Kp(C) = ( 1/Kp)² = ( 1/ 2.9 x 10⁻⁵ )² = 1.2 x 10⁹

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3 years ago
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What mass is contributed by Br-79?
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Calculate the mass of BR -79? Bromine has 2 naturally occurring isotopes (Br-79 and Br-81) and has an atomic mass of 79.904 amu

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3 years ago
Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o
kkurt [141]

<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of CrO_2 = 480.1 g

Molar mass of CrO_2 = 84 g/mol

Putting values in equation 1, we get:

\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol

For the given chemical equation:

4CrO_2\rightarrow 2Cr_2O_3+O_2

By Stoichiometry of the reaction:

4 moles of CrO_2 produces 2 moles of chromium (III) oxide

So, 5.72 moles of CrO_2 will produce = \frac{2}{4}\times 5.72=2.86mol of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g

To calculate the percentage yield of chromium (III) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

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3 years ago
What is the lowest temperature a human can survive?
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3 0
3 years ago
"11. Barium nitrate reacts with aqueous sodium sulfate to produce solid barium sulfate and aqueous sodium nitrate. Abigail place
Amanda [17]

Answer:

44 mL of Na2SO4

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

Step 2:

Determination of the number of mole of Ba(NO3)2 in 20.00 mL of 0.500 M barium nitrate (Ba(NO3)2). This is illustrated below:

Molarity of Ba(NO3)2 = 0.5 M

Volume of solution = 20 mL = 20/1000 = 0.02 L

Mole of solute (Ba(NO3)2) =?

Molarity = mole /Volume

0.5 = Mole of Ba(NO3)2 / 0.02

Cross multiply to express in linear form

Mole of Ba(NO3)2 = 0.5 x 0.02

Mole of Ba(NO3)2 = 0.01 mole

Step 3:

Determination of the number of mole of Na2SO4 that reacted.

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

From the balanced equation above,

1 mole of Ba(NO3)2 reacted with 1 mole of Na2SO4.

Therefore, 0.01 mole of Ba(NO3)2 will also react with 0.01 mole of Na2SO4.

Step 4:

Determination of the volume of Na2SO4 needed for the reaction. This is illustrated below:

Mole of Na2SO4 = 0.01 mole

Molarity of Na2SO4 = 0.225M

Volume =?

Molarity = mole /Volume

0.225 = 0.01 / volume

Cross multiply to express in linear form

0.225 x Volume = 0.01

Divide both side by 0.225

Volume = 0.01/0.225

Volume of Na2SO4 = 0.044 L

Converting 0.044 L to mL, we have

Volume of Na2SO4 = 0.044 x 1000

Volume of Na2SO4 = 44 mL

Therefore, 44 mL of Na2SO4 is needed for the reaction

6 0
4 years ago
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