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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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mafiozo [28]

Answer:

Option B is correct.

4

Explanation:

We know that an atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example, if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.

In given problem we are given with 2 neutrons of helium. We know that the atomic number of He is 2. Thus Mass number of He is,

Number of neutrons + number of proton

2 + 2 = 4

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7 0
3 years ago
A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62
vesna_86 [32]

Answer: The concentration of H_2SO_4 is 0.234 M

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity H_2SO_4 = 2

M_1 = molarity of H_2SO_4 solution = ?

V_1 = volume of  H_2SO_4 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 0.375 M

V_1 = volume of  NaOH solution = 62.5 ml

Putting in the values we get:

2\times M_1\times 50.0=1\times 0.375\times 62.5

M_1=0.234M

Therefore concentration of H_2SO_4 is 0.234 M

6 0
3 years ago
Q: A 25.5 mL aliquot of HCl (aq) of unknown concentration was titrated with 0.113 M NaOH (aq). It took 51.2 mL of the base to re
blondinia [14]
M ( HCl ) = ?

V ( HCl ) = 25.5 mL in liters : 25.5 / 1000 => 0.0255 L

M ( NaOH ) = 0.113 M

V ( NaOH ) = 51.2 mL / 1000 => 0.0512 L

number of moles NaOH:

n = M x V

n = 0.113 x <span> 0.0512 => 0.0057856 moles of NaOH

mole ratio:

</span><span>HCl + NaOH = NaCl + H2O
</span><span>
1 mole HCl -------------- 1 mole NaOH
( moles HCl ) ----------- </span><span> 0.0057856 moles NaOH
</span>
(moles HCl ) = <span> 0.0057856 x 1 / 1
</span>
= <span> 0.0057856 moles of HCl
</span>
M ( HCl ) = n / V

M =  0.0057856 / <span>0.0255
</span>
= 0.227 M

Answer A

hope this helps!

4 0
3 years ago
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Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ
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Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ. 
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How do you test for starch in green plant
Sunny_sXe [5.5K]

Answer:

1,Iodine solution is used to test leaves for the presence of starch. ...

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3.The leaf on the left is a variegated leaf

Explanation:

4 0
3 years ago
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