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tiny-mole [99]
3 years ago
10

What is the oxidation number of phosphorus (P) in sodium phosphate (Na3PO4)?

Chemistry
2 answers:
joja [24]3 years ago
8 0
Na⁺¹₃P⁺⁵O⁻²₄

+1*3 + (+5) + (-2*4) = 0
marissa [1.9K]3 years ago
3 0

Let the oxidation number of P be x.  

We know that oxidation number of Na= +1, O= -2  

therefore,  3*1 + x+ 4* -2 = 0 (net charge on the molecule)  

3 + x - 8 =0  

x = +5  

therefore, the answer is C)+5

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Consider the following system at equilibrium:A(aq)+B(aq) <---> 2C(aq)Classify each of the following actions by whether it
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Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

  • On addition of reactant at equilibrium shifts the equilibrium in forward direction.
  • On addition of product at equilibrium shifts the equilibrium in backward direction.
  • On removal of reactant at equilibrium shifts the equilibrium in backward direction.
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A(aq)+B(aq)\rightleftharpoons 2C(aq)

Reactants = A , B

Product = C

1. Increase A

On increasing the amount of A at equilibrium will shift the equilibrium in forward or rightward direction.

2. Increase B

On increasing the amount of B at equilibrium will shift the equilibrium in forward or rightward direction.

3. Increase C

On increasing the amount of C at equilibrium will shift the equilibrium in backward or leftward direction.

4. Decease A

On decreasing the amount of A at equilibrium will shift the equilibrium in backward or leftward direction.

5. Decease B

On decreasing the amount of B at equilibrium will shift the equilibrium in backward or leftward direction.

6. Decease C

On decreasing the amount of C at equilibrium will shift the equilibrium in forward or rightward direction.

7. Double A and Halve B

Equilibrium constant of the reaction = K

K=\frac{[C]^2}{[A][B]}

On doubling A and halving B, equilibrium constant of the reaction = K'

K'=\frac{[C]^2}{[2A][\frac{B}{2}]}=\frac{[C]^2}{[A][B]}

The value of equilibrium constant K' is equal to K, which means that equilibrium will not shift in any direction.

8. Double both B and C

Equilibrium constant of the reaction = K

K=\frac{[C]^2}{[A][B]}

On doubling B and C, equilibrium constant of the reaction = K'

K'=\frac{[2C]^2}{[A][2B]}=\frac{4[C]^2}{[A][2B]}=\frac{2[C]^2}{[A][B]}

K' = 2 K

The value of equilibrium constant K' is double the K, which means that product is increasing which means that equilibrium will shift in backward or leftward direction.

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3 years ago
How many atoms are in 5.62 moles of Mercury?
Maslowich

Answer:

3.384364 time 10^24 atoms

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multiply by avogadro's number

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