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Katen [24]
3 years ago
13

Compute 4.659×104−2.14×104. Round the answer appropriately

Chemistry
1 answer:
Vilka [71]3 years ago
7 0
 the answer is 261.976
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The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that
kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is 1.1\times 10^6 M.

HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)

This value indicates that

A. CN^- is a stronger base than ONO^-

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is ONO^-

D. The conjugate acid of CN- is HCN

Answer: A. CN^- is a stronger base than ONO^-

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When K_{p}>1; the reaction is product favoured.

When K_{p}; ; the reaction is reactant favored.

When K_{p}=1; the reaction is in equilibrium.

As, K_p>>1, the reaction will be product favoured and as it is a acid base reaction where HONO acts as acid by donating H^+ ions and CN^- acts as base by accepting H^+

Thus HONO is a strong acid thus ONO^- will be a weak conjugate base and CN^- is a strong base which has weak HCN conjugate acid.

Thus the high value of K indicates that CN^- is a stronger base than ONO^-

7 0
2 years ago
im making a poster for chemistry. the topic is acid and base. i have to make a creative title to go with the poster. any ideas?
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3 years ago
What on earth is the oxidation number of N in N2H4?
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N₂H₄ :

N =  - 2  

H = + 1

 2 * ( -2 ) + ( 1 * 4 ) =  - 4 + 4 = 0

hope this helps!

5 0
2 years ago
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