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Answer: option D. the ability of a base to react with a soluble metal salt.
Justification:
NaOH is a strong base, which means that in water it will dissociate according to this reaction:
- NaOH(aq) → Na⁺ (aq) + OH⁻ (aq)
On the other hand, CuSO₄ is a soluble ionic salt which in water will dissociate into its ions according to this other reaction:
Hence, in solution, the sodium ion (Na⁺) will react with the metal salt in a double replacement reaction, where the highly reactive sodium ion (Na⁺) will substitute the Cu²⁺ in the CuSO₄ to form the sodium sulfate salt, Na₂SO₄ (water soluble), and the copper(II) hydroxide, Cu(OH)₂ (insoluble).
That is what the given reaction represents:
CuSO₄ (aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
↑ ↑ ↑ ↑
soluble metal salt strong base insoluble base solube salt
Conjugate base of Propanoic acid (
is propanoate where -COOH group gets converted to -CO
. The structure of conjugate base of Propanoic acid is shown in the diagram.
The 
above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.
= 
+ log![\frac{[Conjugate base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BConjugate%20base%5D%7D%7B%5Bacid%5D%7D)
=4.9+log
=5.85
As 90% conjugate base is present, so propanoic acid present 10%.
The value of equilibrium constant is equal to the quotient of the products raised to its stoichiometric coefficient over the reaction's reactants raised to its respective stoichiometric coeff. The equation is Kc=[SO2][Cl2]/[SO2Cl2]= [1.3*10^-2][1.3*10^-2]/[2.2*10^-2-<span>1.3*10^-2]=0.0188. The final answer is Kc=0.0188.</span>
Answer:1 3 4 1 im pretty positive
