Explanation:
An acid is a substance that produces excess hydroxonium ions in solution.
An acid based on the pH scale is a substance that has a low pH. Acid lies within a range of 1-7 on the pH scale.
- A pH of 7 is for neutral compounds like water.
- A pH greater that 7 is for basic compounds.
- In order to raise the pH, we are driving at a substance becoming more neutral or basic.
This can be achieved by adding more base to the solution of the substance. When we add more base, hydroxyl ions will neutralize the excess hydroxonium ions and drag the pH towards that of neutrality.
Addition of more base can eventually make the substance basic.
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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
Answer:
The higher the temperature, the faster the particles move, the lower the temperature, the slower.
Explanation:
(a) As the given chemical reaction equation is as follows.
So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.
Hence, equation for rate law of reaction will be as follows.
Rate = ![K \times [OCl^{-}] \times [l^{-}]](https://tex.z-dn.net/?f=K%20%5Ctimes%20%5BOCl%5E%7B-%7D%5D%20%5Ctimes%20%5Bl%5E%7B-%7D%5D)
(b) Since, the rate equation is as follows.
Rate = ![K [OCl^{-}][l^{-}]](https://tex.z-dn.net/?f=K%20%5BOCl%5E%7B-%7D%5D%5Bl%5E%7B-%7D%5D)
Let us assume that (![[OCl^{-}] = [l^{-}]](https://tex.z-dn.net/?f=%5BOCl%5E%7B-%7D%5D%20%3D%20%5Bl%5E%7B-%7D%5D)
)
Putting the given values into the above equation as follows.
K = 
= 
Hence, the value of rate constant for the given reaction is
.
(c) Now, we will calculate the rate as follows.
Rate =
= 
= 
Therefore, rate when ![[OCl^{-}] = 1.8 \times 10^{3}](https://tex.z-dn.net/?f=%5BOCl%5E%7B-%7D%5D%20%3D%201.8%20%5Ctimes%2010%5E%7B3%7D)
M and ![[I^{-}]= 6.0 \times 10^{4}](https://tex.z-dn.net/?f=%5BI%5E%7B-%7D%5D%3D%206.0%20%5Ctimes%2010%5E%7B4%7D)
M is .
The (B) answer (B) is (B) (B) (B)
