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3 years ago

The equation, H2SO4 (aq) + Sr(OH)2 (aq) arrow SrSO4 (aq) + H2O (l), represents what?

2 answers:

7 0

I think the correct answer from the choices listed above is the second option. The equation, H2SO4 (aq) + Sr(OH)2 (aq) ----> SrSO4 (aq) + H2O (l), represents overall equation for a strong acid-strong base neutralization reaction. Hope this answers the question. Have a nice day.

hichkok12 [17]3 years ago

6 0

Answer: The correct answer is overall equation for a strong acid-strong base neutralization reaction.

Explanation:

Neutralization reaction is defined as the reaction in which an acid reacts with a base to produce a salt and water molecule.

Strong acid is defined as the acid which gets completely dissociated into its ions when dissolved in water. For Example: $H_2SO_4$ etc..

Weak acid is defined as the acid which does not get completely dissociated into its ions when dissolved in water. For Example: $HCOOH$ etc..

Strong base is defined as the base which gets completely dissociated into its ions when dissolved in water. For Example: $NaOH$ etc..

Weak base is defined as the base which does not get completely dissociated into its ions when dissolved in water. For Example: $NH_4OH$ etc..

For the given chemical reaction:

$H_2SO_4(aq.)+Sr(OH)_2(aq.)\rightarrow SrSO_4(aq.)+H_2O(l)$

Here, $H_2SO_4$ is a strong acid and $Sr(OH)_2$ is a strong base. So, the salt produced by these two will be a neutral salt and their reaction will be a strong acid-strong base neutralization.

Hence, the correct answer is overall equation for a strong acid-strong base neutralization reaction.

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$\boxed{\text{8 da}}$

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The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"

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After one half-life, half (50 %) of the original amount will remain.

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We can construct a table as follows:

$\begin{array}{cccl}\textbf{No. of} & & \textbf{Fraction} & \\\textbf{half-lives} & \textbf{t/da} & \textbf{remaining} & \\1 & 4 & \dfrac{1}{2} & \\\\2 & 8 & \dfrac{1}{4}& \\\\3 & 12 & \dfrac{1}{8}& \\\end{array}$

$\text{We see that 8 da is two half-lives, and the fraction of the element remaining is $\frac{1}{4}$.}\\\text{It takes $\boxed{\textbf{8 da}}$ for three-fourths of the element to decay}$

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