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3 years ago

7

Which solution has the greatest buffer capacity? Select the correct answer below: 1 mole of acid and 1 mole of base in a 1.0 L s

olution

1 answer:

Lorico [155]3 years ago

8 0

Answer:

The answer is

Explanation:

1 mole of acid.

Hope this helps....

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HELP! 10 pts!! || Which does NOT describe a water molecule?

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Which statement best describes the reaction pathway graph for an exothermic reaction but not an endothermic reaction? It has a h

ruslelena [56]

Answer: "The reactants are higher in energy than the products"

Explanation:

The exothermic reactions are characterized by the release of heat to the surroundings. The reactants lose heat that is delivered to the surroundings which implies that the products will be lower in energy than the reactants.

The hills that you can see in a reaction energy diagram are not related with the final change of energy. The hills are an indication of the activation energy needed to start the reaction, but they do not measure the change of energy from the products to the reactants.

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If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?

ICE Princess25 [194]

<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

$k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}$

where,

k = rate constant = ?

t = time taken = 1.52 hrs

$[A_o]$ = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

$k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}$

To calculate the half life period of first order reaction, we use the equation:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life period of first order reaction = ?

k = rate constant = $0.2098hr^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs$

Hence, the half life of the sample of silver-112 is 3.303 hours.

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