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goblinko [34]
3 years ago
7

How do you do this problem?

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Answer:

B

Explanation:

If the temperature of the water goes up, the reaction is exothermic (heat is being given away by the equation -- more precisely the reactants of the equation). Only A and B can be true. In order for the reaction to occur, the water has to absorb the heat. It's temperature goes up. Remember that minus sign. It is almost the key fact for this question.

The question is not as hard as it looks, but that is easy for me to say.

m = 100 g

c = 4.2 J/(g * oC)

deltaT = 21 - 20 degrees = 1 degree.

Heat = 100 * 4.2 * 1

Heat = 420 J

Heat = 420 * [1 kJ/1000 J]

Heat = -0.42 kJ

B

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What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? a: 0.24 b: 13.1 c: 0.076 d: 6.55 e: 14.3
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Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:

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where V represents the volume of solution and M the molar concentration of said solution.

In this case:

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  • M acid= 0.129 M
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Replacing:

0.0137 L* 0.129 M= V base* 0.135 M

Solving:

V base=\frac{0.0137 L*0.129 M}{0.135 M}

V base=0.0131 L = 13.1 mL

<u><em> The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>

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