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erastova [34]
3 years ago
6

Thomas poured 25 mL of 0.60 M HCl into a large bottle. He then added enough water so that the new (diluted) concentration was 0.

10 M HCl. How much water did Thomas add
Chemistry
1 answer:
Molodets [167]3 years ago
5 0

Answer:

The answer to your question is 150 ml

Explanation:

Data

Volume 1 = 25 ml

Concentration 1 = 0.6 M

Volume 2 = ?

Concentration 2 = 0.1 M

Formula

            Volume 1 x Concentration 1 = Volume 2 x Concentration 2

Solve for Volume 2

             Volume 2 = (Volume 1 x Concentration 1)/Concentration 2

Substitution

             Volume 2 = (25 x 0.6) / 0.1

Simplification

            Volume 2 = 15 / 0.1

Result

            Volume 2 = 150 ml

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8 0
3 years ago
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
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Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

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The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

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