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3 years ago

How many milliliters of 0.50 M KOH are needed

Chemistry

1 answer:

spayn [35]3 years ago

3 0

Answer:

Option D. 30 mL.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above,

The mole ratio of the acid, nA = 1

The mole ratio of the base, nB = 1

Step 2:

Data obtained from the question. This include the following:

Volume of base, KOH (Vb) =.?

Molarity of base, KOH (Mb) = 0.5M

Volume of acid, HNO3 (Va) = 10mL

Molarity of acid, HNO3 (Ma) = 1.5M

Step 3:

Determination of the volume of the base, KOH needed for the reaction. This can be obtained as follow:

MaVa / MbVb = nA/nB

1.5 x 10 / 0.5 x Vb = 1

Cross multiply

0.5 x Vb = 1.5 x 10

Divide both side by 0.5

Vb = (1.5 x 10) /0.5

Vb = 30mL

Therefore, the volume of the base, KOH needed for the reaction is 30mL.

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A sample of 508.4 grams of copper completely reacted with oxygen to form 572.4 grams of a copper oxide product. how many grams o

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According to law of conservation of mass, mass can neither be destroyed nor created in a chemical reaction. Thus, sum of masses of reactants must be equal to sum of masses of products in a reaction.

The chemical reaction is as follows:

$2Cu+O_{2}\rightarrow 2CuO$

Here, sum of masses of Cu and oxygen gas should be equal to CuO formed.

$2m_{Cu}+m_{O_{2}}=2m_{CuO}$

Thus, mass of oxygen will be:

$m_{O_{2}}=2(572.4-508.4)g=128 g$

This can be further proved as follows:

The balanced chemical reaction is as follows:

$2Cu+O_{2}\rightarrow 2CuO$

Here, 2 moles of Cu completely reacts with 1 mole of $O_{2}$ to give 2 moles of $CuO$ .

Thus, 1 mole of Cu reacts with 0.5 moles of $O_{2}$ .

The mass of Cu is 508.4 and molar mass is 63.546 g/mol, number of moles can be calculated as follows:

$n=\frac{m}{M}=\frac{508.4 g}{63.546 g/mol}=8 mol$

Thus, number of moles of $O_{2}$ reacting will be:

$n_{O_{2}}=8\times 0.5 mol=4 mol$

Molar mass of oxygen molecule is 32 g/mol thus, mass can be calculated as follows:

m=n×M=4 mol×32 g/mol=128 g/mol

This satisfies the law of conservation of mass.

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Answer:

freezing point (°C) of the solution = - 3.34° C

Explanation:

From the given information:

The freezing point (°C) of a solution can be prepared by using the formula:

$\Delta T = iK_fm$

where;

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the vant Hoff factor is the totality of the number of ions in the solution

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which can now be re-written as :

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