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Ksivusya [100]
3 years ago
9

Discribe how to prepare 50.00mL of a 3.00M hydrochloric acid solution using 12.0 M HCl.

Chemistry
1 answer:
Leokris [45]3 years ago
6 0

Answer:

The answer to your question is given below.

Explanation:

To prepare 50mL of 3M HCl, we must calculate the volume of the stock solution needed. This can obtained as follow:

Molarity of stock solution (M1) = 12M

Volume of stock solution needed (V1) =?

Molarity of diluted solution (M2) = 3M

Volume of diluted solution (V2) = 50mL

The volume of the stock solution needed can be obtained by using the dilution formula as shown below:

M1V1 = M2V2

12 x V1 = 3 x 50

Divide both side by 12

V1 = (3 x 50)/12

V1 = 12.5mL

The volume of the stock solution needed is 12.5mL

Therefore, to prepare 50mL of 3M HCl, we must measure 12.5mL of the stock solution i.e 12M HCl and then, add water to the mark in a 1L volumetric flask. Now we can measure out 50mL of the solution.

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N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol
docker41 [41]

Answer:

A

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:


\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)

Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

5 0
2 years ago
Francine makes several measurements of the mass of a metal block. The data set is shown in the table below.
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As<u> there are 4 measurements</u>, the mean is:

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3 0
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The equilibrium constant, Kc, for the following reaction is 5.10×10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) Calculate the equilibri
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<u>Answer:</u> The equilibrium concentration of HCl is 2.26\times 10^{-3}M

<u>Explanation:</u>

We are given:

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Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}

Molarity of NH_4Cl=\frac{0.564}{1}=0.564M

The given chemical equation follows:

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<u>Initial:</u>         0.564

<u>At eqllm:</u>     0.564-x          x              x

The expression of K_c for above equation follows:

K_c=[NH_3][HCl]

The concentration of pure solid and pure liquid is taken as 1.

We are given:

K_c=5.10\times 10^{-6}

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So, [HCl]=2.26\times 10^{-3}M

Hence, the equilibrium concentration of HCl is 2.26\times 10^{-3}M

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