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DENIUS [597]
3 years ago
14

What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 5

.20 L of an HCl solution with a pH of 1.70
Chemistry
1 answer:
charle [14.2K]3 years ago
8 0

Answer:

The volume we need is 8.89 mL

Explanation:

We analyse data:

36 % by mass → 36 g of HCl in 100 g of solution

Solution's density = 1.179 g/mL

5.20L → Volume of diluted density

pH = 1.70 → [H⁺] = 10⁻¹'⁷⁰ = 0.0199 M

 HCl    →       H⁺     +    Cl⁻

                 0.0199

pH can gives the information of protons concentrations, so as ratio is 1:1, 0.0199 M is also the molar concentration of HCl

Let's verify the molar concentration of the concentrated solution:

We convert the mass to moles: 36 g / 36.45 g/mol = 0.987 moles

As the solution mass is 100 g, we apply density to find out the volume:

Density = Mass / volume → Volume = Mass / Density

Volume = 100 g / 1.179 g/mL → 84.8 mL

Let's convert the volume from mL to L in order to define molarity

84.8 mL . 1L/ 1000mL = 0.0848 L

Molarity → 0.987 mol / 0.0848L = 11.6M

Let's apply the dilution formula:

M concentrated . V concentrated = M diluted . V diluted

11.6 M . V concentrated = 0.0199M . 5.20L

V concentrated = (0.0199M . 5.20L) / 11.6M → 8.89×10⁻³L

We can say, that the volume we need is 8.89 mL

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What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
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Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

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