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Harrizon [31]
3 years ago
11

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 4

.80 L of an HCl solution with a pH of 1.9
Chemistry
1 answer:
geniusboy [140]3 years ago
7 0

Answer:

5.195 ml

Explanation:

Dissociation of HCl yields;

HCl(aq)   ⇄   H⁺(aq)     +    Cl⁻(aq)

∴ [HCl] = [H⁺] = [Cl⁻]

pH = 1.9

= -log[ H⁺]

= 10^{-pH}

= 10^{-1.9}

= 0.01258 M

Moles of HCl = number of mole × volume

= 0.01258 × 4.80 L

= 0.06042 mole

Mass of HCl = number of moles of HCl  ×  Molar mass of HCl

Mass of HCl = 0.06042 mol ×  36.5 g/mol

Mass of HCl = 2.205 grams

Also, given that the density =  1.179 g/mL

Mass of HCl in the solution = 36% of the density

i.e \frac{36}{100}*1.179

= 0.4244 grams

Volume of a concentrated HCl solution needed can now be determined as;

= \frac{2.205}{0.4244}

= 5.195 ml

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Why is it essential for a calorimetry to be an insulated (closed) system?
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B.Thermal insulation minimizes energy loses to the atmosphere.

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Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following
bearhunter [10]

Answer:

210.7~g~MgCO_3

Explanation:

We have to start with the <u>reaction</u>:

MgCO_3~->~MgO~+~CO_2

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:

(12*1)+(16*2)=44~g/mol

In other words: 1~mol~CO_2=~44~g~CO_2. With this in mind, we can calculate the moles:

110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

2.5~mol~CO_2=2.5~mol~MgCO_3

With the molar mass of MgCO_3 ((23.3*1)+(12*1)+(16*3)=84.3~g/mol. With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:

2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3

I hope it helps!

8 0
3 years ago
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