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Rasek [7]
1 year ago
11

9. In a titration experiment 34.7 of a 0.145M solution of barium hydroxide \ Ba(OH) 2 \ is added to 20mL of hydrochloric acid (H

CI) of unknown concentration until the equivalence point is reached. What is a) the molarity of the acid? and b ) How many grams of HCI are in the solution ?
Chemistry
1 answer:
Ksenya-84 [330]1 year ago
8 0

Answer:

Titration reveals that 12 mL of 1.5 M hydrochloric acid are required to neutralize 25 mL calcium hydroxide solution . What is the molarity of the Ca(OH) 2 solution

Explanation:

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What is the % composition of Carbon in Chromium (iii) Carbonate
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Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

\begin{gathered} C\rightarrow3\times12=36 \\  \\ O\rightarrow9\times16=144 \\  \\ Cr\rightarrow2\times52=104 \end{gathered}

The molar mass will be thus:

M=36+104+144=284\text{ g/mol}

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\  \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}

The percent composition of Carbon is thus 12.7 %.

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