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sdas [7]
3 years ago
5

Determine the percent composition of Ca3(PO4)2

Chemistry
1 answer:
Tamiku [17]3 years ago
8 0

determine the percent of composition

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How would you prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0% by mass with a density of 1.19 g/mL
hoa [83]
<h2>Step 1 : Identify the given </h2>

Volume = 250mL

Density = 1.19 g/ML

<h2>Step 2 . Calculate the mass of HCL </h2>

Density = mass/volume

∴Mass = Density * Volume

= 1.19g/mL* 250mL

= 297,5g

<h2>Step 3 : Calculate the total mass of the solution, given that concentration HCL is 38% </h2>

Mass of the total solution can be calculated by the following :

38% = Mc /297.5 * 100

Mc = 38/100 *297.5

= 113.05grams

• Finally, this means that mass of the total solution of 0.125M HCL i,s 113grams, ,you would use this mass to prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0%

6 0
1 year ago
An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

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So..... I believe this is a Convergent boundary and mountains..
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