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vfiekz [6]
3 years ago
15

White vinegar is available as a 5% acetic acid solution and a 10% solution, by volume. How many millilitres of each solution mus

t be added to make 100ml of a 7%solution?
Chemistry
1 answer:
FrozenT [24]3 years ago
7 0

Answer:

The answer to your question is 60 ml of solution 5% and 40 ml of solution 10 %

Explanation:

Data

acetic acid = 5% = concentration 1

acetic acid 2 = 10% = concentration 2

final solution = 7 % =

final volume = 100 ml

Process

1.- Write an equation of the process

x = volume of solution 5%

100 - x = volume of solution 10%

Final concentration x final volume = concentration 1 x volume 1 + concentration 2 x volume 2

      0.07(100) = 0.05x + 0.10(100 - x)

2.- Solve for x

                  7 = 0.05x + 10 - 0.1x

                 7 - 10 = 0.05 - 0.1 x

                   - 3 = -0.05 x

                     x = -3/-0.05

                     x = 60 ml of solution 5%

3.- Calculate the volume of solution 10%

     Volume of 10% = 100 - 60

     Volume of 10% = 40 ml        

                         

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7 0
3 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
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Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

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