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posledela
3 years ago
15

A lead ball is added to a graduated cylinder containing 41.7 ml of water, causing the level of the water to increase to 96.0 mL.

What is the volume in milliliters of the lead ball, Vieras? Viead =
Chemistry
1 answer:
rjkz [21]3 years ago
5 0

Answer:

Vlead ball=54.3 mL

Explanation:

The graduated cylinder contains 41.7mL of water

mL is a volume unit.

Water volume = 41.7 mL

The lead ball caused an increase of volume from 41.7 mL to 96.0 mL

The new volume is the lead ball volume plus the original water volume :

Final volume = Vlead ball+ Water original volume

96.0mL=Vleadball +41.7mL

Vlead ball=96.0mL-41.7mL

Vleadball=54.3mL

This is actually true if we suppose that the lead ball is fully sunken in the water.

We always must consider that the volume difference is the volume that the sunken object is occupying in the water.

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A solution of H 2 SO 4 ( aq ) with a molal concentration of 1.66 m has a density of 1.104 g / mL . What is the molar concentrati
uysha [10]

Answer:

1.58 M

Explanation:

H_2SO_4 is 1.66 m concentration.

Which means that 1.66 moles of  H_2SO_4 are present in 1 kg of the solvent, water.

Mass of water = 1 kg = 1000 g

Moles of H_2SO_4 = 1.66 moles

Molar mass of H_2SO_4 = 98.079 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

1.66\ mole= \frac{Mass}{98.079\ g/mol}

Mass_{H_2SO_4}= 162.81114\ g

Total mass = 1000 g + 162.81114 g = 1162.81114 g

Density = 1.104 g/mL

Volume of the solution = Mass / Density = 1162.81114 / 1.104 mL = 1053.27 mL = 1.05327 L

Considering:-

<u>Molarity = moles/ Volume of solution = 1.66 / 1.05327 M = 1.58 M </u>

4 0
3 years ago
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4 years ago
Calculate the maximum solubility of silver carbonate, Ag2CO3 in g/L when in the presence of 0.057 M AgNO3. The solubility produc
Andreyy89

Answer:

Approximately 4.2 \times 10^{-7}\; \rm g \cdot L^{-1}.

Explanation:

Start by finding the concentration of \rm Ag_2CO_3 at equilibrium. The solubility equilibrium for

\rm Ag_2CO_3 \; (s) \rightleftharpoons 2\, Ag^{+}\; (aq) + {CO_3}^{2-}\; (aq).

The ratio between the coefficient of \rm Ag_2CO_3 and that of \rm Ag^{+} is 1:2. For

Let the increase in \rm {CO_3}^{2-} concentration be +x\; \rm mol \cdot L^{-1}. The increase in \rm Ag^{+} concentration would be +2\,x\; \rm mol \cdot L^{-1}. Note, that because of the 0.057\; \rm mol \cdot L^{-1}of \rm AgNO_3, the concentration of

  • The concentration of \rm Ag^{+} would be (0.057 + 2\, x) \; \rm mol\cdot L^{-1}.
  • The concentration of \rm {CO_3}^{2-} would be x\; \rm mol \cdot L^{-1}.

Apply the solubility product expression (again, note that in the equilibrium, the coefficient of \rm Ag^{+} is two) to obtain:

\begin{aligned}&\rm \left[Ag^{+}\right]^2 \cdot \left[{CO_3}^{2-}\right] = K_{\text{sp}} \\ & \implies (0.057 + x)^2\cdot x = 8.1 \times 10^{-12} \end{aligned}.

Note, that the solubility product of \rm Ag_2CO_3, K_{\text{sp}} = 8.1 \times 10^{-12} is considerably small. Therefore, at equilibrium, the concentration of

Apply this approximation to simplify (0.057 + x)^2\cdot x = 8.1 \times 10^{-12}:

0.057^2\, x \approx (0.057 + x)^2 \cdot x = 8.1 \times 10^{-12}.

\begin{aligned} x &\approx \frac{8.1 \times 10^{-12}}{0.057^2}\end{aligned}.

Calculate solubility (in grams per liter solution) from the concentration. The concentration of \rm Ag_2CO_3 is approximately \displaystyle \frac{8.1 \times 10^{-12}}{0.057^2}\; \rm mol\cdot L^{-1}, meaning that there are approximately \displaystyle n = \frac{8.1 \times 10^{-12}}{0.057^2}\; \rm mol of

\begin{aligned}m &= n \cdot M \\ &\approx \displaystyle \frac{8.1 \times 10^{-12}}{0.057^2} \; \rm mol\times 167.91\; g \cdot mol^{-1} \\ &\approx 4.2 \times 10^{-7}\; \rm g \end{aligned}.

As a result, the maximum solubility of \rm Ag_2CO_3 in this solution would be approximately 4.2 \times 10^{-7}\; \rm g \cdot L^{-1}.

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3 years ago
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