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3 years ago

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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 =

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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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.

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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

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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:

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

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

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

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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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