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riadik2000 [5.3K]

3 years ago

12

When an irregular shaped object was put in a graduated cylinder with an initial volume of 25 mL of water, the water level rose t

o 54.0 mL of water after the object was submerged. If the mass of the object is 7.0 grams, what is the density of the object

1 answer:

alexandr402 [8]3 years ago

7 0

Answer:

0.24 g/mL

Explanation:

The density of an object is given by the ratio between its mass and its volume:

$\rho=\frac{m}{V}$

where

m is the mass of the object

V is its density

In this problem, we have:

m = 7.0 g is the mass of the object

The volume of an irregular shaped object can be measured by putting it into water, and by measuring the difference in water volume.

In this case,

$V_1=25 mL$ is the initial volume of water

$V_2=54 mL$ is the final volume of water

So the volume of the object is

$V=V_2-V_1=54-25=29 mL$

Therefore, the density of the object is:

$\rho = \frac{7.0 g}{29 mL}=0.24 g/mL$

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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

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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 $\rm Ag^{+}$ 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

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

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

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

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

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