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3 years ago

Aluminum boils at 2467°C. Aluminum’s boiling point in Kelvin is 2194. True or False.

Chemistry

2 answers:

Sergeeva-Olga [200]3 years ago

6 0

Answer:

The given statement is false.

Explanation:

Given that aluminum boils at 2,467°C

Boiling point of aluminum in kelvins is 2,194 K

Temperature in degree Celsius can be converted into Kelvins by relation:

$(T)^oC=273 +(T) K$

So, boiling point of aluminium in Kelvins will be:

$2,467 ^oC=273 +(2,467) K=2,740 K$

2,740 K ≠ 2,194 K

But the given value in Kelvins is not coming equal to our calculated value, Hence, the given statement is false.

IceJOKER [234]3 years ago

3 0

False. To get Kelvin add 273 to °C to get your temperature into Kelvin. The correct boiling point of Aluminum in Kelvin is 2740 K.

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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

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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Answer:

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D. Is the best indicator for a chemical change.

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In a five-fold serial dilution of a 20 ug/mL solution, all tube dilutions are 1/5. What is the substance concentration in the th

sergij07 [2.7K]

Answer:

In the third tube, the concentration is 0.16 ug/mL

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