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2 years ago

The____ cation will precipitate after the_ step. Then the _cation will precipitate after the____ step.

Chemistry

1 answer:

Novay_Z [31]2 years ago

8 0

Answer:

f. Sn^4+

c. second

e. Al^3+

d. third

Explanation:

This question comes from a quantitative analysis showing the flowchart of a common scheme for identifying cations.

Now, from the separation scheme, Let's assume that Sn⁴⁺ & Al³⁺ were given; Then, Yes, the separation will work.

However, there will be occurrence of precipitation after the 1st step1.

So, the Sn⁴⁺ cation will precipitate after the second step. Then the Al³⁺ cation will precipitate after the third step.

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2 years ago

indicate whether the entropy of the system increases or decreases. Mixing 10 mL of 90.0 °C water with 10 mL of 10 °C water. The

Ivanshal [37]

Answer:

When the water is mixed with water at lower temperature the effective temperature of the system (i.e the water at lower temperature) will increase, thereby increasing it's entropy

Explanation:

The answer that "the entropy will is increases" is correct as:

The water at 90° C i.e at higher temperature is mixed with the water at 10° C i.e the water at the lower temperature.

The water at lower temperature will have molecules with lower energy while the water with higher temperature will have molecules undergoing high thermal collisions. Thereby, when the water is mixed with water at lower temperature the effective temperature of the system (i.e the water at lower temperature) will increase, thereby increasing it's entropy.

Therefore, the answer is correct with respect to the water at lower temperature.

Meanwhile, for the water at higher temperature , the temperature of the system will decrease. Thus, the entropy of the water at higher level will decrease.

5 0

3 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$