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alexandr1967 [171]
3 years ago
8

For each described change, determine the generally expected impact on a salt's solubility. LABEL No solubility change, SOLUBILIT

Y DECREASE OR SOLUBILITY INCREASE
A- Cooling a saturated solution by 10 oC
B- Placing a saturated solution in a hot water bath
C- Doubling the amount of salt added to a saturated solution
Chemistry
1 answer:
Anna [14]3 years ago
4 0

Answer:

A- Solubility decrease.

B- Solubility increase.

C- No solubility change.

Explanation:

Hello!

In this case, since the solubility of salt stands for the maximum amount of salt that can be added to a specific mass of water, usually 100 g; we need to take into account that for table salt in aqueous solution, the higher the temperature the larger the solubility and the lower the temperature the smaller the solubility; it means that more salt is dissolved in the same mass of water at higher temperatures and vice versa. Therefore, A- would decrease the solubility as the solution is cooled down and B- would increase the solubility as the solution is heated up.

Moreover, since the mass of water is assumed to remain the same, adding more salt do not affect the solubility but increase the degree of saturation of the solution up to supersaturated, yet the solubility remains unchanged.

Best regards!

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aquatic system ; ecosystem in a body of water , communities and organisms that are dependent on each other.
depth ; the distance from the top to the bottom of something .
geography ; the study of the earth s  physical features and the people ,plants , and animals.
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4 0
4 years ago
A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane
son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

                                                                                                   = 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

                                           = 75 - 67.5

                                           = 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

                                                                                                   = 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

                                           = 25 - 21.25

                                           = 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

                       = 64.125 moles

$CO$ formed is = 0.05 x 67.5

                       = 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

                                            = 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

                                            = 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21} $

                                = 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

 = 2 x 64.125

 = 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

                                        = 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3  = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

                                     = 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack  gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b).  The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.  

6 0
3 years ago
Match each lab incident to the item of PPE that will protect you from it. Each item of PPE will only be used once.
andre [41]

Answer:

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Explanation:

wdwdwfefe

5 0
3 years ago
Jacob is doing a report on the galaxy in which Earth is found. Which of the following is the galaxy that Jacob should research f
vichka [17]

Answer:

milky way

Explanation:

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3 years ago
The temperature in a town started out at 55 degrees. By the end of the day the temperature dropped to -6 degrees.
garik1379 [7]

Answer:

The choice of the answer is fourth option that is -61 degrees.

Therefore the temperature drop is -61°Centigrade.

Explanation:

Given:

The temperature in a town started out at 55 degrees

Start temperature = 55°Centigrade. (Initial temperature)

End of the Day      = -6°Centigrade. (Final temperature)

To Find:

How far did the temperature drop?

Solution:

We will have,

\textrm{Temperature drop}=\textrm{Final temperature}-\textrm{Initial temperature}

Substituting the above values in it we get

\textrm{Temperature drop}=-6-55\\\\\therefore \textrm{Temperature drop}=-61\° centigrade

Therefore the temperature drop is -61°Centigrade.

4 0
3 years ago
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