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Gnoma [55]
3 years ago
14

What are the uses of solution?​

Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
5 0

Answer:

The majority of chemical processes are reactions that occur in solution. Important industrial processes often utilize solution chemistry. "Life" is the sum of a series of complex processes occurring in solution. Air, tap water, tincture of iodine, beverages, and household ammonia are common examples of solutions.

four types of solution:

Turpentine as a solvent are used in the production of paints, inks and dyes. ↔Water as a solvent is used in the making of food, textiles, soaps and detergents. ↔Alloys are solid solutions that are used in the manufacture of cars, aerospace and other vehicles.

Explanation:

can you pls make me brainliest

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Be sure to answer all parts. What is the effect of each of the following on the volume of 1 mol of an ideal gas? (a) The pressur
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A)The volume decreases by a factor of 4

B), the volume has increased by factor of 2

Explanation:

A)Given:

P1= 760Kpa

P2 =202Kpa

The temperature changes from37C to155C

There is increase In pressure from P1 to P2

P1= 760torr.

We need to convert to Kpa

But, 1atm= 760torr

Then 760torr 101000pa

Then 101000pa = 101Kpa

We need to convert the temperature from Celsius to Kelvin

T1= 37+273= 310K

But from ideal gas, we know that PV = nRT where nR is constant

Where P= pressure

V= volume

T= temperature

n = number of moles

(P1V1/T1)=(P2V2/T2)

V1/V2 = P2/P1 * T1/T2

V1/V2 = (202/101)*(310/155)

V1/V2=4

V2= V1/4

Therefore, the volume has decreased by factor of 4

B)

Given:

P1= 2atm

P2 =101Kpa

The temperature changes from 305K to 32C

There is increase In pressure from P1 to P2

P1= 2atm

We need to convert to Kpa

But, 1atm= 760torr

Then 760torr 101000pa

Then 101000pa = 101Kpa

P1= 202.65kpa

We need to convert the temperature from Celsius to Kelvin

T2= 32+273= 305K

But from ideal gas, we know that PV = nRT

Where P= pressure

V= volume

T= temperature

n = number of moles

(P1V1/T1)=(P2V2/T2)

V1/V2 = P2/P1 * T1/T2

V1/V2 = (202/101)

V1/V2 = (101/202.65)*(305/305)

V1/V2 = 1/2

V2=2V1

Therefore, the volume has increased by factor of 2

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