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3 years ago

How much water should be added to 85 mL of 0.45 M HCI to reduce the concentration to 0.20 M?

Chemistry

1 answer:

valkas [14]3 years ago

8 0

Answer:

106.25 mL

Explanation:

For this, we can use

C1×V1=C2×V2

C1 = 0.45

V1 = 85

C2= 0.20

V2= ?

0.45 × 85 = 0.20 × V2

V2= (0.45 × 85)/0.20

V2=191.25mL

To find the amount of water added, subtract V1 from V2

191.25 - 85 =106.25mL

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Answer:

The answer to your question is letter E. 288.4 g/mol

Explanation:

Data

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3 years ago

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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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Answer

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

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T= temperature

n = number of moles

(P1V1/T1)=(P2V2/T2)

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Given:

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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

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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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