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1 year ago

200.0 mL of a 0.845 M solution of KCl is diluted to 600.0 mL. What is the new concentration of the solution?

Chemistry

1 answer:

Ipatiy [6.2K]1 year ago

7 0

M be molar concentration
V be volume

M1=0.845M
V1=200mL=0.2L
V2=0.6L
M2=?

$\\ \tt\rightarrowtail M1V1=M2V2$

$\\ \tt\rightarrowtail 0.845(0.2)=0.6M2$

$\\ \tt\rightarrowtail 0.169=0.6M2$

$\\ \tt\rightarrowtail M2=0.281M$

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Learning Task 1 Match column A with column B. Column A Column B 1. Phase Change -2. Freezing 3. Condensation 4. Melting 5. Evapo

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Change of state occurs when heat is supplied or removed from a substance.

<h3>What is change of state?</h3>

Change of state refers to the changes that occur when a substance changes from one physical state to another due to changes in its temperature.

It is also known as phase change.

Phase Change can also be defined as change from one state to another without a change in chemical composition.

Some of the phase changes include:

Freezing: when liquid changes to solid
Condensation: when gas changes to liquid
Melting: when solid changes to liquid
Evaporation: when liquid changes to gas

The other terms associated with phase change include:

Boiling point: the temperature at which vapor pressure becomes high that causes bubbles to form inside the body of the liquid
Freezing point: temperature wherein a liquid solidifies
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2 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

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1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

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