Answer:
E, only one statement is correct
Explanation:
Absorption is favoured by heating the solution containing the colored impurity.
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Answer:
Explanation:
Problem 1
<u>1. Data</u>
<u />
a) P₁ = 3.25atm
b) V₁ = 755mL
c) P₂ = ?
d) V₂ = 1325 mL
r) T = 65ºC
<u>2. Formula</u>
Since the temeperature is constant you can use Boyle's law for idial gases:

<u>3. Solution</u>
Solve, substitute and compute:


Problem 2
<u>1. Data</u>
<u />
a) V₁ = 125 mL
b) P₁ = 548mmHg
c) P₁ = 625mmHg
d) V₂ = ?
<u>2. Formula</u>
You assume that the temperature does not change, and then can use Boyl'es law again.

<u>3. Solution</u>
This time, solve for V₂:

Substitute and compute:

You must round to 3 significant figures:

Problem 3
<u>1. Data</u>
<u />
a) V₁ = 285mL
b) T₁ = 25ºC
c) V₂ = ?
d) T₂ = 35ºC
<u>2. Formula</u>
At constant pressure, Charle's law states that volume and temperature are inversely related:

The temperatures must be in absolute scale.
<u />
<u>3. Solution</u>
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:

You must round to two significant figures: 290 ml
Problem 4
<u>1. Data</u>
<u />
a) P = 865mmHg
b) Convert to atm
<u>2. Formula</u>
You must use a conversion factor.
Divide both sides by 760 mmHg

<u />
<u>3. Solution</u>
Multiply 865 mmHg by the conversion factor:

Answer:
Molality of the solution = 0.7294 M
Explanation:
Given:
Number of magnesium arsenate = 1.24 moles
Mass of solution = 1.74 kg
Find:
Molality of the solution
Computation:
Molality of the solution = Mole of solute / Mass of solution = 1.74 kg
Molality of the solution = 1.24 / 1.7
Molality of the solution = 0.7294 M
Answer:
Solid
Explanation:
Diphenylamine has a melting point of 127.4 F or 53 C so at room temperature ~70 F or 21 C its a solid.
Answer:
17.5 g
Explanation:
Given data
- Mass of solution to be prepared: 50.0 grams
- Concentration of the salt solution: 35.0%
The concentration by mass of NaCl in the solution is 35.0%, that is, there are 35.0 grams of sodium chloride per 100 grams of solution. We will use this ratio to find the mass of sodium chloride required to prepare 50.0 grams of a 35.0% salt solution.
