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3 years ago

QUESTION 13

Chemistry

1 answer:

Tom [10]3 years ago

3 0

Answer:

b. 101.02°C

Explanation:

The elevation of boiling point in a solvent due the addition of a solute is a colligative property called Boiling point Elevation. The formula is:

ΔT = Kb*m*i

Where ΔT is change in boiling point

Kb is boilinig point elevation constant = 0.512°C/m for water

m is molality of the solution = 0.50m

And i is Van't Hoff Factor (4 for the Al(NO₃)₃ because its disolution produce 4 ions)

Replacing:

ΔT = 0.512°C/m*0.50m*4

ΔT = 1.02°C

As the boiling point of pure water is 100°C and the increasing is of 1.02°C, boiling point of the solution is:

101.02°C

Right option is:

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This refers to the most powerful means of communicating confidence and conviction A.Posture B.eye contact C.body movement

skad [1K]

Answer:

Explanation:

the way you move explains how confident you are

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3 years ago

Calculate the mass of aluminum that would have the same number of atoms as 6.35 g of cadmium

Sergeu [11.5K]

Answer:

1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.

Explanation:

Given data:

mass of cadmium = 6.35 g

Number of atoms of aluminum as 6.35 g cadmium contain = ?

Solution:

Number of moles of cadmium = 6.35 g/ 112.4 g/mol

Number of moles of cadmium = 0.06 mol

Number of atoms of cadmium:

1 mole = 6.022×10²³ atoms of cadmium

0.06 mol × 6.022×10²³ atoms of cadmium/ 1mol

0.36×10²³ atoms of cadmium

Number of atoms of Al:

Number of atoms of Al = 0.36×10²³ atoms

1 mole = 6.022×10²³ atoms

0.36×10²³ atoms × 1 mol /6.022×10²³ atoms

0.06 moles

Mass of aluminum:

Number of moles = mass/molar mass

0.06 mol = m/ 27 g/mol

m = 0.06 mol ×27 g/mol

m = 1.62 g

Thus, 1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.

6 0

3 years ago

ANSWER FAST!!

belka [17]

Answer: 800 mL of methyl alcohol should be added to 200 ml of water to make this solution.

Explanation:

Volume of the methyl alcohol = x

Volume of water = y = 200 mL

Volume of the solution ,V = x + y

Volume percentage of solution = 80%

x = 800 mL

800 mL of methyl alcohol should be added to 200 ml of water to make this solution.

4 0

2 years ago

Liquid nitrogen has a density of 0.807 g/ml at –195.8 °c. if 1.00 l of n2(l) is allowed to warm to 25°c at a pressure of 1.00 at

Flauer [41]

Step 1: Change density from g/mL to g/L;

0.807 g/mL = 807 g/L

Step 2: Find Moles of N₂;
As,
Density = Mass / Volume
Or,
Mass = Density × Volume

Putting Values,

Mass = 807 g/L × 1 L

Mass = 807 g
Also,
Moles = Mass / M.mass

Putting values,

Moles = 807 g / 28 g.mol⁻¹

Moles = 28.82 moles

Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,

As,
P V = n R T

V = n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)

V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm

V = 704.76 L

8 0

3 years ago

A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate for

laiz [17]

Answer:

CaF2 will not precipitate

Explanation:

Given

Volume of Ca(NO3)2 $= 10$ ml