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Firdavs [7]
2 years ago
7

Colligative properties include __________ the freezing point, _____________ the vapor pressure, and ___________ the boiling poin

t. Which of the following sequences of words correctly fills in the blanks?
a) raising, lowering, loweringb) raising, raising, loweringc) lowering, raising, loweringd) lowering, lowering, raising
Chemistry
2 answers:
MrMuchimi2 years ago
7 0
Colligative properties include lowering the freezing point, lowering the vapor pressure, and raising the boiling point.

pashok25 [27]2 years ago
3 0

The answer is: d) lowering, lowering, raising.

The vapor pressure depression (lowering), freezing point depression (lowering) and the boiling point elevation (raising) are the colligative properties od solution.

For example pure water has lower boiling point than solution of water and some salt.

The higher is the molarity of the solution, the lower is freezing point.

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Which statement describes the transfer of heat energy that occurs when an ice cube is added to an insulated container with 100 m
nadya68 [22]
Number 3 is the most likely answer
4 0
2 years ago
A mixture of three gases has a pressure at 298 K of 1380 mm Hg. The mixture is analysed and is found to contain 1.27 mol CO2, 3.
enot [183]

Answer:

The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)

Explanation:

<u>Step 1:</u> Data given

A mixture of three gases has a total pressure of 1380 mm Hg (=1.81579 atm) at 298 K

Moles of CO2 = 1.27 moles

Moles of CO = 3.04 moles

Moles of Ar = 1.50 moles

<u>Step 2:</u> Calculate total number of moles

Total number of moles = n(CO2)+ n(CO)+ n(Ar) = 1.27 mol+ 3.04 mol+ 1.50 mol = 5.81 moles

<u>Step 3:</u> Calculate mol fraction Ar

Mol fraction Ar = 1.50 mol/5.81 mol = 0.258

<u>Step 4</u>: Calculate partial pressure

1380 mm Hg * 0.258 moles Ar = 356.04 mm Hg = 0.4685 atm

The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)

8 0
2 years ago
A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and
Ilya [14]

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-

Molarity of the silver nitrate solution = 0.03337 M

Volume of the silver nitrate = 23.28 mL = 0.02328 L

Moles of silver nitrate = n

0.03337 M=\frac{n}{0.02328 L}

n = 0.03337 M\times 0.02328 L=0.0007768 mol

According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.

Then 0.0007768 moles of silver nitrate will react with:

\frac{1}{1}\times 0.0007768 mol=0.0007768 mol chloride ions.

In one mole of aldrin there are 6  moles of chloride ions.

Then moles of aldrin containing 0.0007768 moles chloride ions are:

\frac{0.0007768 mol}{6}=0.0001295 mol

Moles of aldrin present in the sample = 0.0001295 mol

Mass of 0.0001295 moles of aldrin present in the sample :

0.0001295 mol × 364.92 g/mol =0.04726 g

Percentage of an aldrin in the sample:

\frac{0.04726 g}{0.1064 g}\times 100=44.41\%

8 0
2 years ago
Recalling that a beaker of water is two dimensional, what is the three dimensional shape of the micelle
saul85 [17]
A micelle refers to an aggregate of a surfactant molecule, which a dispersed in a liquied colloid. In an aqueous liquid, a micelle is arranged in such a way that its hydrophillic head region will be in contact with the surrounding solvents while the hydrophobic tail region will be embedded in its center. 
The three dimensional shape of a micelle is A SPHERE. 
7 0
2 years ago
The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is __________% by
amm1812

Answer:

The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is 12.90% by mass

Explanation:

2.23 M aqueous solution of NaCl means there are 2.23 moles of NaCl in 1000 mL of solution.

We know that density is equal to ratio of mass to volume.

Here density of solution is 1.01 g/mL.

So mass of 1000 mL solution = (1.01\times 1000) g = 1010 g

molar mass of NaCl = 58.44 g/mol

So mass of 2.23 moles of NaCl = (2.23\times 58.44) g = 130.3 g

% by mass  is ratio of mass of solute to mass of solution and then  multiplied by 100.

Here solute is NaCl.

So % by mass of 2.23 M aqueous solution of NaCl = \frac{130.0}{1010}\times 100% = 12.90%

3 0
2 years ago
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