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Helen [10]
3 years ago
5

Please help

Chemistry
1 answer:
Colt1911 [192]3 years ago
3 0
A i belive is the correct answer

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If more of a liquid evaporates,does that mean it has stronger intermolecular forces
Gwar [14]

Answer:

No

Explanation:

The stronger the intermolecular forces the molecules tend to stay in the liquid phase. Compounds with weak intermolecular forces are more volatile and more molecules overcome those forces to pass to the gas phase.

7 0
3 years ago
A sample in lab displaces 15 mL of water in a graduated cylinder. When placed on a scale the mass is 8.7 g. What is the density
Natasha2012 [34]

Answer:

<h2>0.58 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

From the question

volume = 15 mL

mass of object = 8.7 g

We have

Density =  \frac{8.7}{15}

We have the final answer as

<h3>0.58 g/mL</h3>

Hope this helps you

4 0
3 years ago
Which term names a metamorphic change in rocks over a wide area from tectonic forces in Earth's crust and mantle? A. local metam
Pavlova-9 [17]
D is the answer. A, B,C are absurd. Doing a little word right there.
8 0
3 years ago
Read 2 more answers
A solution containing 20.0 g of an unknown liquid and 110.0 g water has a freezing point of .32 °C. Given Kf 1.86°C/m for water,
hjlf

Answer:

A. 256

Explanation:

In a solution where a liquid is the sovent, we'll use the van't Hoff factor, which is the ratio between the number of moles of particles produced in solution and the number of moles of solute dissolved, will be equal to 1.

ΔTemp.f = i * Kf * b

where,

ΔTemp.f = the freezing-point depression;

i = the van't Hoff factor

Kf = the cryoscopic constant of the solvent;

b = the molality of the solution.

So the freezing-point depression by definition is the difference between the the freezing point of the pure solvent and the freesing point of the solution.

Mathematically,

ΔTemp.f = Temp.f° - Temp.f

where,

Temp.f° = the freezing point of the pure solvent.

Temp.f = the freezin point of the solution.

Freezing point of pure water = 0°C

ΔTemp.f = 0 - (-1.32)

= 1.32°C

i = 1,

Kf = 1.86 °Ckg/mol

Solving for the molality, b = ΔTemp.f/( i * Kf)

= 1.32/(1*1.86)

= 0.71 mol/kg

Converting from mol/kg to mol/g,

0.71 mol/kg * 1kg/1000g

= 0.00071 mol/g.

Mass of solvent = 110g

Number of moles = mass * molality

= 0.00071 * 110

= 0.078 mol.

To calculate molar mass,

Molar mass (g/mol) = mass/number of moles

Mass of solute (liquid) = 20g

Molar mass = 20/0.078

= 256.2 g/mol

4 0
3 years ago
Iupac name for Ag(S2O3)2​
ryzh [129]

Answer:

Dithiosulfatosilver

Explanation:

3 0
3 years ago
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