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PIT_PIT [208]
4 years ago
10

The area from which a river and its tributaries collect water is called the:

Chemistry
1 answer:
prisoha [69]4 years ago
7 0

Answer:

b. Drainage basin

Explanation:

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3. What is the density of a 100 grams (g) box that displaces 20 mL of water?
Zolol [24]

Answer: the density is 997 kg

Explanation:

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3 years ago
Describe what happens to the particles in a chocolate bar as it melts.
Margarita [4]

Answer:

The correct answer is d. As the chocolate bar changes from a solid to a liquid the particles move around more.

Explanation:

When a chocolate bar melts, it suffers a <u>physical</u> change: a change of state from solid to liquid. In the liquid state, the particles move around more, because they have more kinetic energy (more velocity) and more freedom to move than in solid state. Thus, as the solid became into a liquid, the particles move more.

5 0
4 years ago
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SOLVE IT AND HERE'S 30 POINTS In the Northern Hemisphere, the sun is highest in the sky during the _____.
Oksanka [162]
the answer is b) summer solstice
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4 years ago
Match the set of measurements of the boiling point of water with the best
Mashcka [7]

Answer:

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6 0
4 years ago
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
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