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MakcuM [25]
3 years ago
6

What does mass change to and what does it do

Chemistry
1 answer:
Nataly [62]3 years ago
4 0

Answer:

Mass is the amount of matter in an object and does not change with location.

Explanation:

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erik [133]
A: the amount of force would need to be greater.
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3 years ago
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Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0
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How much Oxygen would be required to react with 101 g of Methane?
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Answer:

404g

Explanation:

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The table below compares the radioactive decay rates of two materials. Material Original mass of material (in grams) Mass of mat
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step by step explanation;

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Material 2 disintegrates to half its mass three times in 21.6 s, to go from 200g to 25g. That is,

200g - 50g - 25g - 12.5g.

This means that regardless of their initial masses involved, material 1 and material 2 have equal half-life.

Their half-life is 21.6 ÷ 3 = 7.2 sec

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Every cell is surrounded by a thin membrane. What is the main function of this cell membrane?
zubka84 [21]

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The main function of the cell membrane is to protect the cell from the outer environment.

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