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3 years ago

How many moles of magnesium is 3.01 x 10^22 atoms of magnesium?

Chemistry

1 answer:

Step2247 [10]3 years ago

4 0

3,01 x 10.000.000.000.000.000.000.000=
=30.100.000.000.000.000.000.000

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The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M

anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

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

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0

2 years ago

Read 2 more answers

This

timama [110]

Answer:

4.2 Liters (2 sig-figs)

Explanation:

Apply Boyles Law ... P₁V₁ = P₂V₂

P₁ = 712 Torr P₂ = 3560 Torr

V₁ = 21.0L V₂ = ?

P₁V₁ = P₂V₂ => V₂ = P₁V₁/P₂ = (712Torr)(21.0L)/(3560Torr) = 4.2 Liters (2 sig-figs)

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3 years ago

To make a 1.0L dilute of 0.5M from a stock solution of 12M, how much solution will be needed?

Cerrena [4.2K]

Answer:

$V_{concentrated}=0.042L=4.2mL$

Explanation:

Hello,

In this case, since in a dilution process the moles of the solute must remain unchanged, we use the volumes and molarities as shown below:

$M_{diluted}V_{diluted}=M_{concentrated}V_{concentrated}$

Clearly, the concentrated solution is 12M and the diluted solution is 0.5 M, thus, the volume of the concentrated solution we should take is:

$V_{concentrated}=\frac{M_{diluted}V_{diluted}}{M_{concentrated}} =\frac{0.5M*1.0L}{12M}\\ \\V_{concentrated}=0.042L=4.2mL$

Best regards.

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Alecsey [184]

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