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3 years ago

What is the freezing point of an aqueous solution made by dissolving 0.118 mole of MgCl2 in 100.0 g of water? Kf = 1.86 oC/m

Chemistry

1 answer:

Sauron [17]3 years ago

8 0

Answer:

Freezing point of solution is -4.39°C

Explanation:

Freezing point depression's formula:

ΔT = Kf . m . i

ΔT = Freezing point of pure solvent - Freezing point of solution

0°C - X = 1.86°C/m . m . 2

i = Van't Hoff factor, ions dissolved in solution

MgCl₂ → Mg²⁺ + 2Cl⁻ i = 3 → 1 mol of Mg and 2 mol of chloride

Let's think the molality (mol/kg)

First of all, let's convert the mass of solvent from g to kg

100 g / 1000 = 0.1 kg

Molality (mol/kg) = 0.118 mol / 0.1 kg = 1.18 m

0°C - X = 1.86°C/m . 1.18 m . 2

X = - 4.39°C

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Answer:

Electricity can be present in a static charge, while magnetism's presence is only felt when there are moving charges as a result of electricity. In simple words, electricity can exist without magnetism, but magnetism cannot exist without electricity.

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Explanation:

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2 years ago

Identify the conjugate base in each pairs

DiKsa [7]

Answer: 1) $RCOO^-$

2) $H_2PO_4^-$

3) $RNH_2$

4) $HCO_3^-$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

1) $RCOOH\rightarrow RCOO^-+H^+$

Here, $RCOOH$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $RCOO^-$ which is a conjugate base.

2) $H_3PO_4\rightarrow H_2PO_4^-+H^+$

Here, $H_3PO_4$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $H_2PO_4^-$ which is a conjugate base.

3) $RNH_3^+\rightarrow RNH_2+H^+$

Here, $RNH_3^+$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $RNH_2$ which is a conjugate base.

4) $H_2CO_3\rightarrow HCO_3^-+H^+$

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2 years ago

Research is being carried out on cellulose as a source of chemicals for the production of fibers, coatings, and plastics. Cellul

Luden [163]

Answer:

+ 636 KJ

Explanation:

We want to arrive to the equation

C6H12O6(s) ---------> 6 H2CO(g) ΔH ° rxn = ?

by manipulating algebraically the first four given equations.

We notice the first one has our product H2CO(g) as a reactant. This indicates we must take the inverse of that equation. Also we need 6 mol of H2CO(g), thus it also needs to be multiplied by 6

6 CO2(g) + 6 H2O(g) ------->6H2CO(g) + 6O2(g) ΔH °comb = + 572.9 KJ/x 6

Now we want C6H12O6(s) as a reactant and it is a product in the second one, therefore lets reverse it

C6H12O6(s) -------> 6 C(s) + 6 H2(g) + 3 O2(g) ΔH ° f = + 1274.4 KJ/mol

Now if take the third equation and multiply it by six we will cancel the C(s) with the above equation

6 C(s) + 6O2(g) ---------> 6 CO2(g) ΔH ° f = - 393.5 KJ/mol x 6

Finally by multiplying the last equation by 6 and adding all the equations we will arrive at our desired one

6 H2(g) + 3 O2(g) -----------> 6H2O(g) ΔH ° f = - 285.8 KJ/mol x 6

then lets add them to get ΔH ° rxn:

6 CO2(g) + 6 H2O(g) ------->6H2CO(g) + 6O2(g) ΔH °comb = + 3437.4 KJ

+ C6H12O6(s) -------> 6 C(s) + 6 H2(g) + 3 O2(g) ΔH ° f = + 1274.4 KJ

+ 6C(s) + 6O2(g) ---------> 6 CO2(g) ΔH ° f = - 2361.0 KJ

+6 H2(g) + 3 O2(g) -----------> 6H2O(g) ΔHº f = - 1714.8 KJ

C6H12O6(s) ---------> 6 H2CO(g)

ΔH ° rxn = 3437.4 + 1274.4 - 2361.0 - 1714.8 = 636 KJ

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3 years ago

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Ivan

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2 years ago

A children's liquid medicine contains 100mg of active ingredient in 5 mL. If a child should receive 200 mg of the active ingredi

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3 years ago