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Setler79 [48]
3 years ago
15

Why are Noble gases inert?

Chemistry
1 answer:
sladkih [1.3K]3 years ago
3 0
Correct answer is <span>C. They have a full shell of valence electrons. </span>
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(a) calculate the molarity of a solution made by dissolving 0.0815 mol na2so4 in enough water to form exactly 550. ml of solutio
Vesna [10]
You just need to convert it into moles per Liter (mol/L).

0.0815mol / 0.550L = 0.148mol/L
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A chemist designs a galvanic cell that uses these two half-reactions:
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Answer:

MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)

Explanation:

a)

MnO4⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn(aq)²⁺ + 4H2O (l)

b)

5Fe³⁺ (aq) +5e⁻ → 5Fe²⁺(aq)

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The earth in m is 510
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Be able to identify the correct elements using clues such as numbers of protons,locations (family/period), number of neutrons, n
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an element's name, chemical symbol, atomic number, atomic mass.

IDK what you are even asking for

6 0
3 years ago
Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr
lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Using the equation,

ΔT_{f} = iK_{f}m

where:

ΔT_{f} = change in freezing point (unknown)

i = Van't Hoff factor

K_{f} = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = \frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }

molal concentration = 0.1219m

Now, putting in the values to the equtaion ΔT_{f} = iK_{f}m we get,

ΔT_{f} = 2 × 1.86 × 0.1219

ΔT_{f} = 0.4536°C

So, ΔT_{f} of solution is,

ΔT_{f_{solution} } = 0.00°C - 0.45°C

ΔT_{f_{solution} } =  - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Learn more about freezing point here;

brainly.com/question/3121416

#SPJ4

7 0
1 year ago
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