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Sonja [21]
3 years ago
6

You need to make an aqueous solution of 0.150 M sodium fluoride for an experiment in lab, using a 500 mL volumetric flask. How m

uch solid sodium fluoride should you add?
Chemistry
1 answer:
storchak [24]3 years ago
7 0

Answer:

Explanation:

Solution required is .15 M 500 mL sodium fluoride .

molecular weight of sodium fluoride = 42

one mole of sodium fluoride = 42 grams

No of moles contained in the solution of .15M , 500 mL solution

= .15 x 0.5 moles = .075 moles

1 mole = 42 g

.075 moles = .075 x 42 = 3.15 grams .

So mass of sodium fluoride required = 3.15 g .

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Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2
Mkey [24]

Answer:

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

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ΔH°rxn = 136.5kJ/mol

And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

ΔH°rxn = 0.2875kJ/molK

And replacing in (1) at 298K:

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<em>ΔG°rxn = +50.8 kJ/mol</em>

<em />

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