The Atomic Mass of Al is 26.98154 g/mol. Is it possible to have 5.0 × 10^-25 g of Al? Explain.

1 answer:

8 0

If a mole of aluminum weighs 26.98 grams, that means 1 atom of aluminum weighs = (26.98 g/mole) / (6.023 x 10^23 atoms/mole) = 4.479 x 10^-23 grams,

so, it is not possible because 1 atom weighs that much we calculated which is almost 100 times more than the amount you mentioned

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HCN (aq) + OH- (aq) → CN-(aq) + H2O(l)

So we can see that the molar ratio between HCN: OH-: CN- is 1:1 :1

we need to get number of mmol of HCN = molarity * volume

= 0.2 mmol / mL* 10 mL = 2 mmol

so the number of mmol of NaOH = 2 mmol according to the molar ratio

so, the volume of NaOH = moles/molarity

= 2 mmol / 0.0998mL

= 20 mL

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when we have the value of PKa = 9.31 and we need to get Pkb

so, Pkb= 14 - Pka

= 14 - 9.31 = 4.69

when Pkb = -㏒Kb

4.69 = -㏒ Kb

∴ Kb = 2 x 10^-5

and when the dissociation reaction of CN- is:

CN-(aq) + H2O(l) ↔ HCN(aq) + OH- (aq)

by using the ICE table:

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[CN-] = 0.0662 M

and [HCN] = [OH]- = 0 M

and the equilibrium concentrations are:

[CN-] = (0.0662- X)

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when Kb expression = [HCN][OH-] /[CN-]

by substitution:

2 x 10^-5 = X^2 / (0.0662 - X)

X = 0.00114

∴[OH-] = X = 0.00114

when POH = -㏒[OH]

= -㏒ 0.00114

POH = 2.94

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