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ivanzaharov [21]
3 years ago
11

A hydrobromic acid (HBr) solution has a molar concentration of 0.055 M, Calculate the and pH of the solution. (Remember that Kw=

1.0x10-14M2.)
A) [H3O+]= 5.5x10-2M, [OH-]=1.82x10-13M, pH=1.26
B) [H3O+]= 3.8x10-13, [OH-]=2.6x10-2M, pH=12.42
C) [H3O+]= 4.8x10-3, [OH-]=1.6x10-12M, pH=9.2
D) [H3O+]= 1.0x10-8 [OH-]=1.0x10-6M, pH=8.0
E) cannot be calculated with the information given
Chemistry
1 answer:
seraphim [82]3 years ago
5 0

Answer: Im not sure but i think its B

Explanation: Its hard to explain so im

not gonna do it

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88.98 %.

Explanation:

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  • Now, we need to calculate the calculated yield of lead(II) chloride (PbCl₂).
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  • n of lead(II) nitrate (Pb(NO₃)₂) = mass / molar mass = (870 g) / (331.2 g/mol) = 2.63 mol.
  • Since HCl is in excess, the limiting reactant is lead(II) nitrate (Pb(NO₃)₂).

<u><em>Using cross multiplication:</em></u>

1.0 mole of Pb(NO₃)₂ produces → 1.0 mole of PbCl₂, from the stichiometry.

∴ 2.63 mole of Pb(NO₃)₂ produces → 2.63 mole of PbCl₂.

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∴ The percent yield % of lead(II) chloride (PbCl₂) = [(actual yield) / (calculated yield)] x 100 = [(650 g) / (730.52)] x 100 = 88.98 %.

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