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2 answers:
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Answer:
Approximately .
Explanation:
Lookup Avogadro's Number: (three significant figures.)
Lookup the relative atomic mass of ,
, and
on a modern periodic table:
:
.
:
.
:
.
(For example, the relative atomic mass of is
means that the mass of one mole of
atoms would be approximately
grams on average.)
The question counted the number of molecules without using any unit. Avogadro's Number
helps convert the unit of that count to moles.
Each mole of molecules includes exactly
of these
molecules.
molecules would correspond to
of such molecules.
(Keep more significant figures than required during intermediary steps.)
The formula mass of gives the mass of each mole of
molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:
.
Calculate the mass of approximately of
:
.
(Rounded to three significant figures.)
Answer:
- i) 5.00 mL of 1.00 M NaOH: before the equivalence point
- ii) 50.0 mL of 1.00 M NaOH: before the equivalence point
- iii) 100 mL of 1.00 M NaOH: at the equivalence point
- iv) 150 mL of 1.00 M NaOH: after the equivalence point
- v) 200 mL of 1.00 M NaOH: after the equivalence point
Explanation:
1. First calculate the number of mol acid in the 100 mL of 1.00 M HCl solution.
Equation:
- Molarity = numbrer of moles of solute / volume of the solution in liters.
Thus, you need to convert each volume from mL to liters, which is done dividing by 1,000.
Naming M the molarity, n the number of moles of solute (acid or base), and V the volume in liters:
2. Now calculate the number of moles of NaOH for every condition (addition)
<u>i) 5.00 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.00500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.
<u>ii) 50.0 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.0500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.
<u />
<u>iii) 100 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.100mol) is equal to the number of moles of acid in the solution (0.100mol), this is at the equivalence point.
<u>iv) 150 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.150mol) is greater than the number of moles of acid in the solution (0.100mol), this is after the equivalence point.
<u>v) 200 mL of 1.00 M NaOH</u>
This is more volume of NaOH, then this is also after the equivalence point.