Answer:
1.0190 x 10⁻⁵ mol
Explanation:
We know the titration required 10.19 mL of 0.001000 M KIO₃, from this information we can calculate the number of moles KIO₃ reacted and from there the number of moles of ascorbic acid since it is a monoprotic acid ( 1 equivalent of ascorbic acid to one equivalent KIO₃).
Molarity = mol/V
V KIO₃ = 10.19 mL = 10.19 mL x 1 L/1000 mL = 0.01019 L
⇒ mol KIO₃ = V x M = 0.01019 L x 0.0010 mol / L = 1.0190 x 10⁻⁵ mol KIO₃
# mol ascorbic acid = # mol KIO₃ = 1.0190 x 10⁻⁵ mol
Carbon-6 because it is a neutron number
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Since we already have the balanced equation, we know that the ratio between
is
respectively.
So then we can set up a proportion to find the number of moles produced when 2.90 moles of Na react completely:

Then we cross multiply and solve for x:


Therefore, we know that when 2.90 moles of Na react completely, there are 1.45 moles of
that are produced.
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