Initial moles of CH₃COOH = 1.00 x 0.0100 = 0.0100 mol
Initial moles of CH₃COONa = 1.00 x 0.100 = 0.100 mol
Let a be the volume of HNO₃ that must be added
Moles of HNO₃ added = a/1000 x 10.0 = 0.01a mol
CH₃COONa + HNO₃ => CH₃COOH + NaNO₃
Moles of CH₃COOH = 0.0100 + 0.01a
Moles of CH₃COONa = 0.100 - 0.01a
pH = pKa + log([CH₃COONa]/[CH₃COOH])
= pKa + log(moles of CH₃COONa/moles of CH₃COOH)
5.05 = 4.75 + log((0.100 - 0.01a)/(0.0100 + 0.01a))
log((0.100 - 0.01a)/(0.0100 + 0.01a)) = 0.3
(0.100 - 0.01a)/(0.0100 + 0.01a) = 10^0.3 = 2.0
0.03 a = 0.08
a = 2.67
<span>Volume of HNO3 = a = </span>2.13 mL
Answer:
84.3 g of nitrogen triiodide is the theoretical yield.
Explanation:
Hello there!
In this case, according to the chemical reaction:
It is possible to compute the theoretical yield of nitrogen triiodide by each reactant via stoichiometry as shown below:
Therefore, we infer that the smallest amount is the correct theoretical yield as it comes from the limiting reactant, in this case, diatomic iodine as it yields 84.3 g (three significant figures) of nitrogen triiodide as the theoretical yield; incidentally, nitrogen acts as the excess reactant.
Best regards!
Answer: pH = 7.36. The pKa of H2PO4− is 7.21.
Explanation:
<span>Starch and
cellulose have the same substance but different structures. They are both
polysaccharides. The basic unit of a polysaccharide is the glucose. Glucose,
which contains carbon, hydrogen, and oxygen, have two forms. The alpha-glucose
with an alcohol group attached to carbon 1 is down and the beta-glucose with
the alcohol group attached to carbon 1 is up. Starch is the alpha-glucose while
cellulose is the beta-glucose. Starches are linked into a straight chain whereas
the cellulose are connected like a pile of stack paper. When the human body
eats starch, it can digest the starch but not the cellulose because it has no
enzyme that can break it down. </span>
