The answer is D.sucrose. Sucrose is a disaccharide .
Grams of Chromium(III) nitrate produced : 268.95 g
<h3>Further explanation</h3>
Given
0.85 moles of Lead(IV) nitrate
Required
grams of Chromium(III) nitrate
Solution
Reaction
Balanced equation :
<em>2Cr₂ + 3Pb(NO₃)₄ ⇒ 4Cr(NO₃)₃ + 3Pb </em>
From the equation, mol ratio of Pb(NO₃)₄ : Cr(NO₃)₃ = 3 : 4, so mol Cr(NO₃)₃
mol Cr(NO₃)₃ =

Mass of Chromium(III) nitrate (MW=238.0108 g/mol) :
mass = mol x MW
mass = 1.13 x 238.0108
mass = 268.95 g
Answer:
220.42098 amu
Explanation:
(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu
These are weighted averages.
So, we will take mass of one and multiply by abundance percentage that is provided and add them together.
In order to calculate the average atomic mass, we have to convert the percentages of abundance to decimals. So, you get
(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu
A (they are experimentally determined exponents)