Search found 103 matches

by Aman Sankineni 2L
Thu Oct 03, 2019 11:03 pm
Forum: Limiting Reactant Calculations
Topic: Reaction Stochiometry L.3 [ENDORSED]
Replies: 3
Views: 992

Re: Reaction Stochiometry L.3 [ENDORSED]

a. 8.6 x 10^-5 mol
b. 11.3198 grams
by Aman Sankineni 2L
Wed Oct 02, 2019 3:49 pm
Forum: Molarity, Solutions, Dilutions
Topic: Fundamentals of chem: Dilution G9
Replies: 4
Views: 1736

Re: Fundamentals of chem: Dilution G9

.179 M x .5 L = 0.0895 mol
AgN03 molar mass = (107.87 + 14 + 16*3) = 169.87 g mol^-1
.0895 mol x 169.87 g mol^-1 = 15.034 grams
by Aman Sankineni 2L
Tue Oct 01, 2019 11:32 pm
Forum: Molarity, Solutions, Dilutions
Topic: E 29 part c
Replies: 8
Views: 1122

Re: E 29 part c

For part d, you must figure out the mass of oxygen as part of the formula. Since there is 4 oxygen atoms, the mass of oxygen is 16x4=64. By dividing that by the total mass of the formula, 206.446, you can figure out what fraction of the total mass was due to oxygen. 64/206.446=.310008 or 31%.

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