help question 22-26 please with Agr I the to final igh, Agci of such, the fins!
ID: 524950 • Letter: H
Question
help question 22-26 please
Explanation / Answer
Q22.
find mass of AgCl Collected
Mass of filter + AgCl - Mass of filter = (1.4173-1.0051) = 0.4122 g
best answer is A
Q23.
Cl- in grams in sample:
mol of AgCl = mass/MW = 0.4122/143.32 = 0.0028760 mol of AgCl = 0.0028760 mol of Cl-
mass of Cl- = 0.0028760*35.453 = 0.10196 g of Cl-
Q24.
% Cl- in sample = mas sof Cl- / total mass * 100% = 0.10196/0.334*100 = 30.526 %
Q25
bonds of 90°
A is false, CCl4 has 109
B flase, PCl5 has also 120 angles
choose only CCl4
Q26
linear moleucles are CO2 only
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