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<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
Answer:
The scientific laws have been well proven before they are published so it is difficult to prove mistakes
Explanation:
Answer:
The percent yield of chloro-ethane in the reaction is 82.98%.
Explanation:

Moles of ethane = 
Moles of chlorine gases =
As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.
This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.
According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.
Then 9.1549 moles of chlorien gas will give:
of chloro-ethane
Mass of 9.1549 moles of chloro-ethane:
9.1549 mol × 64.5 g/mol = 590.4910 g
Theoretical yield of chloro-ethane: 590.4910 g
Given experimental yield of chloro-ethane: 490.0 g


The percent yield of chloro-ethane in the reaction is 82.98%.