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1) A 0.464g sample of compound containing the elements carbon, hydrogen, oxygen, and sulphur was burned completely, yielding 0.969 g CO 2 and 0.198 g H 2 O. In a separate experiment, all the sulphur...

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1) A 0.464g sample of compound containing the elements carbon, hydrogen, oxygen, and sulphur was burned completely, yielding 0.969 g CO2 and 0.198 g H2O. In a separate experiment, all the sulphur in another 0.464 g sample was converted to 0.859 g of BaSO4. What is the empirical formula of the unknown compound?
2)Ethyl alcohol, C2H5OH, is a fermentation produce of sugars in the presence of enzymes. How much glucose, C6H12O6, would you need to start with to prepare 4.6 x 104kg of ethyl alcohol if the yield in the reaction is 64.5%?

The equation for this reaction is

C6H12O6 + 3O2->4CO2 + C2H5OH + 3H2O

3)A solution containing 40.7g of calcium nitrate in a volume of 0.500 L was prepared.

a. Calculate the number of moles of calcium nitrate in this solution.

b. Calculate the number of molecules of calcium nitrate in this solution.

c. Calculate the total number of ions from the calcium nitrate in this solution.

d. What is the molarity of this solution?

Answered Same Day Dec 21, 2021

Solution

David answered on Dec 21 2021
122 Votes
Solution:261112_60201_3
1)
Atomic mass of C = 12.011g Molar mass of CO2 = 44.009 g
Atomic mass of H = 1.008 g Molar mass of H2O = 18.015 g
Atomic mass of S = 32.066 g Molar mass of BaSO4 = 233.389 g
Now,
44.009 g CO2 contains 12.011g C
0.969 g CO2 contains 12.011 x 0.969 = 0.2645 g C
44.009
% of C in the sample = 0.2645 x 100 = 57.00
0.464
18.015 g H2O contains 2.016 g H
0.198 g H2O contains 2.016 x 0.198 = 0.0221 g H
18.015
% of H in the sample = 0.0221 x 100 = 4.76
0.464


233.389 g BaSO4 contains 32.066 g S
0.859 g BaSO4 contains 32.066 x 0.859 = 0.1180 g S
233.389
% of S in the sample = 0.1180 x 100 = 25.43
...
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