Asked by anonymous
Please check my answers for the following questions. THANK YOU.
1) The empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen is....
ANSWER: N2O
2) empirical formula of a compound that consists of 4.80 grams of carbon, 1.20 grams of hydrogen, and 2.80 grams of nitrogen is...
Answer: C2H6N
3)empirical formula of a substance composed of 3.05% carbon, 0.26% hydrogen, and 96.69% iodine is...
ANSWER: CHI3
4) Symclosene has an empirical formula of CClNO and a molecular mass of 232.42 u. What is the molecular formula of symclosene?
ANSWER: C3Cl3N3O3
5) Empirical formula of a compound is C2H3O. The gram formula mass of the compound is 172g. What is the molecular formula?
ANSWER: C8H12O4
6) The molecular formula of a given compound is C4H10. The empirical formula of the same compound would be:
ANSWER: C2H5
#2 through #6 are correct. #1 is not. Didn't you mean N2O5?
1) The empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen is....
ANSWER: N2O
2) empirical formula of a compound that consists of 4.80 grams of carbon, 1.20 grams of hydrogen, and 2.80 grams of nitrogen is...
Answer: C2H6N
3)empirical formula of a substance composed of 3.05% carbon, 0.26% hydrogen, and 96.69% iodine is...
ANSWER: CHI3
4) Symclosene has an empirical formula of CClNO and a molecular mass of 232.42 u. What is the molecular formula of symclosene?
ANSWER: C3Cl3N3O3
5) Empirical formula of a compound is C2H3O. The gram formula mass of the compound is 172g. What is the molecular formula?
ANSWER: C8H12O4
6) The molecular formula of a given compound is C4H10. The empirical formula of the same compound would be:
ANSWER: C2H5
#2 through #6 are correct. #1 is not. Didn't you mean N2O5?
Answers
Answered by
Akighir Reuben
Please check my from the question i asked . (1) C=40.0g x
1mole= 12.0g
3.33moles
H= 6.7g x
1mole =1.0g
6.7moles
O=53.3g x
1mole = 16.0g
3.33moles
mole ratio
C = 3.33/3.33
=1.00
= 1
H = 6.7/3.33
= 2. 01
= 2
O = 3.33/3.33
= 1.00
= 1
So, the empirical formular is CH2O
1mole= 12.0g
3.33moles
H= 6.7g x
1mole =1.0g
6.7moles
O=53.3g x
1mole = 16.0g
3.33moles
mole ratio
C = 3.33/3.33
=1.00
= 1
H = 6.7/3.33
= 2. 01
= 2
O = 3.33/3.33
= 1.00
= 1
So, the empirical formular is CH2O
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