Answer:
moles of glucose
<u>2.3166 moles of glucose</u>
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Explanation:
The balance reaction for the formation of glucose is :
here , CO2 = carbon dioxide
H2O = water
C6H12O6 = glucose
O2 = Oxygen
According to this equation :
6 mole of CO2 = 6 mole of H2O = 1 mole of C6H12O6 = 6 mole of O2
We are asked to calculate the mole of Glucose from carbon dioxide.
So,
6 mole of CO2 produce = 1 mole of C6H12O6
1 mole of CO2 will produce =
moles of glucose
13.9 moles of CO2 will produce :
=2.3166 moles of glucose
Note : first , Always calculate for one mole (By dividing)
. After this , multiply the answer with the moles given.
Always write the substance whose amount is asked(glucose) to the right hand side
Answer:
B Genes determine specific traits while the chromosomes contain these genes
Moles=volume*concentration
=0.1*.83
=.083 Moles of HC2H3O2
Mole ratio between HC2H3O2 and CO2 is 1:1
This means .083 Moles of CO2
Mass =Moles*Rfm of CO2
=.083*(12+16+16)
=3.7grams
Answer:
Empirical formula is C₉H₁₅O
Molecular formula = C₈₁H₁₃₅O
₉
Explanation:
Percentage of carbon = 77.60%
Percentage of oxygen = 11.45%
Percentage of hydrogen = 10.95%
Molecular weight = 1253 g/mol
Molecular formula = ?
Empirical formula = ?
Solution:
Number of gram atoms of C = 77.60 g /12g/mol =6.5
Number of gram atoms of O = 11.45 g / 16 g/mol = 0.72
Number of gram atoms of H = 10.95 g / 1.008 g/mol= 10.9
Atomic ratio:
C : H : O
6.5/0.72 : 10.9/0.72 : 0.72/0.72
9 : 15 : 1
C : H : O = 9 : 15 : 1
Empirical formula is C₉H₁₅O
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
n = 1253 / 139
n = 9
Molecular formula = n (empirical formula)
Molecular formula = 9 (C₉H₁₅O
)
Molecular formula = C₈₁H₁₃₅O
₉
