Answer:
moles of glucose
<u>2.3166 moles of glucose</u>
<u></u>
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:
Nucleus,mitochondria,& cell membrane
Abiotic is the right answer
Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
