The 1000 to I think that maybe
There is only a net gain of only two ATP molecules in glycolysis of one six-carbon glucose because it uses the other two to split the glucose.
M(NaCl)=50.0 g
m(H₂O)=150.0 g
m(solution)=m(NaCl)+m(H₂O)
w(NaCl)=100m(NaCl)/m(solution)=100m(NaCl)/{m(NaCl)+m(H₂O)}
w(NaCl)=100*50.0/(50.0+150.0)=25%
We are given the starting amount of Al metal to be used. This will be the starting point of the calculations. We do as follows:
74.00 g ( 1 mol / 26.98 g ) ( 1 mol Al2O3 / 2 mol Al ) ( 101.96 g / 1 mol ) = 139.83 g Al2O3 produced
Hope this answers the question. Have a nice day.
Answer:1.30 x 10 to the 4th
Explanation:
Just know
