Answer:
There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution
Explanation:
Molarity of the solution = 2.20 M
Therefore, there are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution
Answer:
Nothing
Explanation:
I would say this because the parents are picking the baby's traits and they make all the decisions. And i guess the doctor would have to follow the traits, if i'm right i tried
The awnser is D or the 4th one
The first one is 32mL and the second one is 2.62 and I think it’s grams/mL I’m not for sure about the letters on the second one
Answer: a . 152g/mol b. 102g/mol c. 183g/mol
Explanation:
By stating the atomic masses of each element in the questions, we have;
Fe= 56, S= 32, O= 16, Al = 27, C = 12, H =1 , N = 14, therefore
(a). FeSO4 = 56 + 32 + (16 x 4) = 152g/mol
(b). Al2O3 = (27 x 2) + (16 x 3) = 102g/mol
(c). C7H5NO3S ( Saccharin, an artificial Sweetner) =
(12 x 7) + (1 x 5) + 14 + (16 x 3) + 32 = 183g/mol
