17. Calcium phosphate has a MW of 310.18 g/mol. a. How many moles would be in 31

Question

17. Calcium phosphate has a MW of 310.18 g/mol.

a. How many moles would be in 31.018 g calcium phosphate?

b. How many grams would be in 31.018 moles calcium phosphate?

18. In glycolysis under anaerobic conditions, aqueous glucose is made into hydrogen ion and lactate ion.

                                                      C6H12O6(g) à C3H5O3 + 2 H1+(aq)

Each of the following statements would correct a mistake that is present in this equation except one. Which one?

                  a) Changing the subscript (g) for glucose to (aq).

                  b) Changing the charge for lactate ion from nothing to 1-.

                  c) Changing the subscript for lactate from nothing to (aq).

                  d) Changing the coefficient for lactate from 1 to 2.

                  e) You can’t fool me!   All of these are necessary corrections to make a properly-written correct balanced equation.

Explanation / Answer

17 .

a) no of mol of calcium phosphate = w.mwt = 31.018/310.18 = 0.1 mol

b) mass of calcium phosphate = n*Mwt

       = 31.018*310.18

       = 9621.16 g

18. correct equation :      C6H12O6(aq) --->   2C3H5O3^-(aq) + 2H1+(aq)

answer: e) You can’t fool me!   All of these are necessary corrections to make a properly-written correct balanced equation.

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