There are 22,9.10e21 anions.
2.50 g MgBr2 x 1 mole MgBr2/184 g x 2 mole Br/mole MgBr2 = 0.0272 moles Br^-0.0272 moles Br^- x 6.02x10^23 anions/mole = 1.6x10^22 anions
1.33
Total mass of MgBr2- 36.9 g Mass of Br 32.0 / total mass 36.9 = .867 = 86.7% Mass of Mg 4.9 / 36.9 = .133 = 13.3% There is 86.7% of Bromine and 13.3% of Magnesium in the compound MgBr2
392.5 g
350 g C2H6 x 1 mole C2H6/30 g x 2 moles C/mole C2H6 x 12 g/mole C = 280 g carbonDone another way: %C = 24/30 (x100%) = 80% 0.8 x 350 g = 280 grams carbon
2.50 g MgBr2 x 1 mole MgBr2/184 g x 2 mole Br/mole MgBr2 = 0.0272 moles Br^-0.0272 moles Br^- x 6.02x10^23 anions/mole = 1.6x10^22 anions
1.33
The molar mass of MgCl2 = 95.211 g/mol
1.39 moles
molar mass AlF3 = 83.98 g/mole33.4 g AlF3 x 1 mole/83.98 g = 0.398 molesEach mole of AlF3 contains 3 moles of F- anions, because AlF3 ==> Al^3+ + 3F^-.Thus, moles of anions = 3 x 0.398 = 1.19 moles of anions
350 g = 12.3458 oz350 g = 12.3458 oz350 g = 12.3458 oz350 g = 12.3458 oz350 g = 12.3458 oz350 g = 12.3458 oz
350 g sample of CO contain 12,49 moles.
Total mass of MgBr2- 36.9 g Mass of Br 32.0 / total mass 36.9 = .867 = 86.7% Mass of Mg 4.9 / 36.9 = .133 = 13.3% There is 86.7% of Bromine and 13.3% of Magnesium in the compound MgBr2
That is 12.346 ounces
So far I come up with: HBr (aq) + MgSO3 (s) --> H2SO3 (aq) + MgBr2 (??)
The answer is (0.771617 LB) LB Stands for POUNDS.
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