Best Answer

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

Q: How many anions are there in 2.50 g of MgBr2?

Write your answer...

Submit

Still have questions?

Continue Learning about Chemistry

There are 22,9.10e21 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

As you know the density of pure water is 1 g/ml. so the mass of 250 ml water is 250 g. Now the no of molecules of water present in 250 g of water = 250 x 6 x 1023/18 = 8.33 x 1024 Now each molecule of water contains total 10 electron. So total no. of electrons in 250 g of water are 10 x 8.33 x 1024 = 8.33 x 1025 electrons.

250 grams CaCO3 (1 mole CaCO3/100.09 grams) = 2.50 moles of calcium carbonate

Related questions

There are 22,9.10e21 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

250 g = about 0.55116 pounds.

250 g = 0.25 kgTo convert from g to kg, divide by 1000.

250 mg = 0.25 g

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

1 tonne = 1,000,000 g ⇒ 1 tonne ÷ 250 g = 1,000,000 g ÷ 250 g = 4,000

48 250 g = 48.25 kgTo convert from g to kg, divide by 1000.

250 g iron (III) oxide is equal to 1,565 moles.