The molar mass of CsF is 151.9 g/mol. How many grams of CsF are needed to prepare 300.0 mL of 0.0500 M CsF solution?

2.28 g of CsF.

Explanation:

The following data were obtained from the question:

Molar mass of CsF = 151.9 g/mol.

Volume = 300 mL

Molarity of CsF = 0.05 M

Mass of CsF =.?

Next, we shall convert 300 mL to litre (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

300 mL = 300 mL /1000 mL × 1 L

300 mL = 0.3 L

Therefore, 300 mL is equivalent to 0.3 L

Next, we shall determine the number of mole CsF in the solution. This can be obtained as follow:

Volume = 0.3 L

Molarity of CsF = 0.05 M

Mole of CsF =.?

Molarity = mole /Volume

0.05 = mole of CsF /0.3

Cross multiply

Mole of CsF = 0.05 × 0.3

Mole of CsF = 0.015 mole.

Finally, we shall determine the mass of CsF needed to prepare the solution as follow:

Mole of CsF = 0.015 mole.

Molar mass of CsF = 151.9 g/mol.

Mass of CsF =.?

Mole = mass /Molar mass

0.015 = mass of CsF /151.9

Cross multiply

Mass of CsF = 0.015 × 151.9

Mass of CsF = 2.28 g

Therefore, 2.28 g of CsF is needed to prepare the solution.

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

answer; /// i do believe that the correct answer is (c)nuclear bond