What volume (in mls) of 1.50 m na2co3 solution can be prepared from 3.975 g of solid na2co3?

56.25 mL

We are given;

Molarity of Na₂CO₃ = 1.50 M

Mass of Na₂CO₃ = 3.975 g

We are required to calculate the volume of Na₂CO₃

Step by step explanation:

Molar mass of Na₂CO₃ = 106 g/mol

But; moles = Mass ÷ molar mass

Therefore;

Moles of Na₂CO₃ = 3.975 g ÷ 106 g/mol

                             = 0.0375 moles

Relationship between volume, moles and molarity is given by;

Molarity = Moles ÷ Volume

Thus;

Volume = moles ÷ Molarity

             = 0.0375 mole × 1.5 M

             = 0.05625 L

But; 1 L = 1000 mL

Therefore; Volume = 56.25 mL

Hence, the volume of Na₂CO₃ is 56.25 mL

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

the answer is the first option

explanation:

just googled it m8. not that hard

lol