How many grams of K2CO3 are needed to prepare a 400.0 mL of a 4.25M solution?

1.7 g

235 g

81.3 g

138 g

235 g

Explanation:

From the question;

Volume is 400.0 mLMolarity of a solution is 4.25 M

We need to determine the mass of the solute K₂CO₃,

we know that;

Molarity = Number of moles ÷ Volume

Therefore;

First we determine the number of moles of the solute;

Moles = Molarity × volume

Moles of  K₂CO₃ = 4.25 M × 0.4 L

                           = 1.7 moles

Secondly, we determine the mass of  K₂CO₃,

We know that;

Mass = Moles × Molar mass

Molar mass of  K₂CO₃, is 138.205 g/mol

Therefore;

Mass = 1.7 moles × 138.205 g/mol

         = 234.9485 g  

         = 235 g

Thus, the mass of  K₂CO₃ needed is 235 g

Answer:

5.6⋅10−^19  j

do you need the explaination?

Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o