Astandardized solution that is 0.0500 0.0500 m in na + na+ is necessary for a flame photometric determination of the element. how many grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 800.0 ml of this solution?

2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.

Explanation:

Molarity of sodium ions = = 0.0500 M

Moles of sodium ions = n

Volume of the solution = V = 800.0 mL = 0.800 L

1 mole sodium carbonates gives 2 moles of sodium ion and 1 mole of carbonate ions.

Then 0.04 moles of sodium ions will be obtained from:

of sodium m carbonation.

Mass of 0.02 moles of sodium carbonate = 0.02 mol × 106 g/mol= 2.12 g

2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.

he would use the colour black

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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