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.

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

they it depends on if one believes about radioactivity, they make a geiger counter beep but maybe it was going to beep anyway electronics do wierd things all the time or it was programmed to beep near certain metals.