A standardized solution that is 0.0350 M in 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 mL of this solution

1.484 grams.

Explanation:

At first we have to find the number of moles in 0.0350 M of 800.0 mL of Na solution. For this we will use the molarity equation as follows -

Where,

= molarity of the solution

= number of moles of Na

= volume of solution in mL

From the above equation,

= = 0.028 moles.

Now, we know that sodium carbonate dissociates as

⇒

That means 1 mole of sodium carbonate produces 2 moles of Na⁺ ion

So, the moles of sodium carbonate required for 0.028 mole of Na⁺ ion will be

= 0.014 moles.

So, the weight of sodium carbonate required = number of moles × molecular mass

= 0.014 × 105.9888 = 1.484 g

The weight of sodium carbonate required for preparing the solution mentioned in the question is 1.484 grams.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;