It should be for the total solution of 93 plus 20 grams which is 113 grams so 93 divided by 113 grams comes to 82.3% sodium sulfate and this can be checked by multiplying 113 grams by 0.823 which results in 93 grams of sodium sulphate.
Answer:
964ug
Explanation:
The problem here involves converting from one unit to another.
We are to convert from ounces to micrograms.
1ug = 1 x 10⁻⁶g
1oz = 28.35g
So we first convert to grams from oz then take to ug:
Solving:
1oz = 28.35g
3.4 x 10⁻⁵oz will then give 3.4 x 10⁻⁵ x 28.35 = 9.64 x 10⁻⁴g
So;
1 x 10⁻⁶g = 1ug
9.64 x 10⁻⁴g will give
= 9.64 x 10²ug or 964ug
The structure of compound A would be solid that is dense enough for antimicrobial form
The molality of the solution = 17.93 m
<h3>Further explanation</h3>
Given
6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)
Required
The molality
Solution
molality = mol of solute/ 1 kg solvent
mol of solute = mol of ethylene glycol
- mass of ethylene glycol :
= volume x density
= 6 L x 1.1132 kg/L
= 6.6792 kg
= 6679.2 g
- mol of ethylene glycol (MW=62.07 g/mol)
=mass : MW
=6679.2 : 62.07
=107.608
6 L water = 6 kg water(ρ= 1 kg/L)
