The grams of 22.9 % sugar solution that contain 68.5 g of sugar is 299.13 g of solution
<u><em>calculation</em></u>
22.9% means that there are 22.9 g of sugar in 100 g of solution.
what about 68.5 g of sugar
- <em>by cross multiplication</em>
=[(68.5 g sugar x 100 g solution) /22.9 g sugar] =299.13 g of solution
Nb; <em>g sugar cancel each other</em>
Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH
= 48.875 x 10⁻⁴ moles NaOH
It will react with same number of moles of acetic acid
So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴
number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles
= 1.4167 moles
= 1.4167 x 60 gram
= 85 grams .
So 85 grams of acetic acid will be contained in one litre of acetic acid.
Heating up, mostly. Solid can be heated to liquid, then to gas, then eventually to plasma.
