According to the source below, the solubility of sulfanilamide in 95% ethyl alcohol at 78°C is 210 mg/mL. Since 0.1 g = 100 mg, we can set up a proportion:
(210 mg) / (1 mL) = (100 mg) / (x mL) Solving, x = 0.48 mL of 95% ethyl alcohol will be required.
I do not know previously the solubility of sulfanilamide in 95% ethyl alcohol. Let us accept the solubility you quoted here.
100/210 = 0.47619047619.. ≈ 0.48 (ml)
at 0C, the amount of sulfanilamide remains in the solution is: 14*(100/210) = 6.67 (mg), since you only have 0.48 ml solution.
The volume of the solution will change a little by cooling from 78C to 0C. You may also consider this volume change if you have data.
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Answer: What is the mass of disulfur pentoxide (S2O5)?
Explanation:144.14g/mol
The water will started boiling and the pasta is going to start like melting so it can get a soft and smooth texture so you can eat it.
