If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
Is there answer choices ?
Maybe it's [Acid] and [H+ ions]
Ideal Gas Law
Given: P1= 1.25 atm P2= 95 atm
V1= 2.50 L
Unknown: V2=?
Formula: V=nRT/P; n = RT / PV
Solution: n = (0.0821 L x atm / K x mol) (285K) / (1.25 atm) ( 2.50L)
= 23.40 / 3.125
Answer: n = 7.49 moles of gas the balloon holds
Bones, muscles, and tendons