I believe the correct answer from the choices listed above is the last option. The mass of the methanol sample would be 64 grams. It was calculated by multiplying the number of moles to the molar mass of methanol. Hope this answers the question. Have a nice day.
Answer:
CaCO3 (s) → CaO (s) + CO2 (g)
The mass of carbonate that must have reacted was 43.03 grams
Explanation:
CaCO3 → CaO + CO2
Relation between reactant and product is 1:1
Let's apply the Ideal Gas Law to find out the moles of CO2 which were produced.
P . V = n . R . T
1 atm . 23 L = n . 0.082 L.atm/mol.K . 653K
(1atm . 23L) / (0.082 mol.K/L.atm . 653K) = n
0.43 moles = n
0.43 moles of CO2, were produced from 0.43 moles of CaCO3.
Molar weight of CaCO3 = 100.08 g/m
Mass = Molar weight . moles
Mass = 100.08 g/m 0.43 m = 43.03 g
precipitate is the answer
I think c cause if he is in a relationship that’s abusive he needs to speak up before it goes way worst.
Answer:
- Volume = <u>2.0 liter</u> of 1.5 M solution of KOH
Explanation:
<u>1) Data:</u>
a) Solution: KOH
b) M = 1.5 M
c) n = 3.0 mol
d) V = ?
<u>2) Formula:</u>
Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:
<u>3) Calculations:</u>
- Solve for n: M = n / V ⇒ V = n / M
- Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter
You must use 2 significant figures in your answer: <u>2.0 liter.</u>
