441 g CaCO₃ would have to be decomposed to produce 247 g of CaO
<h3>Further explanation</h3>
Reaction
Decomposition of CaCO₃
CaCO₃ ⇒ CaO + CO₂
mass CaO = 247 g
mol of CaO(MW=56 g/mol) :
From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :
mass CaCO₃(MW=100 g/mol) :
Answer:
1.1 mol
Explanation:
n=m/M, where n is moles, m is mass, and M is molar mass.
M of CO2 = 12.01+16.00+16.00 = 44.01g/mol
n=50g/44.01g/mol
n = 1.13610543 mol
n ≈ 1.1 mol
Hope that helps
I believe the answer would be A. Electronegativity increases across a period.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
During the pumping of hot air into the expansion valve.
