Molar mass of oxygen is:
M(O)=16 g/mol
Molar mass of carbon is:
M(C)=12 g/mol
Molar mass of carbon dioxide is:
M(CO2)=M(C)+2*M(O)
M(CO2)=12 g/mol+2*16g/mol
M(CO2)=44 g/mol
<span>Molar mass(M) is the mass of 1 mole of the substance (grams per mole of a compound).</span>
Answer:
The answer to your question is 0.4 moles of Oxygen
Explanation:
Data
Octane (C₈H₈)
Oxygen (O₂)
Carbon dioxide (CO₂)
Water (H₂O)
moles of water = ?
moles of Oxygen = 1
Balanced chemical reaction
C₈H₈ +10O₂ ⇒ 8CO₂ + 4H₂O
Reactant Element Products
8 C 8
8 H 8
20 O 20
Use proportions to solve this problem
10 moles of Oxygen ----------------- 4 moles of water
1 mol of Oxygen ------------------ x
x = (4 x 1) / 10
x = 4 / 10
x = 0.4 moles of water
<h2>Answer:</h2>
He is right that the energy of vaporization of 47 g of water s 106222 j.
<h3>Explanation:</h3>
Enthalpy of vaporization or heat of vaporization is the amount of energy which is used to transform one mole of liquid into gas.
In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.
It means for vaporizing 18 g, 40.65 kJ energy is needed.
So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ
Hence the student is right about the energy of vaporization of 47 g of water.
I don’t see any equal signs to make it an equation. Am I missing something?
Answer:
7
Explanation:
N-14 has 7
i looked it up ye ur probably gonna get it right
