Answer:
11.3 g of H₂O will be produced.
Explanation:
The combustion is:
2C₈H₁₈ + 25O₂→ 16CO₂ + 18H₂O
First of all, we determine the moles of the reactants in order to find out the limiting reactant.
8 g / 114g/mol = 0.0701 moles of octane
37g / 32 g/mol = 1.15 moles of oxygen
The limiting reagent is the octane. Let's see it by this rule of three:
25 moles of oxygen react to 2 moles of octane so
1.15 moles of oxygen will react to ( 1.15 . 2)/ 25 = 0.092 moles of octane.
We do not have enough octane, we need 0.092 moles and we have 0.0701 moles. Now we work with the stoichiometry of the reaction so we make this rule of three:
2 moles of octane produce 18 moles of water
Then 0.0701 moles of octane may produce (0.0701 . 18)/2= 0.631 moles of water.
We convert the moles to mass → 0.631 mol . 18 g/1mol = 11.3 g of H₂O will be produced.
Two moles because one moles would be 6.02×10 23 atoms 2 moles would equal twice that number of atoms
Answer:
k ≈ 9,56x10³ s⁻¹
Explanation:
It is possible to solve this question using Arrhenius formula:
Where:
k1: 1,35x10² s⁻¹
T1: 25,0°C + 273,15 = 298,15K
Ea = 55,5 kJ/mol
R = 8,314472x10⁻³ kJ/molK
k2 : ???
T2: 95,0°C+ 273,15K = 368,15K
Solving:
<em>k ≈ 9,56x10³ s⁻¹</em>
I hope it helps!
7 A lead
B Gold
Csilver
8 Afeathrr
B water
C silver
It would be D since it could be unhealthy for you.
