To solve this problem, we should recall that
the change in enthalpy is calculated by subtracting the total enthalpy of the reactants
from the total enthalpy of the products:
ΔH = Total H of products – Total H of reactants
You did not insert the table in this problem, therefore I
will find other sources to find for the enthalpies of each compound.
ΔHf CO2 (g) = -393.5 kJ/mol
ΔHf CO (g) = -110.5 kJ/mol
ΔHf Fe2O3 (s) = -822.1 kJ/mol
ΔHf Fe(s) = 0.0 kJ/mol
Since the given enthalpies are still in kJ/mol, we have to
multiply that with the number of moles in the formula. Therefore solving for ΔH:
ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0
kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>
ΔH = <span>795.2
kJ</span>
Answer:
It will take 28.5 minutes
Explanation:
<u>Step 1: </u>Data given
Mass of Cu = 4.50 grams
8.00 A of current are used
Molar mass of Cu = 63.5 g/mol
Step 2: Calculate time needed
Cu2+ →Electricity → Cu
we notice a flow of 2 electrons ⇒ This means the Faraday constant = 2F
Since Molar mass of Cu is 63.5 g/mol
63.5 grams of Cu is deposited by 2*96500 C
4.50 grams of Cu ((2*96500)/63.5) * 4.50 = 13677.17 C
Q = It
13677.17 = 8t*60 seconds
t = 28.5 minutes
Answer:
Option B.
Explanation:
As any reaction of combustion, the O₂ is a reactant and the products are CO₂ and H₂O. Combustion reaction for ethane is:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
So 2 moles of ethane react with 7 moles of oxygen to make 4 moles of dioxide and 6 moles of water.
Then 2 moles of ethane will produce 4 moles of CO₂
Because it throws the earth off balance and if it does it often enough then it will soon add up.