Answer:
6.00%
Explanation:
Step 1: Given data
Accepted value for the number of calories in a Bananas Foster: 300 calories
Measured value for the number of calories in a Bananas Foster: 318 calories
Step 2: Calculate the percent error in the measure
We will use the following expression.
%error = |accepted value - experimental value|/ accepted value × 100%
%error = |300 cal - 318 cal|/ 300 cal × 100% = 6.00%
ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
Learn more about ΔG° here:
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0.20 moles of iron will be formed in the reaction.
Explanation:
The balanced chemical equation for the reaction between iron (iii) oxide and carbon monoxide to form Fe is to be known first.
the balanced reaction is :
Fe2O3 + 3CO⇒ 2 Fe + 3 CO2
so from the data given the number of moles of carbon monoxide can be known:
3 moles of CO reacted with Fe2O3 to form 2 moles of iron in the reaction.
Number of moles of CO is 6.20 moles
11.6 gm of iron is formed
so the number of moles of iron formed is calculated as
n = mass of iron ÷ atomic weight of iron
= 11.6 ÷ 55.84
= 0.20 moles of iron will be formed when 11.6 gram of iron is produced.
