<span>C. C4H8
Given that the number of moles of CO2 and H2O produced from the combustion are equal, that means for every carbon atom, there are 2 hydrogen atoms because CO2 has only 1 carbon atom and H2O has 2 hydrogen atoms. So let's look at the available choices and see which one is correct.
A. C2H2
This is a 1 to 1 ratio of carbon to hydrogen. Wrong answer.
B. C2H6
This is a 1 to 3 ratio of carbon to hydrogen. Wrong answer.
C. C4H8
This is a 1 to 2 ratio of carbon to hydrogen. Correct answer.
D. C6H6
This is a 1 to 1 ratio of carbon to hydrogen. Wrong answer.</span>
Answer:
T2 = 29.79°C
Explanation:
Equliibrium signifies that heat loss = heat gained
Heat gained by Ice;
H = ML
Mass, M = Number of moles * Molar mass = 1 * 18 = 18g
l = 6.01 k J m o l = 334 J/g
C = 4.186 J/g
H = 18(334)
H = 6012
Heat lost by water
H = MCΔT
H = 18 * 4.186 * (50 - T2)
H = 3767.4 - 75.348T2
Since H = H, we have;
6012 = 3767.4 - 75.348T2
- 75.348T2 = 3767 - 6012
T2 = 2245 / 75.348
T2 = 29.79°C
You can use the equation ΔS(surr)=q(surr)/T or ΔS(surr)=-q(rxn)/T.
the two equations are equal since we know that the energy the system (reactoin) puts out just goes into the surroundings.
(In other words q(surr)=-q(rxn))
Using the equation, <span>ΔS(surr)=-(-283kJ/298K)=0.9497kJ/K or 949.7J/K
This answer makes sense since the reaction is exothermic which means it released energy into the system which usually causes the entropy to increase.
I hope that helps.</span>
Answer:
Water
Explanation :-
Higher the intermolecular forces between the liquid particles, higher its boiling point.
Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
