Answer:
∴ΔH₂ = - 12,258 KJ
Explanation:
Enthalpy:
Enthalpy is a property of a thermodynamic system. Enthalpy of a system is equal to the sum of internal energy of the system and presser times volume of the system.
The heat absorbes or releases in a closed system is the change of enthalpy of the system.
Given reactions are:
Reaction 1: C₃H₈(g)+5O₂(g)→ 3CO₂(g)+4H₂O, ΔH₁= - 2043 KJ
Reaction 2: 6C₃H₈(g)+30 O₂(g)→ 18 CO₂(g)+24 H₂O, ΔH₂=?
Take a look at reaction 1 and reaction 2, the only difference is that 1 molecule of C₃H₈ is combusted in reaction 1 and 6 molecules of C₃H₈ is combusted in reaction 2.
We can think the reaction 2 as occurring 6 different container and each containers contains 1 molecule of C₃H₈. The enthalpy is an extensive property. Total enthapy of the 6 containers is = 6×(-2043 KJ)
= - 12,258 KJ
∴ΔH₂ = - 12,258 KJ
One
Let's start by stating what we know is wrong. Equilibrium is achieved when the reactants and products have a stable concentration. That makes D incorrect. Equilibrium is not established until about the 6th or 7th second.
The fact that you get any products at all means that the reactants will become products. Just who is favored has to be looked at very carefully. The products start very near 0. They go up until their concentration at equilibrium. When the reach equilibrium, the products have increased to 17. The reactants have dropped from 40 to 27. By a narrow margin, I would say the products are favored.
C is incorrect. There are still reactants left.
E is incorrect. the reactants started out with a concentration of 40. The reaction is not instantaneous. The concentration was highest at 40 or right at the beginning. This assumes that the reactants were mixed and the products were produced and the water/liquid amount has not changed.
B is incorrect. The concentration of the reactants is higher at equilibrium.
A is wrong. It is product favored.
I'm getting none of the above.
Problem Two
AgBr is insoluble (very). You'd have to work very hard to get them to separate into their elemental form. Just putting AgBr in water isn't enough. Lots of heat and lots of electricity are needed to get the elemental form.
I suppose you should pick B. Mass must be preserved. But if you balanced the equation, it would work with heat and electricity.
Answer:
Explanation:
Let's try and figure out the ones you don't keep.
Receipts from the dollar tree. You can't return what you bought and you only paid a dollar. It's not worth it.
Food receipts. You can't return it and unless you want to itemize what you spend / month and on what, they don't serve any purpose.
Now figure out which ones you would want to keep
1. Anything pertaining to medical perscriptions. I don't know how it works in the United States, but in Canada we are allowed to deduct only medications with a DIN number.
2 Anything that you would use for travelling that is not for pleasure (ie it is a business expense). So if you are a salesman and your territory is from A to B you would deduct meals, mileages, motels if you must stay away from home
3 In the states some tax receipts are deductible, so you might want to save those.
D. With the same number of protons and different number of neutrons.
Answer:
Δ S = 26.2 J/K
Explanation:
The change in entropy can be calculated from the formula -
Δ S = m Cp ln ( T₂ / T₁ )
Where ,
Δ S = change in entropy
m = mass = 2.00 kg
Cp =specific heat of lead is 130 J / (kg ∙ K) .
T₂ = final temperature 10.0°C + 273 = 283 K
T₁ = initial temperature , 40.0°C + 273 = 313 K
Applying the above formula ,
The change in entropy is calculated as ,
ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )
ΔS = 26.2 J/K