Answer:
0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.
Explanation:
Step 1: Data given
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = Shows how much the boiling point increases
⇒i = the van't Hoff factor: Says in how many particles the compound will dissociate
⇒ Since all are aqueous solutions Kb for all solutions is the same (0.512 °C/m)
⇒m = the molality
Step 2:
0.20 m glucose
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for glucose = 1
⇒ Kb = 0.512 °C/m
⇒m = 0.20 m
ΔT = 1*0.512 * 0.20
<u>ΔT = 0.1024 °C</u>
0.30 m BaCl2
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for BaCl2 = Ba^2+ + 2Cl- : i = 3
⇒ Kb = 0.512 °C/m
⇒m = 0.30 m
ΔT = 3*0.512 * 0.30
<u>ΔT = 0.4608 °C</u>
0.40 m NaCl
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for NaCl = Na+ + Cl- : i = 2
⇒ Kb = 0.512 °C/m
⇒m = 0.40 m
ΔT = 2*0.512 * 0.40
<u>ΔT = 0.4096 °C</u>
0.50 m Na2SO4.
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for Na2SO4 = 2Na+ + SO4^2- : i =3
⇒ Kb = 0.512 °C/m
⇒m = 0.50 m
ΔT = 3*0.512 * 0.50
<u>ΔT = 0.768 °C</u>
0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.
This is an incomplete question, here is a complete question.
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kJ. You may want to reference (Pages 381 - 385) Section 9.6 while completing this problem. If the change in enthalpy is -5074.2 kJ, how much work is done during the combustion? Express the work in kilojoules to three significant figures.
Answer : The work done during the combustion is, 9.9 kJ
Explanation :
Formula used :
![\Delta H=\Delta E+w](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5CDelta%20E%2Bw)
where,
w = work done = ?
= change in enthalpy = -5074.2 kJ
= change in internal energy = -5084.1 kJ
R = gas constant = 8.314 J/mol.K
Now put all the given values in the above formula, we get:
![-5074.2kJ=-5084.1kJ+w](https://tex.z-dn.net/?f=-5074.2kJ%3D-5084.1kJ%2Bw)
![w=9.9kJ](https://tex.z-dn.net/?f=w%3D9.9kJ)
Thus, the work done during the combustion is, 9.9 kJ
Answer:
Carbon enters the atmosphere as carbon dioxide from respiration and combustion. Carbon dioxide is absorbed by producers to make glucose in photosynthesis. Animals feed on the plant passing the carbon compounds along the food chain. ... The animals and plants eventually die.
30 because it is the highest number of marriage
Answer:
I believe the answer is A: *It is the simplest form of matter" not 100% sure but I think that's correct
Explanation: