Hey there!:
* For 2p subshell :
n = 2, l =1, ml = -1, 0, +1
* for 5d subshell,
n = 5, l = 2, ml = -2, -1, 0, +1, +2
Hope that helps!
Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
Answer:
Relation between , molality and temperature is as follows.
T =
It is also known as depression between freezing point where, i is the Van't Hoff factor.
Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.
i for = 3
i for glucose = 1
i for NaCl = 2
Depression in freezing point will have a negative sign. Therefore, d
depression in freezing point for the given species is as follows.
=
=
=
Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.
Glucose < NaCl <
Explanation:
