The oxidation state is zero for Fe (s)
1) The nucleus of an atom loses 2 protons and 4 neutrons.
2) The nucleus of an atom gains a proton and it's neutrons remain the same.
Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958 = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass
= 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water
= 4.7 / 0.88 = 5.34 moles
∴ moles of the solution = total moles in solu - moles of water
= 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
= 53.8 / 0.64 = 84 g/mole
Q: C
moles of urea (NH2)2 CO = mass weight / molar mass
= 4.49 g / 60 g /mol
= 0.07 mol
moles of methanol = mass weight / molar mass
= 39.9 g / 32 g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg
∴Pv(solu) = 84.55 mmHg
Answer: The standard enthalpy of formation of 
is -252.1 kJ/mol.
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f_%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f_%7Breactantss%7D%5D)
Putting the values we get :
![\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B2%5Ctimes%20H_f%7BFe%7D%2B3%5Ctimes%20H_f%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f%7BFe_2O_3%7D%2B3%5Ctimes%20H_f%7BH_2%7D%5D)
![67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)]](https://tex.z-dn.net/?f=67.9kJ%3D%5B%282%5Ctimes%200%29%2B%283%5Ctimes%20H_f%7BH_2O%7D%29%5D-%5B%281%5Ctimes%20-824.2kJ%2Fmol%29%2B3%5Ctimes%200kJ%2Fmol%29%5D)
Thus standard enthalpy of formation of is -252.1 kJ/mol.
