First we need to find the number of moles of both K and O reacted
K - 0.779 g / 39 g/mol
= 0.02 mol
the mass of O₂ reacted = 1.417 g - 0.779 g = 0.638 g
O₂ moles = 0.638 g / 32 g/mol
= 0.02 mol
the number of both K and O₂ moles reacted are equal
therefore stoichiometry of K to O₂ reacted are 1:1
then the formula of potassium superoxide is KO₂
Answer : The molar concentration of ethanol in the undiluted cognac is 8.44 M
Explanation :
Using neutralization law,
where,
= molar concentration of undiluted cognac = ?
= molar concentration of diluted cognac = 0.0844 M
= volume of undiluted cognac = 5.00 mL = 0.005 L
= volume of diluted cognac = 0.500 L
Now put all the given values in the above law, we get molar concentration of ethanol in the undiluted cognac.
Therefore, the molar concentration of ethanol in the undiluted cognac is 8.44 M
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
First step is to determine the valency of each of x and CaCO3 from the given compounds:
1- As for Li2CO3: we can deduce that the valency of lithium is one while that of CO3 is two
2- As for XCl3: we can deduce that the valency of chlorine is one while that of X is three
Second step is to write the required compound:
X : CO3 (elements involved)
3 : 2 (write the valency of each)
Then write the positive ion (X) followed by the valency of the negative ion (2) and then the negative ion (CO3) followed by the valency of the positive ion (3).
The final x carbonate is written as: X2(CO3)3
The percet of Al will be 100 times the mass of Al in the formula divided by the molar mass of the compound.
This table shows all the calculations involved
Compound: Al2 (OH)5 Cl
element # of atomic mass in the %
atoms mass formula
g/mol
Al 2 27 2*27 = 54 g (54 / 174.5)*100 = 30.9%
O 5 16 5*16 = 80 g
H 5 1 5*1 = 5 g
Cl 1 35.5 1*35.5 = 35.5 g
----------------------
molar mass 174.5 g
Answer: 30.9%
