Answer:
-85 °C
Explanation:
O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.
H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter
molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.
Atomic radius of copper atom (r)= 128 pm
volume of sphere = 4/3 * pi * r^3 = 4/3 * 3.14 * 128^3
volume = 8780076.373 pm^3
When particles are heated they expand
Given:
Moles of H2 = 0.300
Moles of I2 = 0.400
Moles of HI = 0.200
Keq = 870
To determine:
Amounts of the mixture at equilibrium
Explanation:
H2(g) + I2(g) ↔ 2HI(g)
Initial 0.3 0.4 0.2
Change -x -x +2x
Eq (0.3-x) (0.4-x) (0.2+2x)
Keq = [HI]²/[H2][I2]
870 = (0.2+2x)²/(0.3-x)(0.4-x)
x = 0.29 moles
Amounts at equilibrium:
[HI] = 0.2 + 2(0.29) = 0.78 moles
[H2] = 0.3-0.29 = 0.01 moles
[I2] = 0.4-0.29 = 0.11 moles
Answer:
the answer to your question above is D