The answer is
C. condensation
Answer:
A
Explanation:
They will collect samples from different sites and write their observations
Answer:
-1,103.39KJ/mol
Explanation:
We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.
In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.
The standard enthalpies of the molecules above are as follows:
H2S = -20.63KJ/mol
H2O = -285.8KJ/mol
SO2 = -296.84KJ/mol
O2 = 0KJ/mol
ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3
ΔfH⦵(O2)]
ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]
-[ 3 × -20.63)]
= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol
Answer:
b) pH = 9.25
Explanation:
- NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
- NH3 + H2O ↔ NH4+ + OH-
- 2 H2O ↔ H3O+ + OH-
⇒ Kb = [ NH4+ ] * [ OH- ] / [ NH3 ] = 1.86 E-5......from literature
mass balance NH4+:
⇒ M NH4+ = [ NH4+ ] - [ OH- ]
∴ [ NH3 ] ≅ M NH4+ = 0.26 M
⇒ Kb = (( 0.26 + [ OH- ] )) * [ OH- ] / 0.26 = 1.86 E-5
⇒ 0.26 [ OH-] + [ OH- ]² = 4.836 E-6
⇒ [ OH- ]² + 0.26 [ OH- ] - 4.836 E-6 = 0
⇒ [ OH- ] = 1.859 E-5 M
⇒ pOH = - Log ( 1.859 E-5 )
⇒ pOH = 4.7305
⇒ pH = 14 - pOH = 9.269
Answer:
Cyclobutane is a cycloalkane and organic compound with the formula (CH2)4.
Explanation:
