Answer: V2 = 35.54L
Explanation:
Applying
P1= 67.4, V1= 85, T1= 245, P2= 179.6, V2= ?,. T2=273
P1V1/ T1= P2V2/T2
Substitute and simplify
(67.4*85)/245 = (179.6*V2)/273
V2= 35.54L
The electrons are transferred from potassium to sulphur, in this reaction, potassium is being oxidised and sulphur is being reduced. Two potassium atoms each lose one electron forming K+ ions. Sulphur gains two electrons forming S2-/2- ions. The resultant formula will be: K2S, bonding in this compound is ionic.
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: The density of Ammonia is 0.648 g/l
Explanation:
Density = Mass/ Volume
Mass of one mole of Ammonia (NH3) = 17.031g
Volume =?
Using the ideal gas law we can determine the volume.
PV = nRT
P = 0.913 atm, V= ?, n = 1, R = 0.08206 L.atm/K, and T= 293K
Make V the subject of the formular, we then have;
V= nRT/ P = 1 mol x 0.08206 L.atm/ K.mol x 293 / 0.913 atm
V = 24.04358/ 0.913 = 26.3L
Having gotten the value of Volume in this question, we then go back to solve for density.
Density = Mass/ Volume
17.031g/ 26.3L = 0.64756 ≈ 0.648 g/l
