Answer:
the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
Explanation:
for the reaction
PCl₅(g) → PCl₃(g) + Cl₂(g)
where
Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M
and [A] denote concentrations of A
if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M
therefore
Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M
[PCl₅] = 0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M
[PCl₅] = 3.64*10⁻³ M
the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
Answer: Option (A) is the correct answer.
Explanation:
Nitrogen is a non-metal and it is known that non-metals do not conduct electricity. Thus, it will be least conductive out of the given options.
Whereas antimony (Sb) is a metalloid. Metalloid are the substance that show properties of both metals and non-metals. Thus, antimony will conduct electricity.
On the other hand, bismuth (Bi) is a metal hence, it will conduct electricity.
Thus, we can conclude that the order from least conductive to most conductive will be nitrogen (N), antimony (Sb), bismuth (Bi).
Answer:
-0.050 kJ/mol.K
Explanation:
- A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
- The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K
All in all, ΔG° = 0 at 400. K.
We can find ΔS° using the following expression.
ΔG° = ΔH° - T.ΔS°
0 = -20 kJ/mol - 400. K .ΔS°
ΔS° = -0.050 kJ/mol.K
Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol
Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.
W are given a chemical reaction:
To calculate the enthalpy change, we use the formula:
This is the amount of energy released when 0.1326 grams of sample was burned.
So, energy released when 1 gram of sample was burned is = 
Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:
We need to first find the molarity of Ba(OH₂) solution.
A mass of 3.24 mg is dissolved in 1 L solution.
Ba(OH)₂ moles dissolved - 3.24 x 10⁻³ g/171.3 g/mol = 1.90 x 10⁻⁵ mol
dissociaton of Ba(OH)₂ is as follows;
Ba(OH)₂ --> Ba²⁺ + 2OH⁻
1 mol of Ba(OH)₂ dissociates to form 2OH⁻ ions.
Therefore [OH⁻] = (1.90 x 10⁻⁵)x2 = 3.8 x 10⁻⁵ M
pOH = -log[OH⁻]
pOH = -log (3.8 x 10⁻⁵)
pOH = 4.42
pH + pOH = 14
therefore pH = 14 - 4.42
pH = 9.58
