0.0821 L-atm/mol-K
8.314 kPa-L/mol-K
Answer:
.926 moles
Explanation:
Rounding :
H2 0 = 18 gm/mole
50 gm would then be 50 / 18 = 2.7777 moles of water
every two moles of 2 C6H6 produces 6 moles of water
2.7777/6 * 2 = .926 moles
Answer:
NH3(aq)
Explanation:
Gold III hydroxide is an inorganic compound also known as auric acid. It can be dehydrated at about 140°C to yield gold III oxide. Gold III hydroxide is found to form precipitates in alkaline solutions hence it is not soluble in calcium hydroxide.
However, gold III hydroxide forms an inorganic complex with ammonia which makes the insoluble gold III hydroxide to dissolve in ammonia solution. The equation of this complex formation is shown below;
Au(OH)3(s) + 4 NH3(aq) -------> [Au(NH3)4]^3+(aq) + 3OH^-(aq)
Hence the formation of a tetra amine complex of gold III will lead to the dissolution of gold III hydroxide solid in aqueous ammonia.
Answer:
Equilibrium constant Kc for the reaction will be 1.722
Explanation:
O2(g)+NO(g)→CO(g)+ NO2(g)
0.88 3.9 --- ---
0.88x 3.9-x x x
GIVEN:
0.88X-X= 0.11
⇒ X=0.77
CO2(g)+NO(g) → CO(g) + NO2(g)
0.88 3.9 --- ---
0.88-x 3.9-x x x
= 3.13 0.77 0.77
=0.11
Kc = ![\frac{[CO] *[NO2]} {[CO2]*[NO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO%5D%20%2A%5BNO2%5D%7D%20%7B%5BCO2%5D%2A%5BNO%5D%7D%20)
=
= 1.722
