Answer:
the final volume of the gas is = 1311.5 mL
Explanation:
Given that:
a sample gas has an initial volume of 61.5 mL
The workdone = 130.1 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm ; we have
External Pressure :
The workdone W = V
The change in volume ΔV=
ΔV =
ΔV =
ΔV = 1.25 L
ΔV = 1250 mL
Recall that the initial volume = 61.5 mL
The change in volume V is
multiply through by (-), we have:
= 1250 mL + 61.5 mL
= 1311.5 mL
∴ the final volume of the gas is = 1311.5 mL
I don’t know but I think it would be products... that’s the best I can give. I’ll look more into it
Answer:
Cation present- Cr^3+
Cations absent- Al^3+,Fe^3+ and Ni^2+
Explanation:
The cations in group III are; Cr^3+,Al^3+,Fe^3+ and Ni^2+
The first step in the analysis of group III cations is the addition of sodium hypochlorite and sodium hydroxide. Chromium reacts as follows;
2Cr^3+(aq) + 3OCl^-(aq) + 10OH^-(aq) ---------> 2CrO4^2-(aq) + 3Cl^-(aq) + 5H2O(l)
In alkaline solution, Aluminum is separated from Chromium by precipitation. The yellow solution formed is unaffected by treatment with ammonia. However, in acid medium, a blue solution is formed which confirms the presence of Chromium III.