Answer:
— 159.6°C
Explanation:
Data obtained from the question include:
V1 (initial volume) = 960L
T1 (initial temperature) = 38°C = 38 + 273 = 311K
V2 (final volume) = 350L
T2 (final temperature) =?
Since the pressure is constant, then Charles' law is in operation. Using the Charles' law equation V1/T1 = V2/T2, we can easily obtain the final temperature as follow:
V1/T1 = V2/T2
960/311 = 350/T2
Cross multiply to express in linear form.
960 x T2 = 311 x 350
Divide both side by 960
T2 = (311 x 350) /960
T2 = 113.4K
Now let us convert 113.4K to a number in celsius scale. This is illustrated below:
°C = K — 273
°C = 113.4 — 273
°C = — 159.6°C
Therefore, the container will have a volume of 350L at — 159.6°C
Answer:
3.25 moles of H2O
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2HCl + Ca(OH)2 —> CaCl2 +2H2O
Now, we can obtain the number of mole H2O produced from the reaction as follow:
From the balanced equation above,
2 moles of HCl reacted to produce 2 moles of H2O.
Therefore, 3.25 moles of HCl will also react to produce 3.25 moles of H2O.
Therefore, 3.25 moles of H2O were produced from the reaction.
Answer:
Rubidium Acetate
Explanation:
Rb is Rubidium C2H3O2 is acetate
The heat transfer just occurred is mainly conduction.
Conduction happens when two objects are in contact with each other. In the hotter object, the molecules and/or free electrons have a higher kinetic energy, thus they'll travel and collide into other molecules, resulting in spreading the energy to the other object.
The heat transfer happens until thermal equilibrium, where both objects have the same temperature and their molecules have the same kinetic energy rate.
In addition, radiation is also happening since everything that has a higher temperature than the environment is a net emitter. They release electromagnetic waves that turn out to be radiation. These occur even without the presence of air.
Answer:
E° = 0.00 V
E = 0.079 V
Explanation:
We can identify both half-reactions occurring in a concentration cell.
Anode (oxidation): Al(s) → Al³⁺(1.0 × 10⁻⁵ M) + 3 e⁻ E°red = -1.66 V
Cathode (reduction): Al³⁺(0.100 M) + 3 e⁻ → Al(s) E°red = -1.66 V
The global reaction is:
Al(s) + Al³⁺(0.100 M) → Al³⁺(1.0 × 10⁻⁵ M) + Al(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = -1.66 V - (-1.66 V) = 0.00 V
To calculate the cell potential (E) we have to use the Nernst equation.
E = E° - (0.05916/n) .log Q
where,
n: moles of electrons transferred
Q: reaction quotient
E = 0.00 V - (0.05916/3) .log (1.0 × 10⁻⁵/0.100)
E = 0.079 V
