Q = ?
Cp = 0.397 J/ºC
Δt = 40.3 - 21.0<span> => 19.3</span><span> ºC</span>
m = 15.2 g
Q = m x Cp x Δt
Q = 15.2 x 0.397 x 19.3
Q ≈ 116.46 J
<span>hope this helps! </span>
Answer:
This is an oxidation-reduction (redox) reaction:
2 Ni0 - 4 e- → 2 NiII
(oxidation)
2 O0 + 4 e- → 2 O-II
(reduction)
Ni is a reducing agent, O2 is an oxidizing agent.
Answer:
Explanation:
Hello,
In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:
Whereas the change in the internal energy is computed by:
So we compute the initial and final temperatures for one mole of the ideal gas:
Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:
Then, the volume-pressure product in Joules:
Finally, the change in the enthalpy for the process:
Best regards.
The effect of an insoluble impurity, such as sand, on the observed melting point of a compound would be none. It will not depress or elevate the melting point of the compound. Instead, it would affect the reading if you are trying to determine the melting point of the compound. This is because you might be missing the actual melting point of the compound since you will be waiting for the whole sample to liquify. You would not be able to determine exactly that temperature because of the insoluble impurity would have a different melting point than that of the compound.
