Answer:
C: The temperature of the substance increases as it sits in the beaker of water
Explanation:
This question was taken from a video where an attempt was made to investigate the changes in temperature when a substance undergoes change from it's solid phase to its liquid phase.
To do this, as seen in the video online, it shows a solid substance in a test tube being placed in a beaker of water.
From observation, the water in the beaker has a warmer temperature than the solid substance present in the test tube and this in turn makes the test tube gradually increase in temperature.
Thus, the solid substance will as well increase increase in temperature when it is placed in the beaker of water.
Answer:
The answer is in the explanation.
Explanation:
The KHP is an acid used as standard in titrations to find concentration of bases as NaOH.
The reaction that explain this use is:
KHP + NaOH → KNaP + H2O
<em>where 1 mole of KHP reacts per mole of NaOH</em>
That means, at equivalence point of a titration in which titrant is NaOH, the moles of KHP = Moles of NaOH added
With the moles of KHP = Moles of NaOH and the volume used by titrant we can find the molar concentration of NaOH.
The moles of KHP are obtained from the volume and the concentration as follows:
Volume(L)*Concentration (Molarity,M) = moles of KHP
If the concentration is more or less than 0.100M, the moles will be higher or lower. For that reason, we need to know the concentration of KHP but is not necessary to be 0.100M.
1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).
ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).
ΔrH = 180.6 kJ.
2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
ΔS = 24.8 J/K.
3) ΔG = ΔH - TΔS.
55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.
2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.
3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.
