_Cl2 + __KI → __KCI + __12
1 answer:
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Answer:
The boiling point is 308.27 K (35.27°C)
Explanation:
The chemical reaction for the boiling of titanium tetrachloride is shown below:
Ti ⇒ Ti
ΔH° (Ti
) = -804.2 kJ/mol
ΔH° (Ti
) = -763.2 kJ/mol
Therefore,
ΔH° = ΔH°
(Ti
) - ΔH°
(Ti
) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol
Similarly,
s°(Ti) = 221.9 J/(mol*K)
s°(Ti) = 354.9 J/(mol*K)
Therefore,
s° = s° (Ti) - s°(Ti
) = 354.9 - 221.9 = 133 J/(mol*K)
Thus, T = ΔH° /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C
Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.
Answer: 502 Joules
Explanation:
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
Volume of water = 40.0 mL
Putting values in above equation, we get:
When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.
The equation used to calculate heat released or absorbed follows:
q = heat absorbed by water
= mass of water = 40.0 g
= final temperature of water = 20.0°C
= initial temperature of water = 17.0°C
= specific heat of water= 4.186 J/g°C
Putting values in equation 1, we get:
Hence, the joules of heat were re-leased by the lead is 502