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:
D. All of them would have the same kinetic energy
Explanation:
The expression for the kinetic energy of the gas is:-
k is Boltzmann's constant = 
T is the temperature
<u>Since, kinetic energy depends only on the temperature. Thus, at same temperature, at 300 K, all the gases which are 
will posses same value of kinetic energy.</u>
Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
Answer:
Explanation:
The physical methods of separating mixtures are used in sorting a mixture of substances.
It requires no chemical changes occurring between their components and parts in any significant way.
Examples are:
- Decantation
- Filtration
- Sublimation
- Magnetism
- Centrifugation
The methods simply relies on the physical properties of matter.
