Answer:
a) Xbenzene = 0.283
b) Xtoluene = 0.717
Explanation:
At T = 20°C:
⇒ vapor pressure of benzene (P*b) = 75 torr
⇒ vapor pressure toluene (P*t) = 22 torr
Raoult's law:
∴ Pi: partial pressure of i
∴ Xi: mole fraction
∴ P*i: vapor pressure at T
a) solution: benzene (b) + toluene (t)
∴ Psln = 37 torr; at T=20°C
⇒ Psln = Pb + Pt
∴ Pb = (Xb)*(P*b)
∴ Pt = (Xt)*(P*t)
∴ Xb + Xt = 1
⇒ Psln = 37 torr = (Xb)(75 torr) + (1 - Xb)(22 torr)
⇒ 37 torr - 22 torr = (75 torr)Xb - (22 torr)Xb
⇒ 15 torr = 53 torrXb
⇒ Xb = 15 torr / 53 torr
⇒ Xb = 0.283
b) Xb + Xt = 1
⇒ Xt = 1 - Xb
⇒ Xt = 1 - 0.283
⇒ Xt = 0.717
Answer:
2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP? STP is a common abbreviation for "standard temperature and pressure." You have to recognize that five values are given in the problem and the sixth is an x. Also ... 273 1. A gas has a volume of 800.0 mL at minus 23.00 °C and 300.0 torr.
Explanation:
The given question is incomplete. The complete question is as follows.
Which of the following best helps explain why an increase in temperature increases the rate of a chemical reaction?
(a) at higher temperatures, high-energy collisions happen less frequently.
(b) at low temperatures, low-energy collisions happen more frequently.
(c) at higher temperatures, less-energy collisions happen less frequently.
(d) at higher temperatures, high-energy collisions happen more frequently
Explanation:
When we increase the temperature of a chemical reaction then molecules of the reactant species tend to gain kinetic energy. As a result, they come into motion which leads to more number of collisions within the molecules.
Therefore, chemical reaction will take less amount of time in order to reach its end point. This means that there will occur an increase in rate of reaction.
Thus, we can conclude that the statement at higher temperatures, high-energy collisions happen more frequently, best explains why an increase in temperature increases the rate of a chemical reaction.
Chemical change
Hope this helps!
The rate constant of the reaction K we can get it from this formula:
K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B
So when after 10 s and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) = 16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14
