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:
I think the answer is coming up with questions based on observations and reasoning
Answer:
Explanation:
a) For diatomic gas: Translational motion = 3 and rotational motion = 2
∴ Total (internal energy) = 3 + 2 = 5
b) Translational + Rotational + Vibrational = 3 + 2 + 1 = 6
c) Linear molecule
i) Non linear molecule
ii) Monatomic molecule
For the answer to the question above asking, h<span>ow many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
The answer to your question is the the third one among the given choices which is 6.8 mol.
</span><span>moles glucose = 1.5 x 4.5 = 6.8 </span>
Answer:
0.677 moles
Explanation:
Take the atomic mass of K = 39.1, O =16.0, P = 31.0
no. of moles = mass / molar mass
no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)
= 0.02256 mol
From the equation, the mole ratio of KOH : K3PO4 = 3 :1,
meaning every 3 moles of KOH used, produces 1 mole of K3PO4.
So, using this ratio, let the no. of moles of KOH required to be y.
y = 0.02256 x3
y = 0.0677 mol
If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.
