Answer:
+3
Explanation:
Sorry, I don't have one for now. I remember answering this a while ago, but I don't remember the exact reason why it's 3+.
Answer:
(a) Benzene = 0.26; toluene = 0.74
(b) Benzene = 0.55
Explanation:
1. Calculate the composition of the solution
For convenience, let’s call benzene Component 1 and toluene Component 2.
According to Raoult’s Law,

where
p₁ and p₂ are the vapour pressures of the components above the solution
χ₁ and χ₂ are the mole fractions of the components
p₁° and p₂° are the vapour pressures of the pure components.
Note that
χ₁ + χ₂ = 1
So,

χ₁ = 0.26 and χ₂ = 0.74
2. Calculate the mole fraction of benzene in the vapour
In the liquid,
p₁ = χ₁p₁° = 0.26 × 75 mm = 20 mm
∴ In the vapour

Note that the vapour composition diagram below has toluene along the horizontal axis. The purple line is the vapour pressure curve for the vapour. Since χ₂ has dropped to 0.45, χ₁ has increased to 0.55.
Answer:
It is present in third period that's why its valance electrons are present in 3rd energy level.
Its atomic number is greater than lithium when compared in group wise.
There are more electrons in sodium to shield the outer valance electron thus nuclear attraction becomes weak and size increase.
Explanation:
The size of sodium is greater than lithium because atomic number of sodium is 11 and lithium is 3. Both are present in first group but sodium is present down to the lithium. As we move from top to bottom in a group atomic size increases with addition of electrons. The nuclear effect become weaker on valance electrons and atomic size increase. Same time shielding effect is also produces which shield the outer electrons from the influence of nucleus. While in case of lithium less electrons are present to shield the valance electrons.
As we note the position of both elements along period. The sodium is present in third period while lithium is present in second period. So, in case of sodium third energy level is involved. That's why its size is greater than lithium.
The volume of the balloon when the mass of oxygen gas is decreased to 50g is 50L.
We will use the ideal gas equation-
<u>PV=nRT</u>
P=Pressure
V=volume
n=no. of moles
T=temperature
In this question temperature and pressure will remains constant then the above equation can be rewritten as-

where V1= initial volume of the balloon
V2= volume of the balloon when the mass of oxygen gas is decreased to 50g
now, substitute the values in the above equation-
100/3.12=V2/1.56
hence, the volume of the balloon when the mass of oxygen gas is decreased to 50g is 50L.
learn more about the ideal gas equation here:
brainly.com/question/1056445
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Since there is more energy added as heat rises, the particles disperse and have larger movements.