Answer: dispersion forces, dipole-dipole and hydrogen bonds
Explanation:
Answer:
5.55 L
Explanation:
This excersise can be solved by the Boyle's law.
This law for gases states that the pressure of a gas in a vessel is inversely proportional to the volume of the vessel.
P₁ . V₁ = P₂ . V₂
The law comes from the Ideal Gases Law, in the first term.
P . V = n . R . T In this case, n . R . T are all constant.
6.35 L . 88.6 kPa = 101.3 kPa . V₂
V₂ = (6.35 L . 88.6 kPa) / 101.3 kPa
V₂ = 5.55 L
It is inversely proportional because, as it happened in this case, pressure was increased, therefore volume decreased.
Answer:
3 e⁻ transfer has occurred.
Explanation
This is a redox reaction.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
- [Ne]= (1s²) (2s² 2p⁶)
A combination of both the reactions( Half-reactions) leads to a redox reaction.
Let us look at initial configurations of Al and Cl
[Al]= 1s² 2s² 2p⁶ 3s² 3p¹
[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵
Hence, Al can lose 3 electrons to achieve octet config.
and, Cl can gain 1e to achieve nearest noble gas config. [Ar]
This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.
Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃
Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)
Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.
Answer:
Compound B has greater molar mass.
Explanation:
The depression in freezing point is given by ;
..[1]
Where:
i = van't Hoff factor
= Molal depression constant
m = molality of the solution
According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.
The depression in freezing point of solution with A solute: 
Molar mass of A = 
The depression in freezing point of solution with B solute: 
Molar mass of B = 
As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.
This means compound B has greater molar mass than compound A,
