When molality = no.of moles of solute/weight Kg of solvent
and when we have the no. of moles of solute = 8.1 moles
and the weight Kg of the solvent =4847 g /1000 = 4.847 Kg
so by substitution :
Molarity = 8.1m / 4.847Kg = 1.67 M
Answer:- Melting point of [tex[CCl_4[/tex] is higher than
where as the vapor pressure of
.
Explanations:- Since both the molecules are non polar, the forces of attraction for carbon tetrachloride are comparatively stronger as it's a bigger molecule. We know that, stronger are the forces of attraction, higher is the melting point as more heat is required to break the bonds.
Stronger are the forces of attraction, lower is the vapor pressure as the molecules are more tightly bonded and less molecules goes to the gas phase.
So, melting point of carbon tetrachloride is higher but it's vapor pressure is lower as compared to carbon disulfide.
Answer: Option (c) is the correct answer.
Explanation:
A monomial is defined as an algebraic expression in which there is only one term present.
For example, 3x is a monomial.
A binomial is defined as an expression where there will be two terms present. For example,
is a binomial.
A polynomial is defined as an algebraic expression which contains more than two terms.
For example,
is a polynomial.
Thus, we can conclude that the given expression
is a polynomial.
Answer:
The answer to your question is molality = 0.61
Explanation:
Freezing point is the temperature at which a liquid turns into a solid if a solute is added to a solution, the freezing point changes.
Data
Kf = 1.86 °C/m
molality = ?
ΔTc = 1.13°C
Formula
ΔTc = kcm
Solve for m
m = ΔTc/kc
Substitution
m = 1.13 / 1.86
Simplification and result
m = 0.61
The two resonating structures of Carbon Monoxide are shown below.
The movement of electrons are shown by arrow.
Structure A: In structure a A the formal charges of Carbon and Oxygen are zero. As
Formal Charge is calculated as,
= # of valence electron - electrons in lone pairs + 1/2 bonding electrons electrons
For C: = 4 - 2 + 4/2
= 4- 4
=
0For O: = 6 - 4 + 4/2
= 6- 6
=
0Structure B:
Formal Charge on C: = 4 - 2 + 6/2
= 4- 5
=
-1Formal Charge on O: = 6 - 2 + 6/2
= 6 - 5
=
+1