Answer:
Force of attraction = 35.96 N
Explanation:
Given: charge on anion = -2
Charge on cation = +2
Distance = 1 nm = m
To calculate: Force of attraction.
Solution: The force of attraction is calculated by using equation,
---(1)
where, q represents the charge and the subscripts 1 and 2 represents cation and anion.
k =
F = force of attraction
r = distance between ions.
Substituting all the values in the equation (1) the equation becomes
Force of attraction = 35.96 N
Answer:
Average atomic mass = 17.5 amu.
Explanation:
Given data:
X-17 isotope = atomic mass17.2 amu, abundance:78.99%
X-18isotope = atomic mass 18.1 amu, abundance 10.00%
X-19isotope = atomic mass:19.1 amu, abundance: 11.01%
Average atomic mass of X = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100
Average atomic mass = 1358.628 + 181 +210.291 / 100
Average atomic mass = 1749.919 / 100
Average atomic mass = 17.5 amu.
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,
Answer:
n₂ =1.4 mol
Explanation:
Given data:
Mass of nitrogen = 2 g
Initial Volume occupy by nitrogen = 1.25 L
Final volume occupy by nitrogen = 25.0 L
Final number of moles = ?
Solution;
Formula:
V₁ / n₁ = V₂ / n₂
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 2 g/ 28 g/mol
Number of moles = 0.07 mol
Now we will put the values in formula:
V₁ / n₁ = V₂ / n₂
n₂ = V₂× n₁ /V₁
n₂ = 25 L × 0.07 mol / 1.25 L
n₂ = 1.75 L. mol / 1.25 L
n₂ =1.4 mol