The answer is: " 56 g CaCl₂ " .
__________________________________________________________
Explanation:
__________________________________________________________
2.0 M CaCl₂ = 2.0 mol CaCl₂ / L ;
Since: "M" = "Molarity" (measurement of concentration);
= moles of solute per L {"Liter"} of solution.
__________________________________________________________
Note the exact conversion: 1000 mL = 1 L .
Given: 250 mL ;
250 mL = ? L ? ;
250 mL * (1 L / 1000 L) = (250/1000) L = 0.25 L .
___________________________________________________________
(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol = 0.50 mol CaCl₂ ;
We have: 0.50 mol CaCl₂ ; Convert to "g" (grams):
→ 0.50 mol CaCl₂ .
___________________________________________________________
1 mol CaCl₂ = ? g ?
From the Periodic Table of Elements:
1 mol Ca = 40.08 g
1 mol Cl = <span>35.45 g .
</span>
There are 2 atoms of Cl in " CaCl₂ " ;
→ Note the subscript, "2", in the " Cl₂ " ;
__________________________________________________________
So, to calculate the molar mass of "CaCl₂" :
40.08 g + 2(35.45 g) =
40.08 g + 70.90 g = 110.98 g ; round to 4 significant figures;
→ round to 111 g/mol .
__________________________________________________________
So:
→ 0.50 mol CaCl₂ = ? g CaCl₂ ? ;
→ 0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;
= (0.50) * (111 g) CaCl₂ ;
= 55.5 g CaCl₂ ;
→ round to 2 significant figures;
→ 56 g CaCl₂ .
___________________________________________________________
The answer is: " 56 g CaCl₂ " .
___________________________________________________________
The answer is (3) Cu2O. Copper (I) has an oxidation state of +1 (that's what the "I" indicates). You can also think of this as copper (I) having a charge of +1. Oxygen has an oxidation state of -2 (that's just a rule you have to know), and you can think of it as oxygen having a charge of -2. You need oxidation numbers in a neutral compound to add up to 0 (or charges in a neutral compond to add up to 0), so you need two Cu to balance the O, which is Cu2O.
Answer:
Question 1
C) polarizability
Question 2
C) London dispersion forces
Question 3:
D)Kr
Question 4:
E) strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other
Answer:
0.004522 moles of hydrogen peroxide molecules are present.
Explanation:
Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%
Mass of the solution , m= 5.125 g
Mass of the hydrogen peroxide = x



Mass of hydregn pervade in the solution = 0.15375 g
Moles of hydregn pervade in the solution :

0.004522 moles of hydrogen peroxide molecules are present.
Distance travelled will be equal to displacement when the line drawn is completely straight.
Hopefully, this helps.