Boyle law is a gas law stating the pressure and the volume of a gas have an inverse relationship when held at constant temperature.
<span>A ) AB is a compound . Ionic compound because the electronegativity differs a lot in both !
</span><span>
B) Same
C ) Compound . But the bond between them is covalent .
D ) AB is an ionic compound . A is the cation and B is the anion </span>
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 " .
___________________________________________________________
Answer:
the conversion factor is f= 6 mol of glucose/ mol of CO2
Explanation:
First we need to balance the equation:
C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)
the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:
f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction
f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose/ mol of CO2
f = 6 mol of CO2/ mol of glucose
for example, for 2 moles of glucose the number of moles of CO2 produced are
n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2
Answer:
a. 4,00L
b. 16,00L
c. 12,31L
Explanation:
Avogadro's law says:
a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:
<em>V₂ = 4,00L</em>
b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:
<em>V₂ = 16,00L</em>
c. 25,0g of Ne are:
25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne
<em>V₂ = 12,31L</em>
I hope it helps!
