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:
C
Explanation:
polar has unequal sharing of electrons that has the lone pairs which has the electronegativity difference. can be mixed with water.
32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is
of butane reacted
Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol
= 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol
= 99.3 g of CO₂
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced
You first add the manganese and exchange the number of electrons needed with the hydroxide. While the hydroxide needs only 1 electron the manganese needs 4, so after you exchange the electrons the manganese will be just 1 atom while the hydroxide is 4. Mn(OH)4
Answer:
= 72.73%
Explanation:
The percentage by mass of an element is given by;
% element = total mass of element in compounds/molar mass of compound × 100
The mass of oxygen in carbon dioxide = 32 g
Molar mass of CO2 = 44 g
Therefore;
% of O2 = 32/44 × 100%
<u>= 72.73%</u>
