The answer is: " 56 g CaCl₂ " .
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Explanation:
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2.0 M CaCl₂ = 2.0 mol CaCl₂ / L ;
Since: "M" = "Molarity" (measurement of concentration);
= moles of solute per L {"Liter"} of solution.
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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 .
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(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 .
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1 mol CaCl₂ = ? g ?
From the Periodic Table of Elements:
1 mol Ca = 40.08 g
1 mol Cl = <span>35.45 g .
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There are 2 atoms of Cl in " CaCl₂ " ;
→ Note the subscript, "2", in the " Cl₂ " ;
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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 .
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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 .
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The answer is: " 56 g CaCl₂ " .
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Answer:
8.70 liters
Explanation:
First we <u>convert 36.12 g of AI₂O₃ into moles</u>, using its <em>molar mass</em>:
- 36.12 g ÷ 101.96 g/mol = 0.354 mol AI₂O₃
Then we <u>convert AI₂O₃ moles into O₂ moles</u>, using the stoichiometric coefficients of the reaction:
- 0.354 mol AI₂O₃ *
= 0.531 mol O₂
We can now use the <em>PV=nRT equation</em> to <u>calculate the volume</u>, V:
- 1.4 atm * V = 0.531 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 280.0 K
X:5.8g=16:(23+1+12+3*16)
x:5.=16:84
x:=5.8* 16/84
this is approximately 1.1
According to Charles' Law the volume of an ideal gas is directly proportional to its absolute temperature in Kelvin keeping the pressure constant.
V∝ T, P is constant
where V, T and P are volume, temperature and pressure
= 
where V₁, T₁, V₂ and T₂ are initial volume, initial temperature, final volume and final temperature.
