Answer:
Never store used oil in anything other than tanks and storage containers. Used oil may also be stored in units that are permitted to store regulated hazardous waste. Tanks and containers storing used oil do not need to be RCRA permitted, however, as long as they are labeled and in good condition.
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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Explanation:
We have to find the number of moles of N₂ that are present in a sample that has a volume of 40.0 L at STP.
STP means Standard Conditions of Temperature and Pressure. These conditions are 273.15 K and 1 atm. We know that 1 mol of N₂ will occupy 22.4 L. We can use that ratio to find the answer to our problem.
1 mol of N₂ = 22.4 L
moles of N₂ = 40.0 L * 1 mol/(22.4 L)
moles of N₂ = 1.79 mol
Answer: 1.79 moles of nitrogen are present.
Answer:
It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.
Explanation:
A → B
Initial concentration of the reactant = x
Final concentration of reactant = 10% of x = 0.1 x
Time taken by the sample, t = ?
Formula used :
where,
= initial concentration of reactant
A = concentration of reactant left after the time, (t)
= half life of the first order conversion = 56.6 hour
= rate constant
Now put all the given values in this formula, we get
t = 188.06 hour
It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.
