A standard soft drink can has a volume of 0.355 L. An empty can is sealed while
1 answer:
Answer:
0.35 atm
Explanation:
To solve this problem, we use Boyle's Law: , where P is the pressure and V is the volume.
Here, V_1 = 0.355 L, P_1 = 1.0 atm, and V_2 = 0.125 L. So, just plug these values into the equation:
(1.0) * (0.355) = * (0.125) ⇒
≈ 0.35 atm
Thus, the pressure is 0.35 atm.
Hope this helps!
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Answer:
The answer to your question is 25.9 g of KCl
Explanation:
Data
Grams of KCl = ?
Volume = 0.75 l
Molarity = 1 M
Formula
Solve for number of moles
Substitution
Number of moles = 1 x 0.75
Simplification
Number of moles = 0.75 moles
Molecular mass KCl = 39 + 35.5 = 34.5
Use proportions to find the grams of KCl
34.5 g of KCl ---------------- 1 mol
x ---------------- 0.75 moles
x = (0.75 x 34.5) / 1
x = 25.9 g of KCl
<h3>Answer:</h3>
The mass of excessive water (H₂O) is 40.815 kg
<h3>Explanation:</h3>The Equation for the reaction is;
Li₂O(s) + H₂O(l) → 2LiOH(s)
From the question;
Mass of water removed is 80.0 kg
Mass of available Li₂O is 65.0 kg
We are required to calculate the mass of excessive reagent.
<h3>Step 1: Calculating the number of moles of water to be removed</h3>Moles = Mass ÷ Molar mass
Molar mass of water = 18.02 g/mol
Mass of water = 80 kg (but 1000 g = 1kg)
= 80,000 g
Therefore;
= 4.44 × 10³ moles
<h3>Step 2: Moles of Li₂O available </h3>Moles = mass ÷ molar mass
Mass of Li₂O available = 65.0 kg or 65,000 g
Molar mass Li₂O = 29.88 g/mol
Moles of Li₂O = 65,000 g ÷ 29.88 g/mol
= 2.175 × 10³ moles Li₂O
<h3>Step 3: Mass of excess reagent </h3>From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)
1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH
The ratio of Li₂O to H₂O is 1:1
- Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
- However, the number of moles of water to be removed is 4.44 × 10³ moles but only 2.175 × 10³ moles will react with the available Li₂O.
- This means, Li₂O is the limiting reactant while water is the excessive reagent.
Therefore:
Moles of excessive water = 4.44 × 10³ moles - 2.175 × 10³ moles
= 2.265 × 10³ moles
Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol
= 4.0815 × 10⁴ g or
= 40.815 kg
Thus, the mass of excessive water is 40.815 kg
Answer : q = 6020 J, w = -6020 J, Δe = 0
Solution : Given,
Molar heat of fusion of ice = 6020 J/mole
Number of moles = 1 mole
Pressure = 1 atm
Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.
The relation between heat and molar heat of fusion is,
(in terms of mass)
or, (in terms of moles)
Now we have to calculate the value of q.
When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.
So, the value of
Now we have to calculate the value of w.
Formula used :
where, q is heat required, w is work done and is internal energy.
Now put all the given values in above formula, we get
w = -6020 J
Therefore, q = 6020 J, w = -6020 J, Δe = 0