Answer:
The answer is
<h2>0.052 g/cm³</h2>Explanation:
The density of a substance can be found by using the formula
From the question
mass = 2.5 g
volume = 48 cm³
The density is
We have the final answer as
<h3>0.052 g/cm³</h3>Hope this helps you
Calculate the mass of ammonium sulfide (nh4)2s in 3.00 l of a 0.0200 m solution. g (nh4)2s
1 answer:
Taking into account the definition of molarity, 4.08 grams of ammonium sulfide are present in 3.00 L of a 0.0200 M solution.
<h3>Definition of molarity</h3>Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.
The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:
Molarity is expressed in units .
In this case, you know:
- Molarity: 0.0200 M
- number of moles: ?
- volume: 3 L
Replacing in the definition of molarity:
Solving:
0.0200 × 3 L= number of moles
<u><em>0.06 moles= number of moles</em></u>
On the other side, the molar mass is the amount of mass that a substance contains in one mole. In this case, the molar mass of ammonium sulfide is 68 g/mole.
Then, the amount of mass contained in 0.06 moles is calculated as:
0.06 moles×= <u><em>4.08 grams</em></u>
Finally, 4.08 grams of ammonium sulfide are present in 3.00 L of a 0.0200 M solution.
Learn more about
molarity:
<u>brainly.com/question/9324116</u>
<u>brainly.com/question/10608366</u>
<u>brainly.com/question/7429224</u>
molar mass:
<u>brainly.com/question/5216907</u>
<u>brainly.com/question/11209783</u>
<u>brainly.com/question/7132033</u>
<u>brainly.com/question/17249726</u>
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<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