Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L
At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then
PV = c
Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.
It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,
P₁V₁ = P₂V₂
Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.
In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,
1.00 atm x 3.60 L = 2.50 atm x y
We cleared y to find V₂,
V₂ = y = 
= 1.44 L
Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>
Percent strength (% w/w) of a solution is defined as the amount of solute present in 100 g of the solution.
Given data:
Mass of the solute, potassium chloride = 62.5 g
Volume of water (solution) = 187.5 ml
We know that the density of water = 1 g/ml
Therefore, the mass corresponding to the given volume of water
= 187.5 ml * 1 g/1 ml = 187.5 g
We have a solution of 62.5 g of potassium chloride in 187.5 g water
Therefore, amount of solute in 100 g of water= 62.5 * 100/187.5 = 33.33
The percentage strength = 33.33 %
<h3>
Answer:</h3>
9.6724 g MgO
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2Mg + O₂ → 2MgO
[Given] 5.8332 g Mg
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Mg = 2 mol MgO
Molar Mass of Mg - 24.31 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of MgO - 24.31 + 16.00 = 40.31 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 5 sig figs.</em>
9.67241 g MgO ≈ 9.6724 g MgO
Answer:
national
Explanation:
I don't understand, I'm sorry
