Answer:
318 kPa
Explanation:
Step 1: Given data
- Initial volume (V₁): 0.375 L
- Final pressure (P₂): 95.5 kPa
- Final volume (V₂): 1.25 L
Step 2: Calculate the initial pressure of the gas
Assuming constant temperature and ideal behavior, we can calculate the initial pressure of the gas using Boyle's law.
P₁ × V₁ = P₂ × V₂
P₁ = P₂ × V₂ / V₁
P₁ = 95.5 kPa × 1.25 L / 0.375 L = 318 kPa
They are alike because they have the same numbers in a different order. One is reversing the numbers of the original set. They are different because one number is larger in value while the other is lesser in value compared to each other.
Answer:
After the reaction, there will 0.60 g of magnesium oxide and 0.25 g of oxygen gas present in the tube
Explanation:
Equation of the reaction between magnesium and oxygen is given as follows:
2Mg(s) + O₂(g) ---> 2MgO(s)
From the equation of reaction, 2 moles of magnesium reacts with i mole of oxygen gas to produce 1 mole of magnesium oxide
molar mass of magnesium is 24.0 g; molar mass of oxygen gas = 32.0 g; molar mass of magnesium oxide = 40.0 g
Therefore 24 g of magnesium reacts with 32 g of oxygen gas
I.00 g of magnesium will react with (24.0 / 32.0) * 1.00 g of oxygen = 0.75 g of oxygen gas.
Therefore, magnesium is the limiting reagent. Once it is used up, the reaction will stop and the excess oxygen will be left in the tube together with the product, magnesium oxide.
mass of excess oxygen = 1.00 - 0.75 = 0.25 g
mass of magnesium oxide formed = (24.0 / 40.0 g) * 1 = 0.60 g
