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
<span>Use the Arrhenius equation. Use p1 and p2 and T1 and T2 and solve for Ea (actgivation energy) in Joules, then plug that back into the Arrhenius equation and either p1 or p2 to calculate p at 25C.</span>
Make sure that you understand what they are asking you from this question, as it can be confusing, but the solution is quite simple. They are stating that they want you to calculate the final concentration of 6.0M HCl once a dilution has been made from 2.0 mL to 500.0 mL. They have given us three values, the initial concentration, initial volume and the final volume. So, we are able to employ the following equation:
C1V1 = C2V2
(6.0M)(2.0mL) = C2(500.0mL)
Therefore, the final concentration, C2 = 0.024M.
On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ <span>0,082 L</span>·<span>atm/K</span>·<span>mol </span>· 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.
Answer:
im guessing it's the second one
