The second volume : V₂= 0.922 L
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Further explanation
</h3><h3>Given
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7.03 Liters at 31 C and 111 Torr
Required
The second volume
Solution
T₁ = 31 + 273 = 304 K
P₁ = 111 torr = 0,146 atm
V₁ = 7.03 L
At STP :
P₂ = 1 atm
T₂ = 273 K
Use combine gas law :
P₁V₁/T₁ = P₂V₂/T₂
Input the value :
0.146 x 7.03 / 304 = 1 x V₂/273
V₂= 0.922 L
Answer:
Me too. What is this for? A Lab. You are missing some kind of key info bud.
Explanation:
Answer:
We will expect 4 moles of MgO to be formed (option b).
Explanation:
Step 1: The balanced equation
2Mg + O2 → 2MgO
Step 2: Data given
Number of moles of Magnesium = 4 moles
Oxygen = in excess → this means Magnesium is the limiting reactant
Magnesium will completely be consumed ( 4 moles). There will remain 0 moles.
For 2 moles of magnesium consumed, we need 1 mole of oxygen to produce 2 moles of MgO.
For 4 moles of magnesium, we need 4/2 = 2 moles of oxygen.
For 4 moles of magnesium, we will produce 4/1 = 4 moles of MgO
We will expect 4 moles of MgO to be formed (option b).
