<span>187.56 g/mol
That is the answer</span>
Answer: 1, magnesium & mg
2, two energy shells
Explanation:
Use the ideal gas equation PV=nRT. You can compare before and after using P1V1/n1T1=P2V2/n2T2. Since the number of moles remains constant you can disregard moles from the equation and use pressure, volume and temp. Make sure your pressure is converted to atmospheres, your volume is in liters, and your temperature is in kelvins.
Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
I believe it would be 1.660539040 × 10−24 gram.
