The heat flows between the two by the cup. The cup gives the coffee heat to the person and vise versa.
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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.
There are 1.078 x 10²³ molecules
<h3>Further explanation</h3>
Given
4 dm³ = 4 L Nitrogen gas
Required
Number of molecules
Solution
Assumptions on STP (1 atm, 273 K), 1 mol gas = 22.4 L, so for 4 L :
mol = 4 : 22.4
mol = 0.179
1 mol = 6.02 x 10²³ particles(molecules, atoms)
For 0.179 :
= 0.179 x 6.02 x 10²³
= 1.078 x 10²³
