Answer: It increases the reaction rate.
Step-by-step explanation:
The particles gain kinetic energy.
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.
Answer:
35.9 ml
Explanation:
Start with the balanced equation:
3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)
This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-
∴ 1 mole CuCl2 will react with 2/3 moles Na3PO4
We know that concentration = moles/volume i.e:
c= n/v
∴n=c×v
∴nCuCl2=0.107×91.01000=9.737×10−3
I divided by 1000 to convert ml to L
∴nNa3PO4=9.737×10−3×23=6.491×10−3
v=nc=6.491×10−30.181=35.86×10−3L
∴v=35.86ml
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