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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.
Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺
|| Ag⁺
| Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu
⇄ Cu²⁺
+ 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺
+ 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu
⇄ Cu²⁺
+ 2e⁻
Ag⁺
+ e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺
+ 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu
+ 2Ag⁺
+ 2e⁻⇄ Cu²⁺
+ 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu
+ 2Ag⁺
⇄ Cu²⁺
+ 2Ag
Answer:
Option D is good to go!
Explanation: as per the reactivity series more reactive substances will react with the counterpart substance.The most reactive substance here is calcium while the least reactive is aluminium, the magnesium comes in between.As per their reactivity, these substances will react with oxygen.
Explanation:
Then as the electrons in the atoms fall back down, they emit electromagnetic radiation (light). The amount of light emitted at different wavelengths, called the emission spectrum, is shown for a discharge tube filled with hydrogen gas in Figure 12.6 below.