Answer:
Answer is given below.
Explanation:
Anode is that electrode where oxidation occurs. Cathode is that electrode where reduction occurs.
In cell representation, half cell present left to salt-bridge notation is anodic system and another half cell present right to salt-bridge notation is cathodic system.
So anode is Cu and cathode is Ag.
oxidation:
[reduction: ]
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chemical equation:
Oxidizing agent is that species which takes electron from another species. Here takes electron from Cu. Hence is the oxidizing agent.
Reducing agent is that species which gives electron to another species. Here Cu gives electron to . Hence Cu is the reducing agent.
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2. Small dogs and big dogs
That is all I can answer
The answer would be 1.25g
Since you have 10g of a substance, you would need to half your substance by doing this: 42/14=3
Once doing this you would half your substance 3 times like so:
10/2=5
5/2=2.5
2.5/2=1.25
Explanation:
Alkali metals are the metals which belong to group 1 of the periodic table. As these elements have one valence electron so they are highly reactive in nature.
Some characteristics of metals are as follows.
- Usually metals are hard in nature.
- They have shiny surface.
- Metals are malleable and ductile in nature.
- Metals are good conductors of heat and electricity.
When these metals react with water then it leads to the formation of metal hydrides.
For example,
Thus, we can conclude the following for alkali metals:
a. What is the group number? - Group 1.
Are these metals reactive? - Yes
b. Do these metals occur freely in nature? - No
c. How many electrons are in their outer shell? - 1 valence electron
d. What are the three characteristics of ALL metals?
- Usually metals are hard in nature.
- They have shiny surface.
- Metals are malleable and ductile in nature.
e. Are these metals soft or hard? - Both
f. Name the two most reactive elements in this group? - Lithium And Sodium
g. What happens when they are exposed to water? - They form hydrides.
Na₂S(aq) + CuSO₄(aq) = Na₂SO₄(aq) + CuS(s)
2Na⁺(aq) + S²⁻(aq) + Cu²⁺(aq) + SO₄⁻(aq) = 2Na⁺(aq) + SO₄²⁻(aq) + CuS(s)
S²⁻(aq) + Cu²⁺(aq) = CuS(s)
The precipitate of cupric(II) sulfide is formed.