Scope of quantifier (LOGIC DISCRETE MATH) I am finding it quiet understand the scope of a quantifier, and some other concepts of logic. 
For example the below the scenarios below: 
Scenario 1) (∀y H(y) → ∃z W(z, y)) → ∃z G(z)
Question: 
Is the  "y" in the above formula being scoped by universal quantifier ∀y ?
Scenario 2) ∃y(∀y H(y) → ∃z W(z, y)) → ∃z G(z)
Question: 
Can you have 2 quantifiers, quantifying the same variable i.e."y" for scenario 2 ?
Scenario 3) (∀y H(y) → ∃z G(z)) → ∃z W(z, y)
Question :
Is the  "y" in the "∃z W(z, y)" of the above formula considered to be free ?
Scenario 4) (∀z H(y) → ∃y W(z, y)) → ∃z G(z)
Question: 
What would be the scope for the scenario above ? 
Scenario 5) (∀y H(y) → ∃z∃y W(z, y)) → ∃z G(z)
Question: 
What would be the scope for the scenario above ?
Scenario 6) (∀y H(y) → ∃z W(z, y)) → ∃x G(z)
Question: 
What would be the scope for the scenario above ?  
Detail for each question will be highly welcomed. 
Thanks in advance. 
 A: It's often a matter of parentheses:
In a formula like $\forall y P(y)$ the $y$ in $P(y)$ is within the scope of the $\forall y$, but in a formula like $\forall y Q(x) \land P(y)$ it is not, since this formula would be parsed as being a conjunction, whose left conjunct is $\forall y Q(x)$, and right conjunct being $P(y)$, i.e. they get 'separated'.  If you want the $y$ to be within the scope, you'd need to use parentheses: $\forall y (Q(x) \land P(y))$, and indeed now you are dealing with a universal statement.
Let me use colors to indicate the scope in each of your sentences:
Scenario 1) $(\color{red}{∀y }H(\color{red}{y}) → \color{blue}{∃z} W(\color{blue}{z}, y)) → \color{green}{∃z} G(\color{green}{z})$
Question: 
Is the  "y" in the above formula being scoped by universal quantifier ∀y ?
=> first $y$ yes, second one not
Scenario 2) $\color{red}{∃y}(\color{blue}{∀y} H(\color{blue}{y}) → \color{green}{∃z} W(\color{green}{z}, \color{red}{y})) → \color{yellow}{∃z} G(\color{yellow}{z})$
Question: 
Can you have 2 quantifiers, quantifying the same variable i.e."y" for scenario 2 ?
=> You can never have a variable quantified by two quantifiers ... if a variable falls within the scope of multiple quantifiers for the variable, then it is only quantified by the most 'inside' one
Scenario 3) $(\color{red}{∀y} H(\color{red}{y}) → \color{blue}{∃z} G(\color{blue}{z})) → \color{green}{∃z} W(\color{green}{z}, y)$
Question :
Is the  "y" in the "∃z W(z, y)" of the above formula considered to be free ?
=> Correct
Scenario 4) $(\color{red}{∀z} H(y) → \color{blue}{∃y} W(z, \color{blue}{y})) → \color{green}{∃z} G(\color{green}{z})$
Question: 
What would be the scope for the scenario above ? 
=> in all cases, the quantifiers merely quantify the atomic formula right after it. Thus, both the first $y$ and the second $z$ are free. 
Scenario 5) $(\color{red}{∀y} H(\color{red}{y}) → \color{blue}{∃z}\color{green}{∃y} W(\color{blue}{z}, \color{green}{y})) → \color{yellow}{∃z} G(\color{yellow}{z})$
=> Again, in all cases, the quantifiers merely quantify the atomic formula right after it.
Question: 
What would be the scope for the scenario above ?
Scenario 6) $(\color{red}{∀y} H(\color{red}{y}) → \color{blue}{∃z} W(\color{blue}{z}, y)) → \color{green}{∃x} G(\color{green}{z})$
Question: 
What would be the scope for the scenario above ?  
=> And once again, in all cases, the quantifiers merely quantify the atomic formula right after it. So the second $y$ is free.
