The correct answers are
-formation of a precipitate
-bubble formation
-color change
-temperature change
-odor formation
The only one that isn’t correct is change in state of matter. A change in a state of matter does not mean it’s a chemical change. For example, water boiling so it turns into gas is not a chemical change, and is a physical one. Also, water can freeze and turn into ice, which is also still a physical change. If something changes state of matter, it does not necessarily mean it’s a chemical change.
Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂

(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed

The number of significant figures in 369,132,000 is 6
Explanation:
Relation between pressure, latent heat of fusion, and change in volume is as follows.

Also, 
where,
is the difference in specific volumes.
Hence, 
As,
= 22.0 J/mol K
And,
...... (1)
where,
= density of water
= density of ice
M = molar mass of water =
Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.
=
=
Therefore, calculate the required pressure as follows.

=
or, = 145 bar/K
Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP =
= 681.5 bar
Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.
If a solution is saturated, that means it already posses the maximum number of solutes thus have been dissolved in it, and thus the concentration cannot be increased.