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
So,
Formate has a resonating double bond.
In molecular orbital theory, the resonating electrons are actually delocalized and are shared between the two oxygens. So the carbon-oxygen bonds can be described as 1.5-bonds (option B). I'm not sure if option C is correct, however, because the likelihood of both delocalized electrons being in the area of one oxygen atom is less than 50%.<span />
Aluminum? It is a chemical element with the symbol Al and atomic number 13. It is a silvery-white, soft, non-magnetic and ductile metal in the boron group. By mass, aluminium is the most abundant metal in the Earth's crust and the third most abundant element
Answer:
1. How do metals and non-metals react with acids?
Ans : Non metals does not react with acids while metals react with acids and produce hydrogen gas that burns with a 'pop'sound.
2. Write and explain the chemical equation for the reaction of magnesium with sulphuric acid and aluminium with hydrochloric acid.
Magnesium + sulphuric acid = Hydrogen + salt
Mg(s) + H2SO4 (aq) MgSO 4(aq) +H2 (g)
Aluminium + Hydrochloric acid = Hydrogen + Aluminium chloride
2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)
Answer:
7. 3–ethyl–6 –methyldecane
8. 5–ethyl–2,2–dimethyl–4–propyl–4 –heptene
Explanation:
It is important to note that when naming organic compounds having two or more different substituent groups, we simply name them alphabetically.
The name of the compound given in the question above can be written as follow:
7. Obtaining the name of the compound.
Compound contains:
I. Decane.
II. 3–ethyl.
III. 6 –methyl.
Naming alphabetically, we have
3–ethyl–6 –methyldecane
8. Obtaining the name of the compound.
Compound contains:
I. 2,2–dimethyl.
II. 4–propyl.
III. 4 –heptene.
IV. 5–ethyl.
Naming alphabetically, we have
5–ethyl–2,2–dimethyl–4–propyl–4 –heptene
