Answer:
how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?
(A) n=m/M,
n(Al)=5.4/27=0.2 moles
n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles
Number of oxygen atoms= n(O2)*Avogadro's number
=0.15*6.02*10^23=9.03*10^22 oxgyen atoms
(B)
n=m/M
n(Al)=0.6/27=0.02222 moles
n(O2)=n(Al)*3/4=0.016666 moles
m=n*M
m(O2)=0.0166666*32=0.53333 grams
Answer : The molar concentration of sucrose in the tea is, 0.0549 M
Explanation : Given,
Mass of sucrose = 3.765 g
Volume of solution = 0.200 L
Molar mass of sucrose = 342.3 g/mole
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :

Now put all the given values in this formula, we get:

Therefore, the molar concentration of sucrose in the tea is, 0.0549 M
Answer:
Option b. Decomposition
Followed by a reduction process using charcoal
Explanation:
Lead can be obtained from lead nitrate by thermal decomposition of lead nitrate as shown below:
2Pb(NO3)2 —> 2PbO + 4NO2 + O2
The PbO obtained is reduced by charcoal(C) to obtain the metallic Pb as shown below:
2PbO + C —> Pb + CO2
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

Answer:
Density (ρ) = 5 kilogram/cubic meter
Explanation:
Steps:
ρ =
m
V
=
10 kilogram
2 cubic meter
= 5 kilogram/cubic meter