Maybe to not get rained on.
Hahhahahaha I ain't sure tho
Answer:
Nothing
Explanation:
I would say this because the parents are picking the baby's traits and they make all the decisions. And i guess the doctor would have to follow the traits, if i'm right i tried
C magnetic repulsion, forgive me if I’m wrong
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
![\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }](https://tex.z-dn.net/?f=%5Ctext%7BMoles%7D%20%3D%20%5Ctext%7B1.5%20L%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1.0%20mol%7D%7D%7B%5Ctext%7B1%20L%7D%7D%20%3D%20%5Ctext%7B1.5%20mol%20%7D)
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
![\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }](https://tex.z-dn.net/?f=%5Ctext%7BMoles%7D%20%3D%20%5Ctext%7B1.5%20L%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B0.22%20mol%7D%7D%7B%5Ctext%7B1%20L%7D%7D%20%3D%20%5Ctext%7B0.33%20mol%20%7D)
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed
![\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20O%7D_%7B2%7D%20%3D%20%5Ctext%7B1.33%20mol%20H%24_%7B2%7D%24O%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20O%7D_%7B2%7D%7D%7B%5Ctext%7B2%20mol%20H%24_%7B2%7D%24O%7D_%7B2%7D%7D%20%3D%20%5Ctextbf%7B0.58%20mol%20O%7D_%7B2%7D%5C%5C%5C%5C%5Ctext%7BThe%20amount%20of%20oxygen%20formed%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.58%20mol%7D%7D%24%7D)
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)
The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)
ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa