The density of the rock is 3.314g/mL
CALCULATE DENSITY:
- According to this question, a rock weighs 23.2g. After dropping the rock into a graduated cylinder containing 55mL of water, the level changes to 62mL.
- This means that the volume of the rock can be calculated as follows:
Volume of rock = 62mL - 55mL
Volume of rock = 7mL
Density can be calculated using the formula as follows:
Density = mass ÷ volume
Density = 23.2 ÷ 7
Density = 3.314g/mL
Therefore, the density of the rock is 3.314g/mL
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Following reaction is involved in above system
HOCl(aq) ↔ H+(aq) + OCl-<span>(aq)
</span>OCl-(aq) + H2O(l) ↔ HOCl(aq) + OH-<span>(aq)
</span>
Now, if the system is obeys 1st order kinetics we have
K = [OCl-][H+<span>]/[HOCl] ............. (1)
</span>∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8<span>) ............. (2)
</span>
and now considering that system is obeying 2nd order kinetics, we have
K = [OH-][HOCl-] / [OCl-] ................. (3<span>)
</span>Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8<span>)
</span>we know that, [OH-][H+] = 10<span>-14
</span>∴K = 3.3 * 10<span>-7
</span>
Thus, correct answer is e i.e none of these
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
