Hello!
Use the following equation to solve for the average acceleration of the motorcycle:
Plug in the given final, initial velocities, and the time:
If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m
Initial molarity of Mn₂ = 0.30 M
Final molarity of Mn₂ = 4.6 x 10⁻¹¹
pH = ?
Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)
Write the ionic equation
Mn(OH)₂ → Mn⁺² + 2OH⁻
[Mn⁺²] = 4.6 x 10⁻¹¹
We will calculate the concentration of OH⁻ by using Ksp expression
Ksp = [Mn⁺²][OH-]²
[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴
[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹
[OH⁻]² = 10⁻³
[OH⁻] = (10⁻³)¹⁽²
[OH⁻] = 0.0316 M
Calculate the pOH
pOH = -log [OH⁻]
pOH = -log [0.0316]
pOH = 1.5
Now calculate pH
pH = 14 - pOH
pH = 14 - 1.5
pH = 12.5
You can also learn about molarity from the following question:
brainly.com/question/14782315
#SPJ4
The haze is called smog, hope this helps, pls mark brainliest!
The glow emitted by a substance exposed to external radiation is called 'fluorescence'. In fluorescence, a fluorophore is exposed to external radiation, absorbs the energy and emits a form of light or glow. The answer to this question is 'fluorescence'. I hope this helps.
Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ
