The knowledge of plate tectonics help people living in areas prone to earthquakes and volcanic eruptions can help by knowing the chances of the earthquake or volcanic eruption
Answer:
Volcanologists use many different kinds of tools including instruments that detect and record earthquakes (seismometers and seimographs), instruments that measure ground deformation (EDM, Leveling, GPS, tilt), instruments that detect and measure volcanic gases (COSPEC), instruments that determine how much lava is moving underground (VLF, EM-31), video and still cameras, infrared cameras, satellite imagers, webcams, etc!
Explanation:
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Answer:
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓
Ksp = [2s]² . [s] → 4s³
Explanation:
Ag₂CrO₄ → 2Ag⁺ + CrO₄⁻²
Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓ Ksp
That's the expression for the precipitation equilibrium.
To determine the solubility product expression, we work with the Ksp
Ag₂CrO₄ (s) ⇄ 2Ag⁺ (aq) + CrO₄⁻² (aq) Ksp
2 s s
Look the stoichiometry is 1:2, between the salt and the silver.
Ksp = [2s]² . [s] → 4s³
Answer:
Explanation:
Given that:
From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:
Multiplying (2) with equation (4) ; we have:
From equation (1) ; multiplying (-1) with equation (1); we have:
From equation (2); multiplying (3) with equation (2); we have:
Now; Adding up equation (5), (6) & (7) ; we get:
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(According to Hess Law)
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
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