We need to increase the concentration of common ion first, in order to promote the common ion effect
<h3>What is the Common ion effect?</h3>
It is an effect that suppresses the dissociation of salt due to the addition of another salt having common ions.
For example, a saturated solution of silver chloride in equilibrium has Ag⁺ and Cl⁻ . Sodium Chloride is added to the solution and has a common ion Cl⁻. As a result, the equilibrium shifts to the left to form more silver chloride. Thus, solubility of AgCl decreases.
The Equilibrium law states that if a process is in equilibrium and is subjected to a change
- in temperature,
- pressure,
- the concentration of reactant or product,
then the equilibrium shifts in a particular direction, according to the condition.
Thus, an increase in the concentration of common ion promotes the common ion effect.
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It uses the voltages and sound freq. in the air to measure the wave lengthths
Answer: V = 33.9 L
Explanation: We will use Charles Law to solve for the new volume.
Charles Law is expressed in the following formula. Temperatures must be converted in Kelvin.
V1 / T1 = V2 / T2 then derive for V2
V2 = V1 T2 / T1
= 35 L ( 308 K ) / 318 K
= 33.9 L
3Mg + N₂ = Mg₃N₂
Mg₃N₂
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This is a straightforward question related to the surface energy of the droplet.
<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>
<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>
<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>
<span>The five smaller droplets need to have the same volume as the original. Therefore </span>
<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>
<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>
<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>
<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>
<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
