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Allisa [31]
3 years ago
15

The particle diagram shown above represents the dissolution of CuCl(s) assuming an equilibrium concentration for Cu+ ions of abo

ut 4×10^-4M in a saturated solution at 25°C. The equilibrium being represented is shown in the following chemical equation.
CuCl(s)⇄Cu+(aq)+Cl−(aq)

Which of the following changes to the particle diagram will best represent the effect of adding 1.0mL of 4MNaCl to the solution?

A. Some of the Cu+ and Cl− ions combine to form CuCl(s) because the Ksp will be lower than 1.6×10−7.
B. Some of the Cu+ and Cl− ions combine to form CuCl(s) because the molar solubility will be lower than 4×10−4M.
C. More Cu+ and Cl− ions will be in solution because the molar solubility will be higher than 4×10−4M.
D. More Cu+ and Cl− ions will be in solution because the Ksp will be higher than 1.6×10−7.

Chemistry
1 answer:
enot [183]3 years ago
5 0

Due to common ion effect, addition of 1.0mL of 4MNaCl  will cause some of the Cu+ and Cl− ions combine to form CuCl(s) because the Ksp will be lower than 1.6×10−7.

<h3>What is solubility product?</h3>

The term solubility product refers to the equilibrium constant that is set up when an ionic substance dissolve in water. Since the solution already contains chloride ions, addition of more chloride ions from NaCl will cause more CuCl(s) to separate from solution.

Hence, when 1.0mL of 4MNaCl is added to the solution, some of the Cu+ and Cl− ions combine to form CuCl(s) because the Ksp will be lower than 1.6×10−7.

Learn more about solubility product: brainly.com/question/857770

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Bob measured out 1.60 grams of sodium. He calculates that 1.60 g of
saul85 [17]

Answer:

84.8%

Explanation:

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Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.

Na + 1/2 Cl₂ ⇒ NaCl

According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.

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6 0
3 years ago
12. What is the frequency of a photon with an energy of 3.03 x 10-19 J?
bazaltina [42]

Answer:

u=4.57x10^5GHz

Explanation:

Hello.

In this case, given the formula:

E=h*u

Whereas E is the energy, h the Planck's constant and u the frequency of the photon. Thus, solving for it, we obtain:

u=\frac{E}{h}=\frac{3.03x10^{-19}J}{6.63x10^{-34}J*s}\\  \\u=4.57x10^{14}s^{-1}

Or also:

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