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3 years ago

Add a nail in CuSO4 Sodium, the colour of CuSO4 disappears why?

2 answers:

5 0

The iron nail displaces the Cu from the copper sulfate solution. This results in the color of CuSO4's disappearance. Copper is less reactive than iron.

Ugo [173]3 years ago

5 0

<h2>Answer and Explanation </h2>

Add a nail in CuSO4 Sodium, the color of CuSO4 disappears because iron (Fe) is more reactive than the copper (Cu). Hence, it replaces copper. The color of copper sulphate (CuSO4) is blue but when iron (Fe) replaces its place, it becomes iron sulphate (FeSO4) which is the light green color.

When we add an iron nail in copper sulfate solution (CuSO4) the reaction takes place between them which is as Fe+CuSO4->FeSO4+Cu. Due to more reactivity of iron, replacement happens and FeSO4 forms which is light green in color. Therefore, in this reaction color changes from blue to green.

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Explanation:

atom changes from a ground state to an excited state by taking on energy from its surroundings in a process called absorption. The electron absorbs the energy and jumps to a higher energy level. In the reverse process, emission, the electron returns to the ground state by releasing the extra energy it absorbed

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3 years ago

Calculate the number of moles in 136g of ammonia

Lesechka [4]

Answer:

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3 years ago

If the temperature of a gas confined to such a cylinder is raised from 287°C to 587°C, what is the ratio of the initial volume t

jenyasd209 [6]

Answer:

1 : 2

Explanation:

From the question given, we obtained the following information:

T1 = 287°C = 287 +273 = 560K

T1 = 587°C = 587 +273 =860K

V1 = V

V2 =?

V1 /T1 = V2 /T2

V/560 = V2/860

Cross multiply

V2 x 560 = 860 x V

Divide both side by 560

V2 = (860 x V) / 560

V2 = 1.54V = 2V

The ratio of initial to final volume =

V1 : V2

= V : 2V

= 1 : 2

8 0

3 years ago

The nonvolatile, nonelectrolyte estrogen (estradiol), C18H24O2 (272.40 g/mol), is soluble in chloroform CHCl3.

Artist 52 [7]

<u>Answer:</u> The molarity of solution is 0.274 M and the osmotic pressure of the solution is 6.70 atm

<u>Explanation:</u>

To calculate the molarity of the solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of estrogen = 13.5 g

Molar mass of estrogen = 272.40 g/mol

Volume of solution = 181 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{13.5\times 1000}{272.40\times 181}\\\\\text{Molarity of solution}=0.274M$

Hence, the molarity of solution is 0.274 M

To calculate the osmotic pressure of the solution, we use the equation:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = ?

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = 0.274 M

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$\pi=1\times 0.274mol/L\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\pi=6.70atm$

Hence, the osmotic pressure of the solution is 6.70 atm

8 0

3 years ago

Ca(OH)₂(s) precipitates when a 1.0 g sample of CaC₂(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g sa

Alex

Answer:

D) Ca(OH)₂ will not precipitate because Q < Ksp

Explanation:

This is a question in which we will employ the reaction quotient Q to determine whether a precipitate will form.

Here we have first a chemical reaction in which Ca(OH)₂ is produced:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂ is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product constant for Ca(OH)₂ is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

Thus the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M = 0.002 M ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q= [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is (D)

6 0

3 years ago