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timurjin [86]
3 years ago
6

g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each

of the following reactions: a. The yellow precipitate formed in the reaction between KI and Pb(NO3)2 is . b. The white precipitate formed in the reaction between BaCl2 and H2SO4 is . c. The brown precipitate formed in the reaction between NaOH and FeCl3 is . d. The blue precipitate formed in the reaction between CuSO4 and NaOH is .
Chemistry
1 answer:
RUDIKE [14]3 years ago
6 0

<u>Answer:</u>

<u>For a:</u> Lead iodide is a yellow precipitate.

<u>For b:</u> Barium sulfate is a white precipitate.

<u>For c:</u> Ferric hydroxide is a brown precipitate.

<u>For d:</u> Copper (II) hydroxide is a blue precipitate.

<u>Explanation:</u>

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given options:

  • <u>For (a):</u>

The chemical reaction between KI and lead (II) nitrate follows:

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)

The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.

  • <u>For b:</u>

The chemical reaction between barium chloride and sulfuric acid follows:

BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)

The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.

  • <u>For c:</u>

The chemical reaction between NaOH and ferric chloride follows:

3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)

The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.

  • <u>For d:</u>

The chemical reaction between NaOH and copper sulfate follows:

CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4

The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.

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3 years ago
The equilibrium constant Kc for the reaction below is 0.00584 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentra
Vinil7 [7]

Answer:

Explanation:

Given that:

The chemical equation for the reaction is:

             Br2(g)    ⇌  2Br(g)

Initially  0.0345M  0.0416M

Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016

Q_C =0.05016 >>> K_c(0.00584)

Thus, the given reaction will proceed in the  backward direction

 The I.C.E table is as follows:

                       Br2(g)    ⇌  2Br(g)

I               0.0345                 0.0416

C                 +x                        -2x

E             (0.0345+x)            (0.0416 -2x)

K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584

= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)

= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x

= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908  - 0.17184x + 4x²

Solving by using Quadratic formula

x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] =  (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] =  (0.0416 -2(0.0126)) = 0.0164 M

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Bonds formed between atoms can be classified as ionic and covalent

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In H2S, the S atom is bonded to 2 H atoms. The electronegativity of H = 2.2 and S= 2.56. Since the difference is not high the bond formed will be covalent (polar covalent).

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What is the solution to the problem to the correct number of significant figures (102,900/12)+(170•1.27)
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As per as the Multiplication rules of the significant figures, whenever any numbers in the decimals forms are multiplied or divided then result in mentioned in such a way so that the significant figures after the decimal will be same as that in the given least condition.


_______________________________


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170 × 1.27 = 215.9


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= 8790.9


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