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timurjin [86]
1 year 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]1 year 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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Explanation:

It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.

             \frac{1 g}{1000 mg} or \frac{1000 mg}{1 g}

So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.

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2 years ago
What is the density ρh of hot air inside the balloon? assume that this density is uniform throughout the balloon. express the de
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Answer 1) : The density of the hot air inside the balloon can be found out by using ideal gas equation;


PV = nRT;


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here, density (ρ) = mass (m) / volume (V); substituting in the ideal gas equation we get,

ρ = mP / RT


Answer 2) ρ (hot air) = ρ (cold air) X \frac{T_{h}}{T_{c}}

Here according to the formula because T(hot air) >T(cold air),


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The relationship between the ρ (h) = ρ(c) X \frac{T_{h}}{T_{c}}

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2 years ago
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Draw the Lewis structure (including resonance structures) for diazomethane (CH2N2)(CH2N2). For each resonance structure, assign
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Answer : The Lewis-dot structure and resonating structure of CH_2N_2 is shown below.

Explanation :

Resonance structure : Resonance structure is an alternating method or way of drawing a Lewis-dot structure for a compound.

Resonance structure is defined as any of two or more possible structures of the compound. These structures have the identical geometry but have different arrangements of the paired electrons. Thus, we can say that the resonating structure are just the way of representing the same molecule.

First we have to determine the Lewis-dot structure of CH_2N_2.

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, CH_2N_2

As we know that carbon has '4' valence electrons, nitrogen has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in CH_2N_2 = 4 + 2(1) + 2(5) = 16

Now we have to determine the formal charge for each atom.

Formula for formal charge :

\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}

For structure 1 :

\text{Formal charge on H}=1-0-\frac{2}{2}=0

\text{Formal charge on H}=1-0-\frac{2}{2}=0

\text{Formal charge on C}=4-2-\frac{6}{2}=-1

For structure 2 :

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\text{Formal charge on C}=4-0-\frac{8}{2}=0

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2 years ago
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Answer:

B

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To the question, options A-D are diamagnetic but configurations with d8 are the ones that form square planar complexes

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when the volume of a gas is changed from 3.75 L to 6.52 L the temperature will change from 100k to _K
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The temperature will change from 100K to 173.87 K

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