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4 years ago

Did Pb2+ react with KI? Yes or No? State the reason of your answer.

Chemistry

2 answers:

Snowcat [4.5K]4 years ago

8 0

Answer: Yes, lead ion will react with potassium iodide.

Explanation: When lead ion reacts with KI, a bright yellow colored precipitate is formed.

The reaction of these two follows:

$Pb^{2+}(aq.)+2KI(aq.)\rightarrow PbI_2(s)+2K^+(aq.)$

Hence, $Pb^{2+}$ ion reacts with KI to yield $PbI_2$

This test is the confirmatory test for lead ion which is present in Group I.

madam [21]4 years ago

4 0

Yes, Pb2+ react with KI.

Pb2+ is lead (II) ion
KI is a compound. It is a combination of potassium (K) and iodine (I). It is called potassium iodide.

Both potassium and iodine are highly reactive elements. Thus, when they are combined with another element like Pb2+, Pb2+ reacts with KI

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3 years ago

Give an example of a homogeneous mixture and an example of a heterogeneous mixture.

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4 years ago

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A sample of seaweed contains 1 liter of water and has 50 grams of salt dissolved in its cells. The seaweed is placed in a soluti

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3 years ago

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An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in jou

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Question:

Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.

Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

Answer:

D) 85 J/K

E) - 50 J/K

F) 62.5 J/K

G) 12.5 J/K

Explanation:

Let's make use of the entropy equation: ΔS = $\frac{Q}{T}$

Part D)

Given:

T = 20°C = 20 +273 = 293K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{293}$

= 85 J/K

Part E)

Given:

T = 500K

Q = -25.0 kJ

Entropy change will be:

ΔS = $\frac{-25*1000}{500}$

= - 50 J/K

Part F)

Given:

T = 400K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{400}$

= 62.5 J/K

Part G:

Given:

T1 = 400K

T2 = 500K

Q = 25.0 kJ

The net entropy change will be:

ΔS = $(\frac{25*1000}{400}) + (\frac{-25*1000}{500}$

= 12.5 J/K

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