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Gekata [30.6K]
3 years ago
9

Reema took 5ml of Lead Nitrate solution in a beaker and added approximately 4ml of Potassium Iodide solution to it. What would s

he observe?
Chemistry
2 answers:
Reptile [31]3 years ago
8 0

Answer:

The relation between lead nitrate and potassium iodide is an example of precipitation reaction.

Pb(No3)2 + 2 KI   ⟶ ⟶ PbI2 + 2KNO3 (yellow ppt)

The yellow ppt of load iodide can be recovered by filtration.

Explanation:

Brrunno [24]3 years ago
4 0

She would observe a yellowish  solid precipitate which is the lead iodide and a white solid precipitate which is the potassium nitrate.

This is because the lead nitrate solution which contains particles of lead will mix with the potassium iodide solution containing  particles of iodide. Upon mixing,the  lead particles  from the Lead nitrate solution combines with the  iodide particles from Potassium iodide and form two compounds, a yellowish solid precipitate called lead iodide and a white solid  precipitate called Potassium nitrate.

The formation of entirely two new compounds is known as the double displacement reaction and can be written in a chemical equation as  

2KI(aq.)+Pb(NO₃)₂(aq.)------>2KNO₃(aq.)+PbI ₂(s)

See similar answer here  :https://brainly.in/question/46262462

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Calculate the entropy change in the surroundings associated with this reaction occurring at 25∘C. Express the entropy change to
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Answer:

That means that if you are calculating entropy change, you must multiply the enthalpy change value by 1000. So if, say, you have an enthalpy change of -92.2 kJ mol-1, the value you must put into the equation is -92200 J mol-1

8 0
3 years ago
44.8% of a 250. mL acid solution is used in an experiment, what volume of acid (in mL) was used?
postnew [5]

Answer:

112 mL  

Explanation:

The formula for percent by volume is

\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%

If you have 250 mL of a solution that is 44.8 % v/v,

\begin{array}{rcl}44.8\, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\44.8 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{44.8 \times 250\text{ mL}}{100}\\\\ & = & \textbf{112 mL}\\\end{array}

7 0
3 years ago
Which of the following describes a solution of 1 gram of salt dissolved in 500 mL of water?
Greeley [361]

Answer:

B

Explanation:

Homogeneous= all particles are dissolved thoroughly

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8 0
3 years ago
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Explanation:

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2 years ago
What is the average yearly rate of change of carbon-14 during the first 5000 years?
erica [24]

Answer:

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year

Explanation:

Given that the mass of the carbon 14 at the start = 5 gram

At the end of 5,000 years we will have;

A = A_0 \times e^{-\lambda \times t}

Where

A = The amount of carbon 14 left

A₀ = The starting amount of carbon 14

e = Constant = 2.71828

T_{1/2} = The half life

\lambda = 0.693/T_{1/2}

t = The time elapsed = 5000 years

λ = 0.693/T_{1/2} = 0.693/5730 = 0.0001209424

Therefore;

A = 5 × e^(-0.0001209424×5000) = 2.7312 grams

Therefore, the amount of carbon 14 decayed in the 5000 years is the difference in mass between the starting amount and the amount left

The amount of carbon 14 decayed = 5 - 2.7312 = 2.2688 grams

The average yearly rate of change of carbon-14 during the first 5000 years  is therefore;

2.2688 grams/(5000 years) = 0.0004538 grams per year

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year.

5 0
3 years ago
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