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3 years ago

What is the definition of physical and chemical changes?

2 answers:

6 0

Physical change, a change that occurs physically, and can also be seen physically. But chemically, it's a change that occurs within the chemical, and most likely not be seen within the eyes, but, has a different charater afterward.

BaLLatris [955]3 years ago

3 0

A physical change in a substance doesn't change what the substance is. In a chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed. For example, if a piece of paper is cut up into small pieces it still is paper.

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Suppose you have 100 grams of radioactive plutonium-239 with a half-life of 24,000 years. how many grams of plutonium-239 will r

Alexandra [31]

To solve this problem, let us first calculate for the rate constant k using the half life formula:

t1/2 = ln 2 / k

where t1/2 = half life period = 24,000 years, therefore k is:

k = ln 2 / 24,000

k = 2.89 x 10^-5 / yr

Now we use the rate equation:

A = Ao e^(-k t)

where,

A = mass of Plutonium-239 after number of years

Ao = initial mass of Plutonium-239

t = number of years

A. t = 12,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 12,000)

A = 70.7 g

B. t = 24,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 24,000)

A = 50 g

C. t = 96,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 96,000)

5 0

3 years ago

15 POINTS PLEASE HELP What volume of water must be added to 35mL of 2.6m KCl to reduce its concentration to 1.2m? Please explain

BartSMP [9]

First, find the volume the solution needs to be diluted to in order to have the desired molarity:
You have to use the equation M₁V₁=M₂V₂ when ever dealing with dilutions.

M₁=the starting concentration of the solution (in this case 2.6M)
V₁=the starting volume of the solution (in this case 0.035L)
M₂=the concentration we want to dilute to (in this case 1.2M)
V₂=the volume of solution needed for the dilution (not given)

Explaining the reasoning behind the above equation:
MV=moles of solute (in this case KCl) because molarity is the moles of solute per Liter of solution so by multiplying the molarity by the volume you are left with the moles of solute. The moles of solute is a constant since by adding solvent (in this case water) the amount of solute does not change. That means that M₁V₁=moles of solute=M₂V₂ and that relationship will always be true in any dilution.

Solving for the above equation:
V₂=M₁V₁/M₂
V₂=(2.6M×0.035L)/1.2M
V₂=0.0758 L
That means that the solution needs to be diluted to 75.8mL to have a final concentration of 1.2M.

Second, Finding the amount of water needed to be added:
Since we know that the volume of the solution was originally 35mL and needed to be diluted to 75.8mL to reach the desired molarity, to find the amount of solvent needed to be added all you do is V₂-V₁ since the difference in the starting volume and final volume is equal to the volume of solvent added.
75.8mL-35mL=40.8mL
40.8mL of water needs to be added

I hope this helps. Let me know if anything is unclear.
Good luck on your quiz!

5 0

3 years ago

Vitamins are food.<br> True OR False

Serga [27]

Answer: Hello!

False

Explanation:

hope you do good!they are in ur food though

6 0

3 years ago

Read 2 more answers

Complete the mechanism for the base-catalyzed opening of the epoxide below by adding any missing atoms, bonds, charges, nonbondi

worty [1.4K]

Complete Question:

check the first image for complete part of the question

Answer and Explanation:

Epoxide is a three membered ring made up of two carbon atoms and one oxygen atom. Epoxides are cyclic ethers. Due to its ring size, it is highly strained and very reactive. Epoxide ring opening takes place with respect to addition of acid and base.

Ring opening of epoxide with acid:

In the presence of base, the nucleophile attacks the epoxide ring at more substituted site and inverse stereochemistry takes place.(check file 2 attached)

Ring opening of epoxide with base:

The backside attack of nucleophile takes place in less substituted site and then it undergoes protonation to form a product.

(check file 2 attached)