What fraction of a Sr-90 sample remains unchanged after 87.3 years

Answers

Answer 1
Answer: The answer is 1/8.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. (1/2)^(n) = x,
where:
n - a number of half-lives
x - a remained fraction of a sample

2. t_(1/2) = (t)/(n)
where:
t_(1/2) - half-life
t - total time elapsed
n - a number of half-lives

The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
t_(1/2) = 28.8 years

We need:
n = ?
x = ?

We could first use the second equation, to calculate n:
If:
t_(1/2) = (t)/(n),
Then: 
n = (t)/( t_(1/2) )
⇒ n = (87.3 years)/(28.8 years)
⇒ n=3.03
⇒ n ≈ 3

Now we can use the first equation to calculate the remained amount of the sample.
(1/2)^(n) = x
⇒ x=(1/2)^3
x= (1)/(8)

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Which of the following would not affect the rate of dissolution of salt in a beaker of water?a. stirring the solution
b. increasing the pressure of the solution
c. decreasing the size of the salt particles
d. increasing the temperature of the solution

Answers

The following which would not affect the rate of dissolution of salt in a beaker of water is:

  • B. increasing the pressure of the solution

The dissolution of salt in a beaker of water increases when an external force stirs it which makes it move from its initial position to a new position.

As a result of this, we can see that if there is a decrease in the size of salt in the beaker of water, there would also be a change in the dissolution of salt in the water.

However, increasing the pressure of the solution would not affect the rate of dissolution of salt in the beaker of water.

Therefore, the correct answer is option B

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Answer: Option (b) is the correct answer.

Explanation:

When we stir a solution then its particles start to move from their initial position leading to more number of collisions between them.

As a result, rate of dissolution increases with stirring.

When we increase the pressure then no change in dissolution occurs because in liquids and solids pressure does not play any role is solubility or dissolution.

When we decrease the size of solute particles then there occurs an increase in surface area of solute. Hence, due to this more number of solute particles are able to come in contact with the solvent.

This will lead to more number of collisions. Therefore, rate of dissolution will increase.

Also, when we increase the temperature then molecules of solution will gain kinetic energy leading to more number of collisions which will eventually lead to increase in rate of dissolution.

Thus, we can conclude that out of the given options increasing the pressure of the solution would not affect the rate of dissolution of salt in a beaker of water.

Given the balanced equation: 3fe3+(aq) + al(s) → 3fe2+(aq) + al3+(aq)what is the total number of moles of electrons lost by 2 moles of al(s)?

Answers

Answer is: the total number of moles of electrons lost by two moles of aluminium are six moles (6 mol).
Oxidation half reaction: Al° → Al³⁺ + 3e⁻, lost of electrons.
Reduction half reaction: Fe³⁺ + e⁻→ Fe²⁺ /·3; 3Fe³⁺ + 3e⁻→ 3Fe²⁺; gain of electrons.
One mole of aluminium lost 3 moles of electrons, so 2 moles of alumiunim:
2 · 3 mol = 6 mol.

Methane, CH4, is an organic compound and a hydrocarbon.
a. True
b. False

Answers

Methane (CH4) is an organic compound an a hydrocarbon. The statement above is a fact and therefore the answer to the question is letter "A" or "A.True". I hope this helps you on your assignment. 

Answer:

true

Explanation:

got it right on odyssey

Which of the following would most likely happen if water did not form hydrogen bonds?A) Water would exist only as a solid.

B) Water would be able to dissolve all ionic compounds.

C) Water would not expand when it freezes.

D) Water would not exist on Earth.

Answers

The hydrogen bond is a secondary bond formed between a hydrogen (attached to an highly electronegative element ,: F, O or N) and another atom of an highly electronegative element ,: F, O or N.

In case of water hydrogen bond helps water

a) to dissolve certain substances which can make hydrogen bond

b) it makes it liquid and room temperature unlike H2S which is a gas at room temperature due to absence of hydrogen bond

c) higher volume and low density of ice as compared to liquid water. Due to hydrogen bond ice forms open cage like structure which increases its volume and decreases its density. Thus water expands on freezing

So in absence of hydrogen bond Water would not expand when it freezes.

Water would not expand when it freezes. 

What sample is a pure substance?A. a test tube of zinc oxide.b. a container of zinc and oxygen.
c. a flask of pond water.
d. a jar of soil.

Answers

The correct answer is A. A test tube of zinc oxide. Zinc oxide is a compound which is a pure substance composed of two or more elements. They can only be separated or broken down into its individual elements or smaller compounds by chemical means. Therefore, the answer is a test tube of zinc oxide.

The sample that is a pure substance is:A. a test tube of zinc oxide.

What sample is a pure substance?

A pure substance consists of only one type of substance with uniform properties throughout. In this case, a test tube of zinc oxide would be a pure substance. Zinc oxide (ZnO) is a chemical compound made up of zinc (Zn) and oxygen (O) atoms in a fixed ratio. It has a specific chemical formula and a consistent composition, making it a pure substance.

On the other hand:

B. A container of zinc and oxygen would be a mixture since it contains two different substances, zinc and oxygen, in their elemental forms.

C. A flask of pond water is a mixture because it contains a combination of different substances, such as water, dissolved minerals, microorganisms, and organic matter.

D. A jar of soil is also a mixture since it consists of a combination of various components, including minerals, organic matter, water, air, and possibly microorganisms.

Therefore, the correct answer is A. a test tube of zinc oxide, which represents a pure substance.

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How do you find the volume occupied by 454 grams of copper if the density equals 8.92 g/mL?

Answers

V=M/P
M=454g
p=8.92
v=50.896860986
3 significant figures
v= 50.9

Final answer:

The volume occupied by 454 grams of copper, given a density of 8.92g/mL, can be calculated using the formula for density (Density = Mass / Volume). When rearranged to find volume (Volume = Mass / Density) and substituting the given values, the volume occupied is approximately 50.9 mL.

Explanation:

When calculating volume in chemistry, we use the formula for density, which is: Density = Mass / Volume. In this case, the given values are mass (454g) and density (8.92g/mL). If we rearrange the formula to solve the volume, we get: Volume = Mass / Density. So, if we substitute the given values into the equation, we obtain: Volume = 454g / 8.92g/mL. In performing the operation, we get approximately 50.9 mL. Hence 454 grams of copper occupy 50.9 mL of volume.

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