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Misha Larkins [42]
2 years ago
9

The half-life of cesium-137 is 30 years. Suppose we have a 150 mg sample. (a) Find the mass (in mg) that remains after t years.

Chemistry
1 answer:
11Alexandr11 [23.1K]2 years ago
4 0

The half-life of cesium-137 is 30 years. Suppose we have a 150 mg sample. The masses (in mg) that remains after t years A=150/2^t/30yrs

<h3>what do you mean by half-life?</h3>

A substance's half-life is the amount of time it takes for half of it to decompose.

<h3>What is a half-life example?</h3>

Half-life is the length of time it takes for half of an unstable nucleus to go through its decay process. A radioactive element's half-life decay time varies depending on the element. For instance, carbon-10 has a half-life of only 19 seconds, making it impossible to discover in nature. On the other hand, uranium-233 has a half-life of almost 160000 years.

When n half-lives have passed, the formula for estimating the amount still left is:-

A=A°/2^n

where,

A=initial amount

A°=remaining amount

n=t/t_{1/2}

A=150/2^t/30yrs

Learn more about half-life here:-

brainly.com/question/28001741

#SPJ1

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nataly862011 [7]

Answer: 3.96L

Explanation:

PV = nRT

The way I like to do it, it's I get rid of whatever I do not need.

We have to do an equation for sure to make both sides equal and find our desired result.

The pressure is constant, so it would be constant in both sides, therefore, there is no point in using it.

The R is a constant used in both sides so there is no point on using it either.

We don't need to work with moles in this case, so let's forget about the moles.

Therefore we are left with only V (volume ) and T(Temperature)

Which would logically make this:

V1/T1 = V2/T2

OR

T1/V1 = T2/V2

Both of these would work, we always use whatever makes our calculations easier without complicating our lives with the algebra.

That should remind you of Charles' Law

Transform degrees into kelvins

20 + 273.15 = 293.15K

-15 + 273.15 = 258.15K

So, 4.5L/293.15K = V2/258.15K

V2 = (4.5L/293.15K) x 258.15K

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The Gateway Arch in St. Louis, MO is approximately 630 ft tall. How many U.S. dimes would be in a stack of the same
miskamm [114]

Answer:

142240

Explanation:

We are told in the question:

Height of Gateway Arch in St. Louis, MO = 630ft

We are asked, how many U.S. dimes would be in a stack of the same

height when 1 dime is 1.35 mm thick.

Step 1

Convert height in ft to mm

1 ft = 304.8 mm

630ft =

Cross Multiply

630ft × 304.8mm/1ft

= 192024 mm.

Step 2

To find how many US dimes would be in a stack of the same height

= Total thickness/ Thickness of 1 US dime

= 192024 mm/1.35mm

= 142240

Therefore, the number of dimes that would be in a stack of the same

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4 years ago
Og is the noble gas after Rn. To go from [Rn] to [Og], you must fill four subshells (s, p, d, and f) with a total of 32 electron
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Answer:

See explanation

Explanation:

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