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Nikolay [14]
3 years ago
7

the half life of a certain element is 100 days. how many half-lives will it be before only one-fourth of this element reamains

Chemistry
1 answer:
Rama09 [41]3 years ago
4 0

Answer:

200 Days

Explanation:

Usually not so good in chemistry, but this is math!

Half life means in this time half goes away.  What happens when two half lives pass?  half of what was remaining goes away.  Maybe an example will make it more clear.

Say we start with 100 grams.  After the amount of time for the half life to pass completes, we have 50, or half of the original amount.  The half life time passes again and THAT gets cut in half to 25 grams.  this is 1/4 of the original (Hey, what we're looking for.)  Just to make it clear what is happening after another half life 1/8 remains, so it goes from The starting amount to 1/2 to 1/4 to 1/8 and so on, it keeps getting cut in half.

So how many times do we have to cut 1 in half until we get to 1/4?  Twice as was shown before.  Now, two half lives for this element is what?  100+100 days.  So 200 Days.

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balu736 [363]

Answer:

The pressure is 5.62 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= ?
  • V= 5.005 L
  • n= 1.255 mol
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 273.5 K

Replacing:

P* 5.005 L= 1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K

Solving:

P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}

P= 5.62 atm

<u><em>The pressure is 5.62 atm.</em></u>

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Draw the line-bond structure of oleic acid (cis-9-octadecenoic acid), CH3(CH2)7CH=CH(CH2)7COOH, at physiological pH. You do not
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Answer:

Explanation:

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The line-bond structure of the given oleic acid in the question can be found in the attached below.

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Answer:

C) 107 g

Explanation:

Step 1: Calculate the molar mass of Ca(NO₃)₂

We can calculate the molar mass of Ca(NO₃)₂ by adding the masses of its elements.

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M = 1 × 40.08 g/mol + 2 × 14.01 g/mol + 6 × 16.00 g/mol

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Step 2: Calculate the mass corresponding to 0.650 moles of Ca(NO₃)₂

We multiply the number of moles by the molar mass.

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