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iogann1982 [59]
3 years ago
7

I'm deaf and help 6th grader

Chemistry
2 answers:
Katena32 [7]3 years ago
7 0

Answer:

first choose which ones that u think is best it may be wrong

Explanation: next time add description please so that i can see

Aleks04 [339]3 years ago
7 0
Answer:
The anterior end of an earthworm contains an organ that detects smell. Evidence of this is the way the earthworms respond to the ammonia. The earthworms backed away from the ammonia when it was waved neither anterior end, But they did not respond when the ammonia was waved near their posterior end
Your Welcome. Sorry i'm late-
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Why do you think the molarity relationship is so important to the chemist?
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There are two big advantages of using molarity to express concentration. The first advantage is that it's easy and convenient to use because the solute may be measured in grams, converted into moles, and mixed with a volume.


The second advantage is that the sum of the molar concentrations is the total molar concentration. This permits calculations of density and ionic strength

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Where did water become gas?
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What happens to the solar energy absorbed by plants during photosynthesis ?
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Photosynthesis in plants converts solar energy into chemical energy using electrons and protons from water. in plants involves a series of steps and reactions that use solar energy, water, and carbon dioxide to produce organic compounds and oxygen.

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Question 3 of 10
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Phosphorus-32 (32P) is an isotope that is commonly used for medical and biological research. Phosphorus-32 has a half-life of 14
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The activity of the sample when it was shipped from the manufacturer is 4.54 mCi

<h3>How to determine the number of half-lives that has elapsed </h3>

From the question given above, the following data were obtained:

  • Time (t) = 48 hours
  • Half-life (t½) = 14.28 days = 14.28 × 24 = 342.72 hours
  • Number of half-lives (n) =?

n = t / t½

n = 48 / 342.72

n = 0.14

<h3>How to determine the activity of the sample during shipping </h3>
  • Number of half-lives (n) = 0.14
  • Original activity (N₀) = 5.0 mCi
  • Activity remaining (N) =?

N = N₀ / 2ⁿ

N = 5 / 2^0.14

N = 4.54 mCi

Thus, the activity of the sample during shipping is 4.54 mCi

Learn more about half life:

brainly.com/question/2674699

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3 years ago
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