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Sunny_sXe [5.5K]
3 years ago
12

2 points

Chemistry
1 answer:
Taya2010 [7]3 years ago
7 0
Single Replacement bc Al replaces H and displaces H making it hydrogen gas H2
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Read the interview with Tom Zambrano.
NISA [10]

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3 years ago
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Molecules in all organisms are composed in part of radioactive carbon-14. Which statement best explains the presence of carbon-1
lana66690 [7]

Answer:

The correct answer is option b, that is, the radioactive isotope reacts in the same way as stable isotopes in the chemical reactions in the body.

Explanation:

The carbon-14 atoms, which are created by the cosmic rays associates with oxygen to produce carbon dioxide that plants captivate naturally and comprise it into plant fibers by the process of photosynthesis. Afterward, the plants are consumed by animals and human beings, thus, they take in carbon-14 as well.  

The ratio of usual carbon-12 to carbon-14 in the air and in all the living species at a particular period is almost the same. These radioactive isotopes react in a similar manner as stable isotopes in the chemical reactions in the body.  

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4 years ago
Which of the following isotopes is not commonly used for dating objects? a. carbon-14 b. phosphorus-32 c. potassium-40 d. rubidi
Alekssandra [29.7K]
<span> a. carbon-14 b. phosphorus-32 c. potassium-40 d. rubidium-87
d</span>
3 0
3 years ago
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I Bleached my white shirt for the first time yesterday. I thought it would get the stain out but it turned the spot yellow. Why?
Alex17521 [72]

Answer:

The base layer of the shirt ( ORIGINAL COLOR) before dyed white was  most likely yellow so the bleach lightened the colors to their original form

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5 0
4 years ago
The dipole measured for HI is 0.380 D. The bond distance is 161 pm. What is the percent ionic character of the HI bond?
Feliz [49]

Answer:

Percent ionic character of HI bond is 4.91%.

Explanation:

<h3>Given Data:</h3>

Measured Dipole = 0.380D

bond distance = d = 161pm = 1.61*10^-8 cm

<h3>Calculation:</h3>

% ionic character is determined by following equation:

% ionic= (dipole measured/dipole calculated)*100

Now,

Dipole(calc)=qd

Dipole(calc)= (1.6*10^{-19}*3*10^{9})esu  *1.61*10^{-8}cm

(In above step 3*10^8 is multiplied to convert coulomb into esu)

Dipole(calc)=7.728*10^{-18} esu*cm

As,

10^{-18}esu*cm= 1D

So,

Dipole(calc)=7.728D

Now we can % ionic character using above equation:

%ionic=(0.380D/7.728D)*100

% ionic character=4.91%

5 0
4 years ago
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