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seropon [69]
3 years ago
6

4 multicellular yes autotrophic It is unable to move around its environment. A. Kingdom animalia B. kingdom fungi C. kingdom pla

ntae D. kingdom protista
Chemistry
1 answer:
Alina [70]3 years ago
5 0
Hello there.
<span>
4 multicellular yes autotrophic It is unable to move around its environment.

</span><span>D. kingdom protista </span>
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A bacterium is in a nasty environment. What sort of reproduction would it use in this situation? Why? How else would it protect
lana [24]

The bacteria in nasty environment undergoes multiple fission.

<h3><u>Explanation</u>:</h3>

The bacteria is a unicellular prokaryotic organisms that are found in each and every places of the world. They can survive in extremes of temperatures and pH. They can save themselves through special processes in the extreme climates.

The bacteria undergoes multiple fission in these climates. They cover themselves up with a strong and tough capsule inside which they undergo several Binary fissions. This leads to the formation of multiple cells enclosed with a capsule.

With the return of the favourable climate, the capsule rupture and these newly formed cells come out.

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3 years ago
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Which word equation represents a neutralization reaction
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8 0
2 years ago
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Ammonia, NH3, can be made by reacting nitrogen and hydrogen and the equation is N2 + 3H2 --&gt; 2NH3 How many moles of NH3 can b
tino4ka555 [31]
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5 0
2 years ago
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)
inn [45]

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

4 0
3 years ago
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