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Viefleur [7K]
3 years ago
11

Why is there a threshold in response to some toxins?

Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
5 0
Everything is deadly at a certain amount. Sometimes you accidently incest a small amount of something that's toxic, but it doesn't really do any harm because of how little it is. The body just gets rid of it during the its normal biological processes.

There's a certain amount of the toxin that you have to reach before the body begins to really feel it's effects and begin to get rid of it (vomiting, clamminess, all that jazz).

Of course it depends on the substance in question, some things are way more dangerous in much less amounts than others (e.g. ricin vs. poinsettia flowers).

Hope this helps!
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What is the term for photosynthesis<br>​
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PLEASE HELP ME SOLVE THIS.Thank you so much!
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Answer: The coefficients for the given reaction species are 1, 6, 2, 3.

Explanation:

The given reaction equation is as follows.

Cr_{2}O^{2-}_{7} + Cl^{-} \rightarrow Cr^{3+} + Cl_{2}

Now, the two half-reactions can be written as follows.

Reduction half-reaction: Cr_{2}O^{2-}_{7} + 3e^{-} \rightarrow Cr^{3+}

This will be balanced as follows.

Cr_{2}O^{2-}_{7} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O ... (1)

Oxidation half-reaction: Cl^{-} \rightarrow Cl_{2} + 1e^{-}

This will be balanced as follows.

6Cl^{-} \rightarrow 3Cl_{2} + 6e^{-} ... (2)

Adding both equation (1) and (2) we will get the resulting equation as follows.

Cr_{2}O^{2-}_{7} + 14H^{+} + 6Cl^{-} \rightarrow 2Cr^{3+} + 3Cl_{2} + 7H_{2}O

Thus, we can conclude that coefficients for the given reaction species are 1, 6, 2, 3.

6 0
3 years ago
The percent of remaining parent isotope in a radioactive decay process is 40 percent. How many half-lives have elapsed since the
muminat

Answer: Between 1 and 2.

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

a=\frac{a_o}{2^n}        ............(1)

where,

a = amount of reactant left after n-half lives  = 40

a_o = Initial amount of the reactant  = 100

n = number of half lives

Putting in the values we get:

40=\frac{100}{2^n}  

2^n=2.5

taking log on both sides

nlog(2)=log(2.5)

n=1.32

Thus half-lives that have elapsed is between 1 and 2

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