It's a base because it increases the concentration of hydroxid ions
Answer : Any natural sources of CFC's are not known only the major sources like aerosols, propellants, refrigerants,etc are known. So, if any natural sources are given then it cannot be called as a major source for emitting CFC into environment.
What is the empirical formula for ribose (C5H10O5)?
C. CH20
The answer is 3.
<span>The relation between number of half-lives (n) and decimal amount remaining (x) can be expressed as:
</span>

We need to calculate n, but we need x to do that. To calculate what p<span>ercentage of a radioactive species would be found as daughter material, we must calculate what amount remained:
1.28 -</span> 1.12 = 0.16
If 1.28 is 100%, how much percent is 0.16:
1.28 : 100% = 0.16 : x
x = 12.5%
Presented as decimal amount:
x = 0.125
Now, let's implement this in the equation:
<span>

</span>
Because of the exponent, we will log both sides of the equation:


<span>

</span>


Therefore, 3 half-lives have passed <span> since the sample originally formed.</span>