<span>Petroleum is found in sedimentary rock.
</span><span>Petroleum was once living organisms, which were compressed due to high pressure.
</span><span>Petroleum is used to make plastics.
Petroleum is used to make petrochemicals.
It is a yellow-black liquid compound found in various geologic sites and used in numerous products, and as a fuel. </span>
Single displacement, double displacement
Answer:
B----->2 upper C l plus 2 e superscript minus right arrow 2 upper C l superscript minus.
Explanation:
I took the test and checked the answers, this is for edg2020
Answer:
The fraction of Sr-90 left after 73 years is 0.1726
Explanation:
Initial mass of the isotope = ![N_o](https://tex.z-dn.net/?f=N_o)
Time taken by the sample, t = ![t_{\frac{1}{2}}](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D)
Formula used :
![N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}](https://tex.z-dn.net/?f=N%3DN_o%5Ctimes%20e%5E%7B-%5Clambda%20t%7D%5C%5C%5C%5C%5Clambda%20%3D%5Cfrac%7B0.693%7D%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D)
where,
= initial mass of isotope
N = mass of the parent isotope left after the time, (t)
= half life of the isotope
= rate constant
We have:
![t_{\frac{1}{2}}=28.8 years](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D28.8%20years)
t = (2018 - 1945 )years = 73 years
So, on substituting the values:
![N=x\times e^{-(\frac{0.693}{28.8 years})\times 73 years}](https://tex.z-dn.net/?f=N%3Dx%5Ctimes%20e%5E%7B-%28%5Cfrac%7B0.693%7D%7B28.8%20years%7D%29%5Ctimes%2073%20years%7D)
Now put all the given values in this formula, we get
![N=N_o\times e^{-1.7566}](https://tex.z-dn.net/?f=N%3DN_o%5Ctimes%20e%5E%7B-1.7566%7D)
![N=N_o\times 0.1726](https://tex.z-dn.net/?f=N%3DN_o%5Ctimes%200.1726)
![\frac{N}{N_o}=0.1726](https://tex.z-dn.net/?f=%5Cfrac%7BN%7D%7BN_o%7D%3D0.1726)
The fraction of Sr-90 left after 73 years = 0.1726