Answer:
1.
Explanation:
Hello,
In this case, for the given reaction we first assign the oxidation state for each species:
![Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-](https://tex.z-dn.net/?f=Sn%5E0%20%2B%20H%5E%2BN%5E%7B5%2B%7DO%5E%7B-2%7D_3%20%5Crightarrow%20Sn%5E%7B4%2B%7DO_2%20%2B%20N%5E%7B4%2B%7DO%5E%7B2-%7D_2%20%2B%20H%5E%2B_2O%5E-)
Whereas the half reactions are:
![Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O](https://tex.z-dn.net/?f=Sn%5E0%2B2H_2O%20%5Crightarrow%20Sn%5E%7B4%2B%7DO_2%20%2B4H%5E%2B%2B4e%5E-%5C%5C%5C%5CH%5E%2B%2BH%5E%2BN%5E%7B5%2B%7DO%5E%7B-2%7D_3%20%2B1e%5E-%5Crightarrow%20%20N%5E%7B4%2B%7DO%5E%7B2-%7D_2%2BH_2O)
Next, we exchange the transferred electrons:
![1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O](https://tex.z-dn.net/?f=1%5Ctimes%28Sn%5E0%2B2H_2O%20%5Crightarrow%20Sn%5E%7B4%2B%7DO_2%20%2B4H%5E%2B%2B4e%5E-%29%5C%5C%5C%5C4%5Ctimes%20%28H%5E%2B%2BH%5E%2BN%5E%7B5%2B%7DO%5E%7B-2%7D_3%20%2B1e%5E-%5Crightarrow%20%20N%5E%7B4%2B%7DO%5E%7B2-%7D_2%2BH_2O%29%5C%5C%5C%5C%5C%5CSn%5E0%2B2H_2O%20%5Crightarrow%20Sn%5E%7B4%2B%7DO_2%20%2B4H%5E%2B%2B4e%5E-%5C%5C%5C%5C4H%5E%2B%2B4H%5E%2BN%5E%7B5%2B%7DO%5E%7B-2%7D_3%20%2B4e%5E-%5Crightarrow%20%204N%5E%7B4%2B%7DO%5E%7B2-%7D_2%2B4H_2O)
Afterwards, we add them to obtain:
![Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O](https://tex.z-dn.net/?f=Sn%5E0%2B2H_2O%2B4H%5E%2B%2B4H%5E%2BN%5E%7B5%2B%7DO%5E%7B-2%7D_3%20%20%5Crightarrow%20Sn%5E%7B4%2B%7DO_2%20%2B4H%5E%2B%2B4N%5E%7B4%2B%7DO%5E%7B2-%7D_2%2B4H_2O)
By adding and subtracting common terms we obtain:
![Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O](https://tex.z-dn.net/?f=Sn%5E0%2B4H%5E%2BN%5E%7B5%2B%7DO%5E%7B-2%7D_3%20%20%5Crightarrow%20Sn%5E%7B4%2B%7DO_2%20%2B4N%5E%7B4%2B%7DO%5E%7B2-%7D_2%2B2H_2O)
Finally, by removing the oxidation states we have:
![Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O](https://tex.z-dn.net/?f=Sn%20%2B%204HNO_3%20%5Crightarrow%20SnO_2%20%2B%204NO_2%20%2B%202H_2O)
Therefore, the smallest whole-number coefficient for Sn is 1.
Regards.
This would be c as for the amswer
Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.
We have the following equation representing the half-life decay:
![A=A_o\times2^{(-\frac{t}{t_{half}})_{}_{}}](https://tex.z-dn.net/?f=A%3DA_o%5Ctimes2%5E%7B%28-%5Cfrac%7Bt%7D%7Bt_%7Bhalf%7D%7D%29_%7B%7D_%7B%7D%7D)
A is the resulting amount after t time
Ao is the initial amount = 50 mg
t= Elapsed time
t half is the half-life of the substance = 14.3 days
We replace the know values into the equation to have an exponential decay function for a 50mg sample
![A=\text{ 50 }\times2^{\frac{-t}{14.3}}](https://tex.z-dn.net/?f=A%3D%5Ctext%7B%2050%20%7D%5Ctimes2%5E%7B%5Cfrac%7B-t%7D%7B14.3%7D%7D)
That would be the answer for a)
To know the P-32 remaining after 84 days we have to replace this value in the equation:
![\begin{gathered} A=\text{ 50 }\times2^{\frac{-84}{14.3}} \\ A=0.85\text{ mg} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D%5Ctext%7B%2050%20%7D%5Ctimes2%5E%7B%5Cfrac%7B-84%7D%7B14.3%7D%7D%20%5C%5C%20A%3D0.85%5Ctext%7B%20mg%7D%20%5Cend%7Bgathered%7D)
So, after 84 days the P-32 remaining will be 0.85 mg
Answer:
There are approximately 200 amino acids.
Explanation:
Proteins are usually larger than peptides. In addition, proteins contains amino acids which are about 50 and more than while the amino acids in peptides are just between 2 and 50. On the other hand, growth hormone which is also known as somatotropin is a type of peptide cell regeneration, growth, and reproduction. The number of amino acids in the growth hormones are approximately 200.
200 volts of current will flow through the conductor.