Answer:
1.
Explanation:
Hello,
In this case, for the given reaction we first assign the oxidation state for each species:

Whereas the half reactions are:

Next, we exchange the transferred electrons:

Afterwards, we add them to obtain:

By adding and subtracting common terms we obtain:

Finally, by removing the oxidation states we have:

Therefore, the smallest whole-number coefficient for Sn is 1.
Regards.
Answer:
C) In[reactant] vs. time
Explanation:
For a first order reaction the integrated rate law equation is:

where A(0) = initial concentration of the reactant
A = concentration after time 't'
k = rate constant
Taking ln on both sides gives:
![ln[A] = ln[A]_{0}-kt](https://tex.z-dn.net/?f=ln%5BA%5D%20%3D%20ln%5BA%5D_%7B0%7D-kt)
Therefore a plot of ln[A] vs t should give a straight line with a slope = -k
Hence, ln[reactant] vs time should be plotted for a first order reaction.
Look at the liter man it’s a great way to learn how much every liquid measurement is
Answer:
protons : 10
electron : 10
neutron : 10
Explanation:
Protons will usually be the same as the electrons when its a <u>Atom</u> (when its a ion or covalent bond or simple bond they will most likely be different)
the atomic number represents protons and electrons
the mass number - the atomic number = neutron
Answer:
i think it was lll,lV ㄸㄸㅇㅊㅎㅍㅋ