Answer:
Ni + Sn^2+ —> Sn + Ni^2+
Explanation:
First let us generate an elemental equation for the reaction. This is illustrated below:
Ni + Sn(NO3)2 —> Sn + Ni(NO3)2
From the equation above, a solid metal Sn is formed.
Now we can generate a net ionic equation as follows:
Ni + Sn^2+ —> Sn + Ni^2+
There are two N≡N bonds and three H–H bonds are in reactants.
Given:
The reaction between nitrogen gas and hydrogen gas.
To find:
Bonds on the reactant side
Solution:
Reactants in the reaction = 
The bond between nitrogen atoms in single 
molecule = N≡N (triple bond)
Then in two molecules = 2 N≡N (triple bonds)
The bond between hydrogen atoms in single 
molecule = H-H (single bond)
Then in three molecules = 3 H-H (single bonds)
Product in the reaction =
The bonds between nitrogen and hydrogen atoms in single molecule = 3 N-H (single bond)
Then in two molecules = 6 N-H (single bonds)
So, there are two N≡N bonds and three H–H bonds are in reactants.
Learn more about reactants and products here:
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<span>a) 7.9x10^9
b) 1.5x10^9
c) 3.9x10^4
To determine what percentage of an isotope remains after a given length of time, you can use the formula
p = 2^(-x)
where
p = percentage remaining
x = number of half lives expired.
The number of half lives expired is simply
x = t/h
where
x = number of half lives expired
t = time spent
h = length of half life.
So the overall formula becomes
p = 2^(-t/h)
And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are:
a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9
b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9
c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
