How do you balance C3H5(NO3)3 -->CO2 + H20 + N2 + O2?
Because of the large molecule on your left hand side and the potential effects of placing a coefficient ahead of it on all those atoms, this is a little tricky to balance.
Totaling all the atoms involved is the first step.
According to the subscipts, we possess
left side
right side
I'm going to choose a component from the right side that I believe will be the key to balancing the entire equation because the molecules there are far simpler than the molecules on the left.
left side
right side
left side
right side
Now let's balance the remainder of the equation, and once more note that all of the atoms on the left side are multiplied by 2 because they are bonded to one another.
left side
Numbers C, H, and N are equivalent to 3 x 2 = 6, 5 x 2 = 10, and 9 x 2 = 18.
right side
left side
Numbers C, H, and N are equivalent to 3 x 2 = 6, 5 x 2 = 10, and 9 x 2 = 18.
right side
At this point, the equation is balanced.
But you can always multiply the WHOLE equation by two if you prefer whole number coefficients to fractions.
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The balanced equation for the given reaction is:
[ C_3H_5(NO_3)_3 \rightarrow 3CO_2 + 2H_2O + 3N_2 + 5O_2 ]
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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