Answer:
Ba(OH)2 + H2SO4 ------> BaSO4 + 2H2O
1) Moles of Ba(OH)2 = moles of H2SO4 = 0.025L x 2)0.02M = 5.0 x 10^-4M
Concn of Ba(OH)2 in g/L = 5.0 x 10^-4M x 171.33g/mol = 0.086g/mol
Answer:
![[A]_0=0.400M](https://tex.z-dn.net/?f=%5BA%5D_0%3D0.400M)
Explanation:
Hello.
In this case, since the first-order reaction is said to be linearly related to the rate of reaction:
![r=-k[A]](https://tex.z-dn.net/?f=r%3D-k%5BA%5D)
Whereas [A] is the concentration of hydrogen peroxide, when writing it as a differential equation we have:
![\frac{d[A]}{dt} =-k[A]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3D-k%5BA%5D)
Which integrated is:
![ln(\frac{[A]}{[A]_0} )=-kt](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B%5BA%5D%7D%7B%5BA%5D_0%7D%20%29%3D-kt)
And we can calculate the initial concentration of the hydrogen peroxide as follows:
![[A]_0=\frac{[A]}{exp(-kt)}](https://tex.z-dn.net/?f=%5BA%5D_0%3D%5Cfrac%7B%5BA%5D%7D%7Bexp%28-kt%29%7D)
Thus, for the given data, we obtain:
![[A]_0=\frac{0.321M}{exp(-2.54x10^{-4}s^{-1}*855s)}](https://tex.z-dn.net/?f=%5BA%5D_0%3D%5Cfrac%7B0.321M%7D%7Bexp%28-2.54x10%5E%7B-4%7Ds%5E%7B-1%7D%2A855s%29%7D)
![[A]_0=0.400M](https://tex.z-dn.net/?f=%5BA%5D_0%3D0.400M)
Best regards!
Finding percent composition is fairly easy. You only need to divide the mass of an element by the total mass of the compound. We can do this one element at a time.
First, let's find the total mass by using the masses of the elements given on the periodic table.
7 x 12.011 (mass of Carbon) = 84.077
5 x 1.008 (mass of Hydrogen) = 5.04
3 x 14.007 (mass of Nitrogen) = 42.021
6 x 15.999 (mass of Oxygen) = 95.994
Add all of those pieces together.
84.077 + 5.04 + 42.021 + 95.994 = 227.132 g/mol is your total. Since we also just found the mass of each individual element, the next step will be very easy.
Carbon: 84.077 / 227.132 = 0.37016 ≈ 37.01 %
Hydrogen: 5.04 / 227.132 = 0.022189 ≈ 2.22 %
Nitrogen: 42.021 / 227.132 = 0.185 ≈ 18.5 %
Oxygen: 95.994 / 227.132 = 0.42263 ≈ 42.26 %
You can check your work by making sure they add up to 100%. The ones I just found add up to 99.99, which is close enough. A small difference (no more than 0.03 in my experience) is just a matter of where you rounded your numbers.
Answer:
Is this the full question
Explanation:
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Answer:
Approximately
.
Explanation:
Make use of the molar mass data (
) to calculate the number of moles of molecules in that
of
:
.
Make sure that the equation for this reaction is balanced.
Coefficient of
in this equation:
.
Coefficient of
in this equation:
.
In other words, for every two moles of
that this reaction consumes, two moles of
would be produced.
Equivalently, for every mole of
that this reaction consumes, one mole of
would be produced.
Hence the ratio:
.
Apply this ratio to find the number of moles of
that this reaction would have produced:
.