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:
Equilibrium concentrations of the gases are
Explanation:
We are given that for the equilibrium
Temperature, 
Initial concentration of
We have to find the equilibrium concentration of gases.
After certain time
2x number of moles of reactant reduced and form product
Concentration of
At equilibrium
Equilibrium constant
![K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7Bproduct%7D%7BReactant%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Substitute the values
By solving we get
Now, equilibrium concentration of gases
The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C
Answer:
Explanation:
The actual boiling point is probably between 34C and 40C.
