Answer:
Explanation:
first, you calculate the amount of O2 in moles:
98.0 ÷ 32 = 3.0625
second, the ratio if O2/C3H8 is 5 so you need to calculate O2 in moles with that:
3.0625 ÷ 5 = 0.6125
third, the amount of CO2 in moles also can be calculate by the ratio of C3H8/CO2 which is 3
0.6125 × 3 = 1.8375
then multiply CO2 in moles by its molar mass which is 44 g/mol
1.8375 × 44 = 80.85g
Chlorine is a halogen and is very reactive and unstable. If released in an elemental form (Cl2), it would react with other substances immediately. However, <span>chlorofluorocarbons (CFCs) which contain chlorine are unreactive and when released they eventually end up in the upper atmosphere still "intact". In the upper atmosphere, sunlight is more intense and is able to break apart CFC, releasing the highly reactive chlorine which in turns destroys ozone which is more abundant in the upper atmosphere (stratosphere). </span>
Answer:
Radium-226 is a radioactive decay product in the uranium-238 decay series and is the precursor of radon-222. Radium-228 is a radioactive decay product in the thorium-232 decay series. Both isotopes give rise to many additional short-lived radionuclides, resulting in a wide spectrum of alpha, beta and gamma radiations.
Answer:
I don't think so
Explanation:
The equation doesn't look balanced
Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
