Answer:
There is a mass of 154 Grams of Carbon Dioxide.
Explanation:
One mole is equal to 6.02 × 10^23 particles.
This means we have 1.05 X 10^24 total particles of Ethane.
Each ethane particle contains 2 carbon atoms.
If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)
Carbon Dioxide has a molar mass of 44.01 g/mol
So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.
Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.
What you should end up with is 154 Grams of Carbon Dioxide.
Hope this helps (And more importantly I hope I didn't make any errors in my math lol)
As a side note this is all assuming that this takes place at STP conditions.
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
The correct answer is b :)
Answer: C2H4 + 3O2 → 2CO2 + 2H2O
Explanation:
2Al + 2O2 → 2AlO + O2 Not Balanced Properly: 2Al + O2 = 2AlO
C2H4 + 3O2 → 2CO2 + 2H2O Looks Good
2CH4 + O2 → 2CO + 4H2 Not Correct: CO should be CO2
Ca + O2 → CaOH Not Balanced and No source for the H
i mean technically, no. only because water is water and water makes things wet. you know? unless you pour water onto water then idk honestly, truly...
