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.
Answer:
<h3>
<u>A). react with acid that is added and make a base.</u></h3>
explanation:
<em>Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added.</em>
Would it be lack of water and food?
<u>Answer:</u> The time taken by the reaction is 84.5 seconds
<u>Explanation:</u>
The equation used to calculate half life for first order kinetics:
where,
= half-life of the reaction = 9.0 s
k = rate constant = ?
Putting values in above equation, we get:
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
......(1)
where,
k = rate constant = 
t = time taken for decay process = 50.7 sec
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the reactant = ?
[A] = amount left after decay process = 0.0741 M
Putting values in equation 1, we get:
![0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}](https://tex.z-dn.net/?f=0.077%3D%5Cfrac%7B2.303%7D%7B50.7%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B0.0741%7D)
![[A_o]=3.67M](https://tex.z-dn.net/?f=%5BA_o%5D%3D3.67M)
Now, calculating the time taken by using equation 1:
![[A]=0.0055M](https://tex.z-dn.net/?f=%5BA%5D%3D0.0055M)
Putting values in equation 1, we get:
Hence, the time taken by the reaction is 84.5 seconds
