The answer is D=Making new fossil fuels, hope this helps:)
Answer:
Explanation:
Let the number of half lives be x
<u>Solve this equation to find the value of x:</u>
- 125*(1/2)ˣ = 3.90625
- (0.5)ˣ = 3.90625 / 125
- (0.5)ˣ = 0.03125
- log (0.5)ˣ = log 0.03125
- x = log 0.03125 / log 0.5
- x = 5
<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
Beryllium, Magnesium, Calcium....etc have two valence electrons
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
