Answer:
Q = 2.60 • 
J
Explanation:
Our specific heat capacity equation is:
Q = mC∆T
Q is the energy in joules.
m is the mass of the substance.
∆T is the temperature chance.
Let's plug in what we know.
- We have 76.0 g of octane
- The specific heat capacity of octane is 2.22 J/(g•K)
- The temperature increases from 10.6º to 26.0º (a 15.4º increase)
Q = 76.0(2.22)(15.4)
Multiply.
Q = 2598.288
We want three significant figures.
Q = 2.60 •
or
Q = 2590 J
Hope this helps!
Answer:
they have differnet properties that repeat across the next period
Explanation:
Answer:
The second option
Explanation:
I took the quiz and got it correct
In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol 
.
(a) For first order reaction, rate constant and half life time are related to each other as follows:
Thus, rate constant of the reaction is 
.
(b) Rate equation for first order reaction is as follows:
![k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt_%7B1%2F2%7D%7Dlog%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D)
now, 75% of the compound is decomposed, if initial concentration ![[A_{0} ]](https://tex.z-dn.net/?f=%5BA_%7B0%7D%20%5D)
is 100 then concentration at time t ![[A_{t} ]](https://tex.z-dn.net/?f=%5BA_%7Bt%7D%20%5D)
will be 100-75=25.
Putting the values,
On rearranging,
Thus, time required for 75% decomposition is 21 min.
