Answer:
4180J
Explanation:
(25.0g)(4.184J/g°C)(75°C-35.0°C)
(25.0g)(40.0°C)(4.184J/g°C)
(1.00*10³g°C)(4.184J/g°C) = 4184J
use sig figs:
4180J
I farted it tickled my butt cheeks jingled it was hot it was squishy so I did it again
Answer:
[IBr] = 0.049 M.
Explanation:
Hello there!
In this case, according to the balanced chemical reaction:
It is possible to set up the following equilibrium expression:
![K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BIBr%5D%5E2%7D%7B%5BI_2%5D%5BBr_2%5D%7D%20%3D0.0110)
Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of 
(reaction extent) would be:
Which can be solved for to obtain two possible results:
Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:
![[IBr]=2x=2*0.0249M=0.049M](https://tex.z-dn.net/?f=%5BIBr%5D%3D2x%3D2%2A0.0249M%3D0.049M)
Regards!
Answer:
The final pressure is 90.1 atm.
Explanation:
Assuming constant temperature, we can solve this problem by using <em>Boyle's Law</em>, which states:
Where in this case:
We <u>input the given data</u>:
- 159 atm * 463 L = P₂ * 817 L
And <u>solve for P₂</u>:
The final pressure is 90.1 atm.
Answer:
0.158 moles KMnO4
Explanation:
According to the Periodic Table,
K = 39.10 g/mol
Mn = 54.94 g/mol
O = 16.00 g/mol
KMnO4 = 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol) = 158.04 g/mol
25.0 grams KMnO4 1 mole
----------------------------- x -------------------------- = 0.158 moles KMnO4
158.04 grams
