Answer: 
for this reaction at this temperature is 0.029
Explanation:
Moles of 
= 2.00 mole
Volume of solution = 4.00 L
Initial concentration of 
The given balanced equilibrium reaction is,
Initial conc. 0.500 M 0 M 0 M
At eqm. conc. (0.500-2x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5Ctimes%20%5BBr_2%5D%7D%7B%5BHBr%5D%5E2%7D)
Equilibrium concentration of ![[Br_2]](https://tex.z-dn.net/?f=%5BBr_2%5D)
= x = 0.0955 M
Now put all the given values in this expression, we get :
Thus for this reaction at this temperature is 0.029
-OH is elctron donating -C=-N is electron withdrawing -O-CO-CH3 is electron withdrawing -N(CH3)2 is electron donating -C(CH3)3 is electron donating -CO-O-CH3 is electron withdrawing -CH(CH3)2 is electron donating -NO2 is electrong withdrawing -CH2
Br2 == 2Br
24% dissociated => n total moles, 0.24 mol*n of Br, and 0.76*n mol of Br2
=> partial pressure of Br, P Br = 0.24 bar, and
partical pressure of Br2, P Br2 = 0.76 bar
kp = (P Br)^2 / P Br2 = (0.24)^2 / 0.76 = 0.0758
Answer:
Explanation:
Hello!
In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:
Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:
Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:
Best regards!
Answer:
The awnser is A.
Explanation:
I got it right on edgenuity. If im wrong sorry ;-;
