Answer:
H^+(aq) + OH^-(aq) —> H2O(l)
Explanation:
We'll begin by writing the balanced equation for the reaction.
2HCl(aq) + Ca(OH)2(aq) —> CaCl2(aq) + 2H2O(l)
Ca(OH)2 is a strong base and will dissociates as follow:
Ca(OH)2(aq) —> Ca^2+(aq) + 2OH^-(aq)
HCl is a strong acid and will dissociates as follow:
HCl(aq) —> H^+(aq) + Cl^-(aq)
Thus, In solution a double displacement reaction occurs as shown below:
2H^+(aq) + 2Cl^-(aq) + Ca^2+(aq) + 2OH^-(aq) —> Ca^2+(aq) + 2Cl^-(aq) + 2H2O(l)
To get the net ionic equation, cancel out Ca^2+ and 2Cl^-
2H^+(aq) + 2OH^-(aq) —> 2H2O(l)
H^+(aq) + OH^-(aq) —> H2O(l)
So for the first one It's 784.56 J (GPE)
For the second one it's KE is equal to 100 J. If you can give me a few more minutes I'll try to get the GPE
Answer:
hi
Explanation:
i am smart but i need this app cuz some are realy hard
Answer: 
at the temperature of the experiment is 0.56.
Explanation:
Moles of 
= 0.35 mole
Moles of 
= 0.40 mole
Volume of solution = 1.00 L
Initial concentration of = 
Initial concentration of = 
Equilibrium concentration of = 
The given balanced equilibrium reaction is,
Initial conc. 0.35 M 0.40 M 0 M 0M
At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M
Given: (0.35-x) = 0.19
x= 0.16 M
The expression for equilibrium constant for this reaction will be,
![K_{eq}=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BCO_2%5D%5Ctimes%20%5BH_2%5D%7D%7B%5BCO%5D%5Ctimes%20%5BH_2O%5D%7D)
Now put all the given values in this expression, we get :
Thus at the temperature of the experiment is 0.56.
PI3, but make sure it is a subscript of 3, not a large number.
