Answer: KMnO4-
Explanation:
You're looking at one potassium plus a polyatomic ion.
So K plus MnO4, equals:
KMnO4-
It also has a molar mass of 158.04 g/mol, I don't know if you need that, but I thought it would be nice to include it.
Answer:
Enthalpy change for the reaction is -67716 J/mol.
Explanation:
Number of moles of
in 50.0 mL of 0.100 M of 
= Number of moles of HCl in 50.0 mL of 0.100 M of HCl
=
moles
= 0.00500 moles
According to balanced equation, 1 mol of
reacts with 1 mol of HCl to form 1 mol of AgCl.
So, 0.00500 moles of
react with 0.00500 moles of HCl to form 0.00500 moles of AgCl
Total volume of solution = (50.0+50.0) mL = 100.0 mL
So, mass of solution = (
) g = 100 g
Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)
=
= ![\frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}](https://tex.z-dn.net/?f=%5Cfrac%7B-100g%5Ctimes%204.18%5Cfrac%7BJ%7D%7Bg.%5E%7B0%7D%5Ctextrm%7BC%7D%7D%5Ctimes%20%5B24.21-23.40%5D%5E%7B0%7D%5Ctextrm%7BC%7D%7D%7B0.00500mol%7D)
= -67716 J/mol
[m = mass, c = specific heat capacity,
= change in temperature and negative sign is included as it is an exothermic reaction]
I would say developed technologies which somehow were lost
like the stone age: the stone hammer, who used that afterwards in the roman empire era and stuff?
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
Answer:
The pH of the solution is 5.31.
Explanation:
Let "
is the dissociation of weak acid - HCN.
The dissociation reaction of HCN is as follows.

Initial C 0 0
Equilibrium c(1-
) c
c
Dissociation constant = 

In this case weak acids
is very small so, (1-
) is taken as 1.


From the given the concentration = 0.050 M
Substitute the given value.

![[H_{3}O^{+}]=c\alpha](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3Dc%5Calpha)
![[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D0.05%5Ctimes%209.8%5Ctimes%2010%5E%7B-4%7D%3D%204.9%5Ctimes10%5E%7B-6%7D)
![pH= -log[H_{3}O^{+}]](https://tex.z-dn.net/?f=pH%3D%20-log%5BH_%7B3%7DO%5E%7B%2B%7D%5D)
![=-log[4.9\times10^{-6}]](https://tex.z-dn.net/?f=%3D-log%5B4.9%5Ctimes10%5E%7B-6%7D%5D)

Therefore, The pH of the solution is 5.31.