Answer : The enthalpy of the reaction = -1839.6 KJ
Solution : Given,
= -520.0 KJ/mole
= -1699.8 KJ/mole
The balanced chemical reaction is,

Formula used :


We know that the standard enthalpy of formation of the element is equal to Zero.
Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.
Now, put all the values in above formula, we get
![\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]](https://tex.z-dn.net/?f=%5CDelta%20%28H_%7Bf%7D%29_%7Breaction%7D%3D%5B2moles%5Ctimes%20%28-1699.8%20KJ%2Fmole%29%7D%2B3moles%5Ctimes%20%280%5Ctext%7B%20KJ%2Fmole%7D%7D%29%5D-%5B%283moles%5Ctimes%28-520.0KJ%2Fmole%20%7D%2B4moles%5Ctimes%280%5Ctext%7B%20KJ%2Fmole%7D%29%5D)
= (-3399.6) + (1560)
= -1839.6 KJ
Covalent and hydrogen bonds
Answer:
The pH of the solution will be 7.53.
Explanation:
Dissociation constant of KClO=
Concentration of acid in 1 l= 0.30 M
Then in 200 ml = 
The concentration of acid, HClO=[acid]= 0.006 M
Concentration of salt in 1 L = 0.20 M
Then in 300 ml = 
The concentration of acid, KClO=[salt]= 0.006 M
The pH of the solution will be given by formula :
![pH=pK_{a}^o+\log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%5Eo%2B%5Clog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
![pH=-\log[2.8\times 10^{-8}]+\frac{[0.06 M]}{[0.06 M]}](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B2.8%5Ctimes%2010%5E%7B-8%7D%5D%2B%5Cfrac%7B%5B0.06%20M%5D%7D%7B%5B0.06%20M%5D%7D)
The pH of the solution will be 7.53.
Answer: Ions may be defined as the element that contains either positive or negative charge over them. Two types of ions are cations and anions. The outermost electrons are involved in the formation of ions.
The atomic number of sulfur is 16. Its outermost electronic configuration is K=2, L= 8, M= 6. The sulfur requres two more electrons to complete its orbit and accquire -2 charge.
Explanation:
I believe the charge is positive as losing an electron suggests that the atom lost a negative charge, leaving it with more positive than negative.
I haven't taken this in a few years very sorry if i'm wrong tho.