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
To calculate the pH of a solution that has a [H3O+] of 7.22x10^-7. You would do the following
pH=-log[H3O+]
pH=-log[7.22x10^-7]
pH=?
V = \sqrt{x} 3 * R * T / MW
V = RMS velocity
R = 8.3145 J/K*mole
T = Temperature K
MW = Molecular weigh in Kg
Answer:
Correct option is
B
5 liters of CH
4
(g)NO
2
at STP
No. of molecules=
22.4
5
mol=
22.4
5
×N
A
molecules
A) 5ℊ of H
2
(g)
No. of moles=
2
5
mol=
2
5
×N
A
molecules
B) 5l of CH
4
(g)
No. of moles of CH
4
=
22.4
5
mol=
22.4
5
N
A
molecules
C) 5 mol of O
2
=5N
A
O
2
molecules
D) 5×10
23
molecules of CO
2
(g)
Molecules of 5l NO
2
(g) at STP=5l of CH
4
(g) molecules at STP
Therefore, option B is correct.
