Answer:

Explanation:
N2(g)+O2(g)⇌2NO(g), 
N2(g)+2H2(g)⇌N2H4(g), 
2H2O(g)⇌2H2(g)+O2(g), 
If we add above reaction we will get:
2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)
Equilibrium constant for Eq (1) is 
Divide Eq (1) by 2, it will become:
N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)
Equilibrium constant for Eq (2) is 

Pure uranium is radioactive. It will react with most nonmetallic elements to make compounds. When it comes into contact with air, a thin, black layer of uranium oxide will form on its surface. Uranium-235 is the only naturally occurring isotope that is fissile.
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Answer: Ionic formula will be
.
Explanation:
and
ions will form a ionic compound. Ionic compounds have both metals and non-metals.
Here
is a metal and
is a non-metal.
The net charge on any compound must be 0.
So we need 2 phosphate ions to balance the charge on
ions. Similarly we need 3 Magnesium ions to balance the charge on
ions.
Criss-crossing the charges, we will get the formula as 
Criss-crossing is shown in the image below.
7.86 is the pOH of water at this temperature of 100 degrees celsius.
Option E is the right answer.
Explanation:
Data given:
Kw = 51.3 x 
pOH = ?
we know that pure water is neutral and will have pH pf 7.
The equation for relation between Kw and H+ and OH- ion is given by:
Kw = [H+] [OH-}
here the concentration of H+ ion and OH- ion is equal
so, [H+]= [OH-]
Putting the values in the equation of Kw
pKw = -log[Kw]
pKw = -log [51.3 x
]
pKw = 12.28
since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14
Now, pOH is calculated by using the equation:
14 = pOH + pH
14- 6.14 = pOH
pOH = 7.86
Answer:
The entropy change in the environment is 3.62x10²⁶.
Explanation:
The entropy change can be calculated using the following equation:

Where:
Q: is the energy transferred = 5.0 MJ
: is the Boltzmann constant = 1.38x10⁻²³ J/K
: is the initial temperature = 1000 K
: is the final temperature = 500 K
Hence, the entropy change is:
Therefore, the entropy change in the environment is 3.62x10²⁶.
I hope it helps you!