Answer:
Answer is in image see it..
Answer:
the frequency of photons 
Explanation:
Given: first ionization energy of 1000 kJ/mol.
No. of moles of sulfur = 1 mole

We know that plank's constant

Let the frequency of photons be ν
Also we know that ΔE = hν
this implies ν = ΔE/h


Hence, the frequency of photons 
If you change the subscripts it would change the reactants or products and then you would be solving a different formula, you would change what the chemical is
Answer:

Explanation:
Hello!
In this case, since the energy involved during a heating process is shown below:

Whereas the specific heat of water is 4.184 J/(g°C), we can compute the heated mass of water by the addition of 11.9 kJ (11900 J) of heat as shown below:

Thus, by plugging in, we obtain:

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