Answer:-3463 kJ and -3452kJ
Explanation:
ΔU is the change in internal energy of a system and its formula is;
ΔU = q + w
Where q represents heat transferred into or out of the system. Its value is positive when heat is transfer into the system and negative when heat is produced by the system.
W represents the work done on or by the system. Its value is positive when work is done on the system and negative when it is done by the system.
For the system in this question, we see that it produces heat which means heat is transferred out of the system, therefore the value of q is negative, it can also be seen that work is done by the system which means that w is also negative.
Therefore,
ΔU = -q-w
ΔU = -3452 kJ – 11kJ
= - 3463kJ
ΔH is the change in the enthalpy of a system and its formuls is;
ΔH = ΔU + Δ(PV)
By product rule Δ(PV) becomes ΔPV + PΔV
At constant pressure ΔP = 0. Therefore,
ΔH = -q-w + PΔV
w is equals to PΔV, So:
ΔH = -q
ΔH = -3452kJ
Answer is: the energy of exactly one photon of this light is 4.75·10⁻¹⁹ J.
Photon energy equation: E = h·ν.
E - energy of one photon.
ν- frequency.
h - Planck's constant.
ν = 7.17·10¹⁴ Hz.
h = 6.63·10⁻³⁴ J·s.
E = 6.63·10⁻³⁴ J·s · 7.17·10¹⁴ Hz.
E = 4.75·10⁻¹⁹ J.
I think answer should be b. Please give me brainlest I hope this helps let me know if it’s correct or not okay thanks
Iron rusts when in contact with water.
