When an atom combines chemically with another atom, it either gains, loses, or shares ELECTRONS.
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E = mc^2
<span>where E is the energy in joules, </span>
<span>m is the mass in kilograms, </span>
<span>and c is the speed of light. </span>
<span>E = mc^2 </span>
<span>E = (5.63 x 10^-7 kg)(3 x 10^8 m/s)^2 </span>
<span>E = 5.07 x 10^10 J </span>
The answer is "More Stable."
Answer:
The total energy change, ΔE, in kilojoules = -61.93 kJ
Explanation:
Relationship between ΔH, ΔE and work done is given by first law of thermodynamics.
ΔE = ΔH - PΔV
Where,
ΔH = Change in enthalpy
ΔE = Change in internal energy
PΔV = Work done
Given that,
ΔH = -75.0 kJ = -75000 J
P = 43.0 atm
ΔV = Final volume - initial volume
= (2.00 - 5.00) = -3.00 L
PΔV = 43 × (-3.00) = -129 L atm
1 L atm = 101.325 J
-129 L atm = 129 × 101.325 = -13071 J
So ,
ΔE = ΔH - PΔV
= (-75000 J) - ( -13071 J)
= -75000 J + 13071 J
= -61929 J
Total energy change, ΔE = -61.929 kJ
