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RioNii RioNii
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3 years ago Edited: 3 years ago, RioNii
Given the following two reactions:

(1)Ca(s) + Cl2(g)CaCl2(s)       H(1) = -795.8 kJ

(2)Zn(s) + Cl2(g)ZnCl2(s)       H(2) = -415.1 kJ

Calculate the enthalpy change for the following reaction:

(3) CaCl2(s) + Zn(s) Ca(s) + ZnCl2(s)
Post Merge: 3 years ago

Sir can you also help me with this one please
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#1
Educator
3 years ago
795.8 + (-415.1) = 380.7
RioNii Author
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#2
3 years ago
Thank you very much again sir!
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