The A antibodies will produce A antibodies and the B antibodies will produce B antibodies
(8.314 J/molK)(310K)ln(3E5) = E-EcatE-Ecat
<span>=32504.22 J/mol = 32.504 KJ/mol (pay attention to any requirement on sig fig)
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Answer:
so according to me I think it should be a <em><u>true</u></em>
