The accepted ΔH for a reaction is −42.3 kJ/mol. A student has performed a lab experiment to determine the enthalpy change of the same reaction. Calculate the percent error
Student 1 Exp. ΔH(kJ/mol)
Trial 1 –43.8
Trial 2 –41.0
Trial 3 –42.7

trial 1: % error = 3.546 %

trial 2: % error = 3.073 %

trial 3: % error = 0.946 %

Explanation:

∴ ΔH = - 42.3 KJ/mol

trial 1:

∴ ΔH1 = - 43.8 KJ/mol

⇒ % error = ((-43.8 - ( -42.3 )) / ( -42.3 ))×100

⇒ % error = (0.03546)*100 = 3.546 %

trial 2:

∴ ΔH2 = - 41.0 KJ/mol

⇒ % error = ((-42.3 - ( -41.0 )) / ( -42.3 ))×100

⇒ % error = (0.03073)*100 = 3.073%

trial 3:

∴ ΔH3 = - 42.7 KJ/mol

⇒ % error = ((-42.7 - ( -42.3 )) / ( -42.3 ))×100

⇒ % error = (9.456 E-3)*100 = 0.946 %

i think slash and

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer