Using the table below, calculate the average atomic mass for an unknown 6 points element“J” to the nearest hundredth atomic mass unit. *
Isotope Relative Abundance (%) Atomic Mass (amu)
J-6
7.59
6.02
J-7
92.41
7.02
Your answer

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

confused

confused             hewdejdnjnjsnd

i think is conclusions.i think so

c. 4.54 x 10–12 j happens to be correct