c both of the above the oh ranges from 0 to 14 a ph of 7 is neutral
Q=cmΔT
Q1(steel)=Q2(water)
c1•m1• (t1-t) = c2•m2• (t-t2)
628•0.1•(80-t) = 4180•0.2• (t-10)
5024 - 62.8t = 836t -8360
5024 + 8360 = (836+62.8)t
<span>t = 14.9°</span>
Answer:
Therefore, 2 moles of oxygen will be required to oxidize 2.66 moles of oxygen. 72g of Aluminium will be completely oxidized by 44.8 lit of oxygen at STP.
One mole of any substance contains 6.02 x 10²³ particles
