Answer:
2K + Br2 --> 2KBr is the answer
There would be eighteen bonding electrons
the reaction is
2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)
Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2
Given
moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062
moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012
moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019
moles of H2O = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138
Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54
Answer:
For large rivers the problem is not simply a matter of deduction of consumptive use from runoff: it is more complex and the complexity is related to the changes in .
Explanation:
1.0×10^−15/4.2×10^−7=<span>2.3809524e-23 Hoped I helped!</span>
