The hydrogen ion concentration is calculated as follows
Ph + POh =14
ph= 14-9.85= 4.15
anti -H+
=anti (-4.15) = 7.07 x10^-5 moles per liter
Answer:
The answer to your question is: 1538095.2 kg of NH3
Explanation:
MW HNO3 = 63 kg
MW NO2 = 46 kg
3 NO2(g) + H2O(l)--- 2 HNO3(aq) + NO(g)
3(46) kg-------------- 2(63) kg
x --------------- 7600000 kg
x = 7600000 x 138/126 = 8323809.5 kg og NO2
MW NO = 30
2 NO(g) + O2(g)---2 NO2(g)
2(30) ------------------2(46)
x ---------------- 8323809.5 kg
x = 8323809.5 x 60/92 = 5428571.4 kg of NO
MW NH3 = 17 kg
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
4(17) -------------------- 4(30)
x ----------------------- 5428571.4
x = 5428571.4 x 34 / 120
x = 1538095.2 kg of NH3
