How many liters of ammonia gas can be formed from 54.1 liters of hydrogen
1 answer:
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Explanation:
The given data is as follows.
Volume of lake = =
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake =
= mg
= kg
Flow rate of river is 50
Volume of water in 1 day =
= liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are or
Flow rate of sewage =
Volume of sewage water in 1 day = liter
Concentration of sewage = 300 mg/L
Total amount of pollutants = or
Therefore, total concentration of lake after 1 day =
= 6.8078 mg/l
= 0.2 per day
= 6.8078
Hence, =
=
= 1.234 mg/l
Hence, the remaining concentration = (6.8078 - 1.234) mg/l
= 5.6 mg/l
Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.
Answer:
1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.
2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.
Explanation:
1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).
2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ
The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:
2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).
CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g) ΔH = 2.80 kJ
The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is: