Answer:
The answer to your question is 47.44 g of Oxygen
Explanation:
Data
mass of Ammonia = 14.4 g
mass of Oxygen = ?
Balanced chemical reaction
4NH₃ + 7O₂ ⇒ 4NO₂ + 6H₂O
Process
1.- Calculate the molar mass of Ammonia
NH₃ = 4[(1 x 14) + (3 x 1)] = 4[14 + 3] = 4[17] = 68 g
2.- Calculate the molar mass of Oxygen
O₂ = 7[16 x 2] = 7[32] = 224 g
3.- Use proportions to calculate the mass of Oxygen
68g of NH₃ --------------------- 224 g of O₂
14.4 g of NH₃ ----------------- x
x = (14.4 x 224) / 68
x = 3225.6/ 68
x = 47.44 g
V = maximum capacity of human lung = 3 liter = 3 x 0.001 m³ = 0.003 m³ (Since 1 liter = 0.001 m³)
P = pressure of oxygen = 21.1 kilo pascal = 21.1 x 1000 = 21100 Pa (since 1 kilo = 1000)
T = temperature of air = 295 K
n = number of moles of oxygen
Using the ideal gas equation
PV = n RT
inserting the above values in the equation
(21100) (0.003) = n (8.314) (295)
n = 0.026 moles
Because atomic number equals protons and electrons
8-electrons
8-protons
Protons+neutrons=mass
8+8=16 and 8+10=18
8-Neutrons for 16
10- Neutrons for 18
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Explanation:
<span>There are 5 atoms including 1 hydrogen, 1 nitrogen, and 3 oxygen atoms.</span>
