Answer:
net force is 100 N and direction is right
Explanation:
The given question is incomplete. The complete question is as follows.
Nitric acid is a key industrial chemical, largely used to make fertilizers and explosives. The first step in its synthesis is the oxidation of ammonia. In this reaction, gaseous ammonia reacts with dioxygen gas to produce nitrogen monoxide gas and water.
Suppose a chemical engineer studying a new catalyst for the oxidation of ammonia reaction finds that 645. liters per second of dioxygen are consumed when the reaction is run at 195.oC and 0.88 atm. Calculate the rate at which nitrogen monoxide is being produced. Give your answer in kilograms per second. Be sure your answer has the correct number of significant digits.
Explanation:
Chemical equation for the oxidation of ammonia is as follows.
Then volume of per second consumed is as follows.
V =
As this reaction is taking place at a temperature of (468.15 K) and pressure 0.88 atm. Hence, moles of consumption of are calculated as follows.
n =
=
= 14.77 mol /sec
When 5 moles of produces 4 moles of NO then the amount of NO produced from 14.77 mol
= 354.60 g/s
Therefore, NO formed per second is as follows.
= 0.35 kg/s
Thus, we can conclude that the rate at which nitrogen monoxide is being produced is 0.35 kg/s.
When the dissociation equation of NH3 is:
NH3(aq) + H2O ↔ NH4(aq) + OH-(aq)
when Kb =(1.8 x 10^-5)
and when PH+POH= 14
∴ POH = 14 -11= 3
when POH = -㏒[OH]
∴[OH] = 0.001
when [OH] = [NH4+] = 0.001
when Kb = [NH4+][OH-]/[NH4OH]
by substitution:
1.8 x 10^-5 = (0.001)^2 / [ NH4OH]
∴[NH4OH] = 0.06 M
Answer:
Mass = 2.355 g
Explanation:
Given data:
Mass of K₂O needed = ?
Mass of KNO₃ produced = 5.00 g
Solution:
Chemical equation:
K₂O + Ca(NO₃)₂ → CaO + 2KNO₃
Number of moles of KNO₃:
Number of moles = mass/molar mass
Number of moles = 5.00 g/ 101.1 g/mol
Number of moles = 0.05 mol
now we will compare the moles of KNO₃ and K₂O.
KNO₃ : K₂O
2 : 1
0.05 : 1/2×0.05 = 0.025 mol
Mass of potassium oxide needed in gram:
Mass = number of moles × molar mass
Mass = 0.025 mol × 94.2 g/mol
Mass = 2.355 g