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.
The balanced chemical reactions are:
Further Explanation:
The following reactions will undergo double displacement where the metal cations in each compound are exchanged and form new products.
For reaction 1, the compounds involved are nitrates and chlorides. To determine the states of the products, the solubility rules for nitrates and chlorides must be followed:
- All nitrates are generally soluble.
- Chlorides are generally soluble except AgCl, PbCl2, and Hg2Cl2.
Therefore, the products will have the following characteristics:
- silver chloride (AgCl) is insoluble
- sodium nitrate (NaNO3) is soluble
For reaction 2, the compounds involved are phosphates and chlorides. The solubility rules for phosphates and chlorides are as follows:
- Phosphates are generally insoluble except for Group 1 phosphates.
- Chlorides are generally soluble except for AgCl, PbCl2, and Hg2Cl2.
Hence, the products of the second reaction will have the following characteristics:
- potassium chloride (KCl) is soluble
- magnesium phosphate is insoluble
Insoluble substances are denoted by the symbols (s) in a chemical equation. The soluble substances are denoted as <em>(aq).</em>
Learn More
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Keywords: solubility rules, precipitation reaction
The lone pairs on the oxygen in the following compound are best described as: <u>one localized and one delocalized</u>
<h3>Lone pair electron</h3>
A lone pair electron is valence shell electron pair associated with one atom, and not part of a covalent bond.
Within the context of this question, the oxygen atom of water for instance has two lone pair of which one of them is localized and other one is delocalized
Learn more about lone pair electrons:
brainly.com/question/3915115
When Ksp = [A2+] [S2-]
when A is the metal: Fe, Ni, Pb, and Cu
When we have [S2-] = 0.1 m and we have Ksp for each metal So by substitution in Ksp formula we can get [A2+] for each metal and compare its value with solution concentration 0.01 M, when we have a concentration more than 0.01 M So there are no sulfides precipitates
- [Fe2+] = Ksp/[S2-]
by substitution with Fe2+ Ksp value:
= 6x10^2 / 0.1
= 6x10^3 M
when [Fe2+] > 0.01 M
∴ no precipitate- [Ni2+] = Ksp /[S2-]
by sustitution with Ni Ksp value :
= 8x10^-1 / 0.1
= 8 M
When [Ni2+] > 0.01 M
∴ no precipitate-[Pb2+] = Ksp / [S2-]
by substitution with Pb Ksp value:
= 6x10^-7 / 0.1
= 6 x 10^-6 M
when [Pb2+] < 0.01 M
∴PbS will be precipited-[Cu2+] = Ksp / [S2-]
by substitution with Cu2+ Ksp value:
= 6x10^-16 / 0.1
= 6x10^-15 M
when [Cu2+] < 0.01 M
∴ CuS will be precipited∴The sulfides precipitates are CuS & PbS
First convert 1.26g to moles:
1.26g *(mole/169.87g) = 0.007417mol
Remember that molarity is the same as mole/L:
0.007417mol *(1/250mL) * (1000mL/L) = 0.02966M