1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lubasha [3.4K]
4 years ago
10

Reduction of aqueous nitrous acid hno2 to gaseous nitric oxide no in acidic aqueous solution. be sure to add physical state symb

ols where appropriate.
Chemistry
1 answer:
telo118 [61]4 years ago
5 0
Reduction is only one half of the reaction of a redox (reduction-oxidation reaction). It is characterized by the reduction of oxidation number or the gain of electrons. So, you would expect the reaction to have moles of electrons in the reactant side to depict gaining of electrons. The reduction reaction is as follows:

<em>HNO₂ + e⁻ --> NO</em>

Why only 1 e-? Compute the oxidation number of N in the reactant side.
1+x+2(-2) = 0; x = +3
Then, compute the oxidation number of N in the product side.
x -2 = 0; x = +2

So, there is a difference of 1 electron. Hence, 1e-.
You might be interested in
.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f
Tasya [4]

Answer : The value of K_c for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)

The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

Now we have to calculate the value of K_c for the given reaction.

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

K_c=\frac{(0.6)^2}{(1)\times (1)}

K_c=0.36

Therefore, the value of K_c for the given reaction is, 0.36

6 0
3 years ago
Read 2 more answers
Write the net ionic equation for the precipitation of calcium sulfide from aqueous solution
Fantom [35]

Answer:

Answer is Ca2+(aq)+S2-(aq)=>CaS(s)

Explanation:

I hope it's helpful!

7 0
3 years ago
Neptunium-237 undergoes a series of α-particle and β-particle productions to end up as thallium-205. How many α particles and β
Fantom [35]
<h3>Answer:</h3>

8 alpha particles

4 beta particles

<h3>Explanation:</h3>

<u>We are given;</u>

  • Neptunium-237
  • Thallium-205
  • Neptunium-237 undergoes beta and alpha decay to form Thallium-205.

We are required to determine the number of beta and alpha particles produced to complete the decay series.

  • We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
  • When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.

In this case;

Neptunium-237 has an atomic number 93, while,

Thallium-205 has an atomic number 81.

Therefore;

²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti

We can get x and y

237 = 4x + y(0) + 205

237-205 = 4x

4x = 32

 x = 8

On the other hand;

93 = 2x + (-y) + 81

but x = 8

93 = 16 -y + 81

y = 4

Therefore, the complete decay equation is;

²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti

Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.

5 0
3 years ago
How many moles of oxygen are required to react with methanol in order to produce 13 moles of formaldehyde?
ddd [48]

Answer:

6.5 moles of Oxygen are required

Explanation:

Based on the reaction:

CH3OH + 1/2 O2 → CH2O + H2O

1 mole of methanol reacts with 1/2 moles O2 to produce 1 mole of formaldehyde and 1 mole of water.

Thus, to produe 13 moles of formaldehyde, CH2O, are needed:

13 moles CH2O * (1/2mol O2 / 1mol CH2O) =

<h3>6.5 moles of Oxygen are required</h3>
3 0
3 years ago
Under what conditions of temperature and pressure is a gas most soluble in water?
Valentin [98]

Answer:

A gas is most soluble in water under conditions of high pressure, and low temperature.

8 0
4 years ago
Other questions:
  • Making homemade ice cream is one of life's great pleasures. fresh milk and cream, sugar, and flavorings are churned in a bucket
    15·2 answers
  • Identify the limiting reagent in the reaction mixture shown below (red = a2, blue = b2). the balanced reaction is a2+2b2→2ab2
    10·1 answer
  • Name the intermolecular force in the compound N = O
    6·1 answer
  • The contents of a Helium ballon are in which phase? A. Solid B. Plasma C. Liquid G. Gas
    14·1 answer
  • One isotope of oxygen has 8 protons and 10 neutrons. Which is the correct reference for this isotope? A. oxygen-18 B. oxygen-2 C
    7·1 answer
  • Properties of substances can be classified as physical or chemical properties. Which statements describe chemical properties? (C
    8·1 answer
  • Mole fraction of the solute in a 1.00. What is the molal aqueous eqaution?
    10·1 answer
  • If the pressure of a gas is 100 kPa when the volume is measured to be 500 mL. What pressure would need to be exerted to have the
    12·1 answer
  • Which are characteristics of natural selection? Select three options.
    8·1 answer
  • What is an acid base indicator​
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!