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scoray [572]
4 years ago
5

Determine the oxidation state of nitrogen in lino3.

Chemistry
2 answers:
belka [17]4 years ago
6 0

The oxidation state of N in {\text{LiN}}{{\text{O}}_3}is \boxed{ + 5}

Further Explanation:

Redox reaction:

It is a type of chemical reaction in which the oxidation states of atoms are changed. In this reaction, both reduction and oxidation are carried out at the same time. Such reactions are characterized by the transfer of electrons between the species involved in the reaction.

The process of <em>gain</em> <em>of electrons</em> or the decrease in the oxidation state of the atom is called <em>reduction</em> while that of <em>loss of electrons</em> or the increase in the oxidation number is known as <em>oxidation</em>. In redox reactions, one species lose electrons and the other species gain electrons. The species that lose electrons and itself gets oxidized is called as a reductant or reducing agent. The species that gains electrons and gets reduced is known as an oxidant or oxidizing agent. The presence of a redox pair or redox couple is a must for the redox reaction.

The general representation of a redox reaction is,

{\text{X}}+{\text{Y}}\to{{\text{X}}^+}+{{\text{Y}}^-}

The oxidation half-reaction can be written as:

{\text{X}}\to{{\text{X}}^+}+{e^-}

The reduction half-reaction can be written as:

{\text{Y}}+{e^-}\to{{\text{Y}}^-}

Here, X is getting oxidized and its oxidation state changes from 0to +1 whereas B is getting reduced and its oxidation state changes from 0 to -1. Hence, X acts as the reducing agent whereas Y is an oxidizing agent.

Rules to calculate the oxidation states of elements:

1. The oxidation number of a free element is always zero.

2. The oxidation number of oxygen is generally taken as -2, except for peroxides.

3. The oxidation state of hydrogen is normally taken as +1.

4. The sum of oxidation numbers of all the elements present in a neutral compound is zero.

5. The oxidation numbers of group 1 and group 2 elements are +1 and +2 respectively.

The oxidation state of O is -2 and the oxidation state of Li is +1.

The expression to calculate the oxidation state in {\text{LiN}}{{\text{O}}_3}is:

\left[ {1\left( {{\text{oxidation state of Li}}} \right) + 1\left( {{\text{oxidation state of N}}} \right) + 3\left( {{\text{oxidation state of O}}} \right)} \right] = 0      …… (1)

Rearrange equation (1) for the oxidation state of N.

{\text{Oxidation state of N}} = \left[ { - 1\left( {{\text{oxidation state of Li}}} \right) - 3\left( {{\text{oxidation state of O}}} \right)} \right]          …… (2)  

Substitute -2 for the oxidation {\text{state}} of O and +1 for the oxidation state of Li in equation (2).

\begin{aligned}{\text{Oxidation state}}\;{\text{of N}}&=\left[{-1\left({{\text{+1}}}\right)-3\left({-{\text{2}}}\right)}\right]\\&=\left[{-1+6}\right]\\&=+5\\\end{aligned}

Therefore the oxidation state of N in {\mathbf{LiN}}{{\mathbf{O}}_{\mathbf{3}}} is +5.

Learn more:

1. What is the oxidation state of phosphorus in {\text{PO}}_3^{3-} ion? brainly.com/question/3059040

2. Determine the oxidation state of Cl in {\text{NaCl}}{{\text{O}}_{\text{3}}}: brainly.com/question/1869644

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Redox reactions

Keywords: redox reaction, N, LiNO3, oxidation, reduction, reductant, oxidant, reducing agent, oxidizing agent, electrons, redox pair, redox couple, oxidation state, oxidized, reduced, simultaneously, +5, -2, +1, Li, O.

Andre45 [30]4 years ago
3 0

Answer : The oxidation state of nitrogen in LiNO_3 is, (+5)

Explanation :

Rules for the oxidation numbers :

  • The oxidation number of a free element is always be zero.
  • The oxidation number of a monatomic ion equal to the charge of the ion.
  • The oxidation number of  Hydrogen (H)  is +1.
  • The oxidation number of  oxygen (O)  in compounds is usually -2, but it is -1 in the peroxides.
  • The oxidation number of a Group 1 element in a compound is +1.
  • The oxidation number of a Group 2 element in a compound is +2.
  • The oxidation number of a Group 17 element in a binary compound is -1.
  • The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
  • The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

The given compound is, LiNO_3

Let the oxidation state of 'N' be, 'x'

(+1)+x+3(-2)=0\\\\x=+5

Hence, the oxidation state of N is, (+5)

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4 0
2 years ago
A tablet of Pain Be Gone Aspirin, which had a mass of 1.213 g, was pulverized and 1.159 g were dissolved in 10.0 mL of ethyl alc
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Answer:

a. Moles of NaOH  = 0.001643 moles

b. 0.296 g

c. 0.3098 g

d. Not acceptable

Explanation:

a.

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For NaOH :

Molarity = 0.1052 M

Volume = 15.62 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.62×10⁻³ L

Thus, moles of NaOH :

Moles=0.1052 \times {15.62\times 10^{-3}}\ moles

Moles of NaOH  = 0.001643 moles

b.

The reaction of NaOH with the acetylsalicylic acid is in the ratio of 1:1.

Thus, Moles of NaOH = Moles of acetylsalicylic acid = 0.001643 moles

Molar mass of acetylsalicylic acid = 180.16 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Mass = Moles * Molar mass = 0.001643 moles * 180.16 g/mol = 0.296 g

c.

1.159 g of sample contains 0.296 g of acetylsalicylic acid

1.213 g of sample contains \frac{0.296}{1.159}\times 1.213 g of acetylsalicylic acid

Mass of acetylsalicylic acid = 0.3098 g = 309.8 mg

d. Sample contains = 309.8 mg

Manufacturer claiming = 315 mg to 335 mg

Thus , it is not acceptable.

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Oxygen and hydrogen are most likely to form what type of bond?
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A wooden tool is found at an archaeological site. Estimate the age of the tool using the following information: A 100 gram sampl
lord [1]

Answer:

2578.99 years

Explanation:

Given that:

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Also,

1 g of the wood is emitting 11.20 β-particles per minute

Given, Decay rate = 15.3 % per minute per gram

So,

Concentration left can be calculated as:-

C left = [A_t]=\frac{11.20\ per\ minute}{15.3\ per\ minute\ per\ gram}\times [A_0]= 0.7320[A_0]

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Also, Half life of carbon-14 = 5730 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{5730}\ years^{-1}

The rate constant, k = 0.000120968 year⁻¹

Time =?

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

So,  

\frac {[A_t]}{[A_0]}=e^{-0.000120968\times t}

\frac {0.7320[A_0]}{[A_0]}=e^{-0.000120968\times t}

0.7320=e^{-0.000120968\times t}

ln\ 0.7320=-0.000120968\times t

<u>t = 2578.99 years</u>

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