CHEMISTRY HIGH SCHOOL

What is the oxidation state of nitrogen in thecompound NH4Br?
(1) –1 (3) –3
(2) +2 (4) +4

Answers

Answer 1

Answer:

Answer : The oxidation number of nitrogen (N) is, (-3)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in 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, $NH_4Br$

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

$x+4(+1)+(-1)=0\n\nx+4-1=0\n\nx+3=0\n\nx=-3$

Therefore, the oxidation number of nitrogen (N) is, (-3)

Answer 2

Answer: The answer is (3) -3. We can know that typically Br has the oxidation state of -1. And the H always has the oxidation state of +1. The total formula needs to have zero oxidation state. So the nitrogen is -3.

Related Questions

Life is possible on earth because of suitable air water and temperatures true or false
What is the valency & molecular formula of sulphate?
Which describes the relationship between [H+] and [OH−] ?
The transition metal with the smallest atomic mass
Calculate the molar mass of glycerin, C3H5 (OH)3

The four basic blood types are:
A
B
D
AB
AO
O

Answers

The four basic blood types are: A, B, AB and O.

Answer:

AB, A, O and B.

Use electron transfer or electron shift to identify what is oxidized and what is reduced in each reaction :a) 2Na(s) + Br2(l) ----> 2NaBr(s)
b) H2(g) + Cl2(g) ----> 2HCl(g)
c) 2Li(s) + F2(g) ----> 2LiF(s)
d) S(s) + Cl2(g) ----> SCl2(g)
e)N2(g) + 2O2(g) ----> 2NO2(g)
f) Mg(s) +Cu(NO3)2(aq) = Mg(NO3)2(aq) + Cu(s)

For each reaction above, identify the reducing agent and the oxidizing agent

Answers

Answer :

Oxidation-reduction reaction : It is a type of reaction in which oxidation and reduction reaction occur simultaneously.

Oxidation reaction : It is the reaction in which a substance looses its electrons. In the oxidation reaction, the oxidation state of an element increases.

Reduction reaction : It is the reaction in which a substance gains electrons. In the reduction reaction, the oxidation state of an element decreases.

(a) The balanced chemical reactions is,

$2Na(s)+Br_2(l)\rightarrow 2NaBr(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Na\rightarrow Na^(1+)+1e^-$

Reduction : $Br_2+2e^-\rightarrow 2Br^(1-)$

From this we conclude that, 'Na' is oxidized and $'Br_2'$ is reduced in this reaction. The reducing agent is, 'Na' and oxidizing agent is, $'Br_2'$ .

(b) The balanced chemical reactions is,

$H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$

Half reactions of oxidation and reduction are :

Oxidation : $H_2\rightarrow H^(1+)+1e^-$

Reduction : $Cl_2+2e^-\rightarrow 2Cl^(1-)$

From this we conclude that, $'H_2'$ is oxidized and $'Cl_2'$ is reduced in this reaction. The reducing agent is, $'H_2'$ and oxidizing agent is, $'Cl_2'$ .

(c) The balanced chemical reactions is,

$2Li(s)+F_2(g)\rightarrow 2LiF(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Li\rightarrow Li^(1+)+1e^-$

Reduction : $F_2+2e^-\rightarrow 2F^(1-)$

From this we conclude that, 'Li' is oxidized and $'F_2'$ is reduced in this reaction. The reducing agent is, 'Li' and oxidizing agent is, $'F_2'$ .

(d) The balanced chemical reactions is,

$S(s)+Cl_2(g)\rightarrow SCl_2(g)$

Half reactions of oxidation and reduction are :

Oxidation : $S\rightarrow S^(2+)+2e^-$

Reduction : $Cl_2+2e^-\rightarrow 2Cl^(1-)$

From this we conclude that, 'S' is oxidized and $'Cl_2'$ is reduced in this reaction. The reducing agent is, 'S' and oxidizing agent is, $'Cl_2'$ .

(e) The balanced chemical reactions is,

$N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$

Half reactions of oxidation and reduction are :

Oxidation : $N_2\rightarrow N^(4+)+4e^-$

Reduction : $O_2+4e^-\rightarrow 2O^(2-)$

From this we conclude that, $'N_2'$ is oxidized and $'O_2'$ is reduced in this reaction. The reducing agent is, $'N_2'$ and oxidizing agent is, $'O_2'$ .

(f) The balanced chemical reactions is,

$Mg(s)+Cu(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Cu(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Mg\rightarrow Mg^(2+)+2e^-$

Reduction : $Cu^(2+)+2e^-\rightarrow Cu$

From this we conclude that, $'Mg'$ is oxidized and $'Cu'$ is reduced in this reaction. The reducing agent is, 'Mg' and oxidizing agent is, 'Cu'.

a) Na is oxidised Br is reduced
b) H is oxidised Cl is reduced
c) Li is oxidised F is reduced
d)S is oxidised Cl is reduced
e) N is oxidised O is reduced
f) Mg is oxidised and N is reduced

Remember: Oxidation= loss and Reduction= gains

A logical, systematic approach to the solution of a scientific problem1. Scientific Method
2. Scientific Law
3. Chemistry
4. Experiment

Answers

The answer is (1) Scientific Method. The Scientific Method is the method in which scientists do an experiment. It's a logical series of steps that ensures scientists don't miss anything and keep things organized.

A chemist measures 340 grams of sodium acetate. To convert this measurement to kilograms, the chemist should multiply 340 by _.a) 1 kg / 1 g
b) 1000 g / 1 kg
c) 1000 kg / 1000 g
d) 1 kg / 1000 g

Answers

Answer:

option d = 1 kg/ 1000 g

Explanation:

Given data:

Mass of sodium acetate = 340 g

Mass of sodium acetate in Kg = ?

Solution:

Gram and kilogram both are units of mass. Kilogram is larger unit while gram is smaller unit. The one kilogram equals to the 1000g. In order to convert the given value into kilogram the value must be multiply with 1 Kg/ 1000g.

340 g × 1 Kg/ 1000g

0.34 Kg

What is the molecular formula of caffeine, given that it has an empirical formula of C_4H_5N_2O and a molar mass of 194 g/mol?

Answers

Answer:

$Empirical formula : C4 H 5 N2O \n Empirical \: molar \: mass \: of \: the \n compound \: obtain : \n 12×4+5+2×14+16=97 \n \n Given \: the \: molar \: mass \: of \n Caffeine = 194.2 \n \n n= 194.2 /97=2 \n Thus \: Formula \: of \: \n Caffeine = (C 4H5 N 2 O)2$