Give two areas where the compressible nature of gas is applied

Equation: 3Cu(s) 8HNO3(aq) --> 2NO(g) 3Cu(NO3)2(aq) 4H2O(l) In the above reaction, the element oxidized is ______, the reduci

igomit [66]

Answer:

1. Cu

2. Cu

3. 2 electrons.

Explanation:

Step 1:

The equation for the reaction is given below:

3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)

Step 2:

Determination of the change of oxidation number of each element present.

For Cu:

Cu = 0 (ground state)

Cu(NO3)2 = 0

Cu + 2( N + 3O) = 0

Cu + 2(5 + (3 x -2)) =0

Cu + 2 (5 - 6) = 0

Cu + 2(-1) = 0

Cu - 2 = 0

Cu = 2

The oxidation number of Cu changed from 0 to +2

For N:

HNO3 = 0

H + N + 3O = 0

1 + N + (3 x - 2) = 0

1 + N - 6 = 0

N = 6 - 1

N = 5

NO = 0

N - 2 = 0

N = 2

The oxidation number of N changed from +5 to +2

The oxidation number of oxygen and hydrogen remains the same.

Note:

1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1

2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1

Step 3:

Answers to the questions given above

From the above illustration,

1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.

2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.

3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.

7 0

2 years ago

0.475 g H, 7.557 gS, 15.107 g O. Express your answer as a chemical formula.

max2010maxim [7]

Answer:

H₂SO₄

Explanation:

We have a compound formed by 0.475 g H, 7.557 g S, 15.107 g O. In order to determine the empirical formula, we have to follow a series of steps.

Step 1: Calculate the total mass of the compound

Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g

Total mass = 23.139 g

Step 2: Determine the percent composition.

H: (0.475g/23.139g) × 100% = 2.05%

S: (7.557g/23.139g) × 100% = 32.66%

O: (15.107g/23.139g) × 100% = 65.29%

Step 3: Divide each percentage by the atomic mass of the element

H: 2.05/1.01 = 2.03

S: 32.66/32.07 = 1.018

O: 65.29/16.00 = 4.081

Step 4: Divide all the numbers by the smallest one

H: 2.03/1.018 ≈ 2

S: 1.018/1.018 = 1

O: 4.081/1.018 ≈ 4

The empirical formula of the compound is H₂SO₄.

7 0

2 years ago

In the best Lewis structure for the fulminate ion, CNO–, what is the formal charge on the central nitrogen atom?

Andrews [41]

Answer : The formal charge on the central nitrogen atom is, (+1)

Explanation :

Resonance structure : Resonance structure is an alternating method or way of drawing a Lewis-dot structure for a compound.

Resonance structure is defined as any of two or more possible structures of the compound. These structures have the identical geometry but have different arrangements of the paired electrons. Thus, we can say that the resonating structure are just the way of representing the same molecule.

First we have to determine the Lewis-dot structure of $CNO^-$ .

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $CNO^-$

As we know that carbon has '4' valence electrons, nitrogen has '5' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $CNO^-$ = 4 + 5 + 6 + 1= 16

According to Lewis-dot structure, there are 8 number of bonding electrons and 8 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

For structure 1 :

$\text{Formal charge on O}=6-6-\frac{2}{2}=-1$

$\text{Formal charge on C}=4-2-\frac{6}{2}=-1$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

For structure 2 :

$\text{Formal charge on O}=6-4-\frac{4}{2}=0$

$\text{Formal charge on C}=4-4-\frac{4}{2}=-2$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

For structure 3 :

$\text{Formal charge on O}=6-2-\frac{6}{2}=+1$

$\text{Formal charge on C}=4-6-\frac{2}{2}=-3$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

The best Lewis-dot structure is, structure 1.

Thus, the formal charge on the central nitrogen atom is, (+1)

3 0

2 years ago

A Gas OCCUPIES 525ML AT A PRESSURE OF 85.0 kPa WHAT WOULD THE VOLUME OF THE GAS BE AT THE PRESSURE OF 65.0 kPa

german

Boyle's law of ideal gas: This law states that the volume of a gas is inversely proportional to its pressure at a constant temperature. Acc to this law we can write the relation of pressure and volume as:

$PV=Constant$

That means:

$P_{1}V_{1}=P_{2}V_{2}$

From that equation we can calculate Volume of gas at a certain pressure:

P₁=Initial pressure

V₁=Initial volume

P₂=Final pressure

V₂= Final volume

Here P₁, initial pressure is given as 85.0 kPa

V₁, initial volume is given as 525 mL

P₂, final pressure is 65.0 kPa

$P_{1}V_{1}=P_{2}V_{2}$

so,

V_{2}=85\times 525\div 65

=686 mL

Volume of gas will be 686 mL.

8 0

2 years ago

Read 2 more answers

A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro

Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q

$Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J$

Heat released on combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows :

Heat released on combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q'

$Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J$

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

$q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J$

Heat released on combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0

2 years ago

Give two areas where the compressible nature of gas is applied​

Give two areas where the compressible nature of gas is applied