As we drive an automobile, we dont't think about the chemical consumed and produced. Prepare a list of the principal chemicals consumed and produced during the operation of an automobile.

Answers

Answer 1

Answer:

Answer:

Explanation:

In an internal combustion engine operational in automobiles, fuels are converted into mechanical energy in order to move pistons.

The basic reaction in automobile engines is combustion.

Principal chemicals consumed Chemicals produced

Petroleum Carbon dioxide

Carbon monoxide

Nitrogen oxides

Sulfur oxides

In diesel engines, production of particulate carbon is also produced.

Answer 2

Answer:

Automobiles such as car, truck, motorbikes run on petrol or diesel. While operating these automobiles, combustion of diesel or petrol takes place which in turn requires oxygen for the process to occur.

Operation of the automobiles consumes oxygen, petrol or diesel and releases harmful chemicals like carbon dioxide (CO2), sulphur dioxide (SO2), carbon monoxide (CO) and many more. These chemicals pollutes the air and also affects the survival of living organisms by affecting the respiratory organs.

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How many moles of sulfur atoms are there in 5.0 g of sulfur?

Answers

Answer:

Number of moles = 0.153 mol

Explanation:

Given data:

Mass of sulfur = 5 g

Number of moles of sulfur atom = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of sulfur is 32. 065g/mol.

By putting values,

Number of moles = 5 g/ 32.06 g/mol

Number of moles = 0.153 mol

Which of the following atoms could have an expanded octet when it is the central atom in a covalent compound?a) Bb) O
c) Cl
d) F

Answers

Answer:

only chlorine can expand its octet.

Explanation:

An atom can expand its octet is it has empty d orbital

the electronic configuration of given elements will be:

B : 1s2 2s2 2p1 [Valence shell n =2 no d orbital]

O :1s2 2s2 2p4 [Valence shell n =2 no d orbital]

F : 1s2 2s2 2p5 [Valence shell n =2 no d orbital]

Cl :1s2 2s2 2p6 3s2 3p5 3d0 [Valence shell n =2 no d orbital]

Out of given elements only chlorine has empty d orbitals in its valence shell

Thus only chlorine can expand its octet.

Aqueous solutions of sodium sulfide and copper(II) chloride are mixed together. Which statement is correct? a.Both NaCl and CuS will precipitate from solution b.No precipitate will form c.Only CuS will precipitate from solution d.Only NaCl will precipitate from solution

Answers

C.Only CuS will precipitate from solution

Aqueous solutions of sodium sulfide and copper(II) chloride are mixed together.

According to question theEquation :

Na₂S(aq) + CuCl₂(aq) → Products

The complete balanced reaction is:

Na₂S(aq) + CuCl₂(aq) → CuS(s) ↓ + 2NaCl(aq)

When the sulfide bond to the cation Cu²⁺ it makes a precipitate (s)

Thus,the correct answer is C.

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Answer:

The correct state for the answer is c.

Only CuS will precipitate from solution

Explanation:

We analyse the compounds for the reaction and we write the equation:

Na₂S(aq) + CuCl₂(aq) → Products

For this case, the products are NaCl and CuS. The complete balanced reaction is:

Na₂S(aq) + CuCl₂(aq) → CuS(s) ↓ + 2NaCl(aq)

When the sulfide bond to the cation Cu²⁺ it makes a precipitate (s)

Salts from chlorides are soluble except for the Ag⁺, Pb⁺ or Cu⁺

Salts from S⁻² which are soluble, are found in the group 2 of the Periodic Table (Ca²⁺, Ba²⁺, Mg²⁺)

The correct state for the answer is c.

Find the equilibrium constants, Kp, for the following equilibria, (i) NO(g) + ½ O2(g) ⇄ NO2(g), Kp = ? (ii) NO2(g) ⇄ NO(g) + ½ O2(g), Kp = ?, given the equilibrium constant, Kp, for the reaction: 2NO (g) + O2(g) ⇄ 2NO2(g) Kp= 100 at the same temperature

Answers

Answer :

(i) The value of equilibrium constants for this reaction is, 10

(ii) The value of equilibrium constants for this reaction is, 0.1

Explanation :

The given equilibrium reaction is,

$2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)$ $K_p=100$

Now we have to determine the equilibrium constants for the following equilibrium reactions.

(i) $NO(g)+(1)/(2)O_2(g)\rightleftharpoons NO_2(g)$ $K_p_1=?$

From the given reaction we conclude that, the reaction (i) will takes place when the given main reaction will be multiplied by half (1/2). That means when reaction will be half then the equilibrium constant will be:

$K_p_1=(K_p)^{(1)/(2)}$

$K_p_1=(100)^{(1)/(2)}$

$K_p_1=10$

The value of equilibrium constants for this reaction is, 10

(ii) $NO_2(g)\rightleftharpoons NO(g)+(1)/(2)O_2(g)$ $K_p_2=?$

From the given reaction we conclude that, the reaction (ii) will takes place when the reaction (i) will be reverse. That means when reaction will be reverse then the equilibrium constant will be:

$K_p_2=(1)/((K_p_1))$

$K_p_2=(1)/((10))$

$K_p_2=0.1$

The value of equilibrium constants for this reaction is, 0.1

6.579 rounded to nearest hundreth

Answers

Answer:

6.58

Explanation:

What will be the theoretical yield of tungsten(is) ,W, if 45.0 g of WO3 combines as completely as possible with 1.50 g of H2

Answers

Answer:

35.6 g of W, is the theoretical yield

Explanation:

This is the reaction

WO₃ + 3H₂ → 3H₂O + W

Let's determine the limiting reactant:

Mass / molar mass = moles

45 g / 231.84 g/mol = 0.194 moles

1.50 g / 2 g/mol = 0.75 moles

Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.

Let's make rules of three:

1 mol of tungsten(VI) oxide needs 3 moles of H₂

Then 0.194 moles of tungsten(VI) oxide would need (0.194 .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)

3 moles of H₂ need 1 mol of WO₃ to react

0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles

It's ok. I do not have enough WO₃.

Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.

Let's convert the moles to mass (molar mass . mol)

0.194 mol . 183.84 g/mol = 35.6 g

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