Tags Chemistry

Chemistry

Get insights from 6.9k questions on Chemistry, answered by students, alumni, and experts. You may also ask and answer any question you like about Chemistry

Follow Ask Question

6.9k

Questions

Discussions

Active Users

Followers

New answer posted

10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Sulphuric acid reacts with sodium hydroxide as follows :

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is

(i) 1 mol L^–1

(ii) 10 g

(iii) 025 mol L^–1

(iv) 55 g

0 Follower 13 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B), (C)

0.1 mole H₂SO₄ reacts with 1 mole of NaOH.

0.1 mole of NaOH will react with $\frac{0.1}{2}$ = mole of H₂SO₄

Here, NaOH is the limiting reagent.

2 mole of NaOH produces 1 mole of Na₂SO₄

0.1 mole of NaOH will give $\frac{0.1}{2}$ mole of Na₂SO₄

No. of mole = $\frac{g i v e n m a s s}{m o l a r m a s s ({N a}_{2} {S O}_{4})}$

On substituting the value in the above equation, the mass can be calculated as

0.05 mol = $\frac{g i v e n m a s s}{142 {g m o l -}^{1}}$

given mass = 7.10 g

Volume of solution after mixing is 2 L.

So, the molarity of Na₂SO₄is

Molarity = $\frac{0.05}{2}$ = 0.025mol L^-1

0 0

New answer posted

10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

In the following questions two or more options may be correct.

One mole of oxygen gas at STP is equal to _______.

(i) 6.022 * 10²³molecules of oxygen

(ii) 6.022 * 10²³atoms of oxygen

(iii) 16 g of oxygen

(iv) 32 g of oxygen

0 Follower 1 View

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A), (D)

One mole of oxygen gas at STP is equal to 6.022 x 10²³ molecules of oxygen The number of moles is given by the following formula,

Moles = $\frac{M a s s}{M o l a r m a s s}$ ----------(1)

The number of moles of O₂ is calculated by using equation (1) as follow.

Moles of O₂= $\frac{16 g}{32 g / m o l}$

= 0.5 mol

The number of molecules can be calculated as, number of moles

= $\frac{n u m b e r o f m o l e c u l e s}{A v o g a d r o s n u m b e r}$ ----------(2)

On substituting the values in the above equation:

0.5 m

...more

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A), (D)

One mole of oxygen gas at STP is equal to 6.022 x 10²³ molecules of oxygen The number of moles is given by the following formula,

Moles = $\frac{M a s s}{M o l a r m a s s}$ ----------(1)

The number of moles of O₂ is calculated by using equation (1) as follow.

Moles of O₂= $\frac{16 g}{32 g / m o l}$

= 0.5 mol

The number of molecules can be calculated as, number of moles

= $\frac{n u m b e r o f m o l e c u l e s}{A v o g a d r o s n u m b e r}$ ----------(2)

On substituting the values in the above equation:

0.5 mol = $\frac{n u m b e r o f m o l e c u l e s}{6.022 * 10^{23}}$

number of molecules = 0.5 $*$ 6.022 $*$ 10²³

The number of moles is given by the following formula,

moles = $\frac{M a s s}{M o l a r m a s s}$ ----------(1)

The number of moles of O₂ is calculated by using equation (1) as follows,

Moles of O₂ = $\frac{32 g}{32 g / m o l}$

=1 mol

The number of molecules can be calculated as, number of moles

= $\frac{n u m b e r o f m o l e c u l e s}{A v o g a d r o s n u m b e r}$ ----------(2)

On substituting the values in the above equation:

1 mol = $\frac{n u m b e r o f m o l e c u l e s}{6.022 * 10^{23}}$

Number of molecules = 1 $*$ 6.022 $*$ 10²³

less

0 0

New answer posted

10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Which of the following statements indicates that law of multiple proportion is being followed.

(i) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1:2.

(ii) Carbon forms two oxides namely CO₂ and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1.

(iii) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.

(iv) At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour.

0 Follower 1 View

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B)

Law of multiple proportion: According to the law of multiple proportion, whe form two or more than two chemical compounds, the ratio between different elements combining with a fixed mass of the other is always in the ratio of tin Compounds of carbon and oxygen:

Carbon and oxygen react to form two different compounds CO and CO2. In Carbon react with 16 parts by mass of oxygen.

In CO₂ ,12 parts by mass of Carbon react with 32 parts by mass of oxygen. If the mass of Carbon is fixed at 12 parts of mass then the ratio in the masses with the fixed mass of

...more

0 0

New answer posted

10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Which of the following reactions is not correct according to the law of conservation of mass

(i) 2Mg(s) + O₂(g) → 2MgO(s)

(ii) C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(g)

(iii) P₄(s) + 5O₂(g) → P₄O₁₀(s)

(iv) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O (g)

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B)

In a, c, and d options given reactions, all the elements are balanced and hence, following the law of conservation of mass.

In reaction (B), C₃H₈+ O₂→ CO₂+H₂0 elements are not balanced. Hence product are different. It is against the law of conservation of mass

0 0

New answer posted

10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter Two

Assertion (A) : It is impossible to determine the exact position and exact momentum of an electron simultaneously.

Reason (R) : The path of an electron in an atom is clearly defined.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true and R is not the correct explanation of A.

(iii) A is true and R is false.

(iv) Both A and R are false.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: option (iii)

According to Heisenberg's uncertainty principle it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Thus, it implies that determining the trajectory of

an electron is impossible as it requires exact position and velocity which is not possible as per the uncertainty principle.

0 0

New answer posted

10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Which of the following statements is correct about the reaction given below:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(g)

(i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass.

(ii) Total mass of reactants = total mass of product; therefore, law of multiple proportions is followed.

(iii) Amount of Fe2O3 can be increased by taking any one of the reactants (iron or oxygen) in excess.

(iv) Amount of Fe2O3 produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A)

According to the law of conservation of mass, matter can neither be created nor be destroyed. In the given reaction,

4Fe + 3O₂→ 2Fe₂O₃

The mass of reactant is equal to the mass of product. Hence, it follows the law of conservation of mass

0 0

New answer posted

10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter Two

Assertion (A) : Black body is an ideal body that emits and absorbs radiations of all frequencies.

Reason (R) : The frequency of radiation emitted by a body goes from a lower frequency to higher frequency with an increase in temperature.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the explanation of A.

(iii) A is true and R is false.

(iv) Both A and R are false.

0 Follower 13 Views

Vishal Baghel

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: option (ii)

The black body is the ideal body that absorbs and emits radiation of all frequencies. The radiation that it emits is said to be black body radiation. With an increase of temperature, the frequency of the black body radiation increases.

0 0

New answer posted

10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter Two

Assertion (A) : All isotopes of a given element show the same type of chemical behaviour.

Reason (R) : The chemical properties of an atom are controlled by the number of electrons in the atom.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A

(iii) A is true but R is false.

(iv) Both A and R are false.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: option (i)

As the chemical properties depend upon the number of electrons present in the atom thus the isotopes resemble the same chemical properties as they have the same number of electrons i.e. atomic number.

0 0

New answer posted

10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Which of the following statements about a compound is incorrect?

(i) A molecule of a compound has atoms of different elements.

(ii) A compound cannot be separated into its constituent elements by physical methods of separation.

(iii) A compound retains the physical properties of its constituent elements.

(iv) The ratio of atoms of different elements in a compound is fixed.

0 Follower 1 View

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (C)

(A) The molecule of a compound is obtained when two or more atoms of different elements are combined together.

(B) A compound can be separated into its constituent elements by physical methods of separation.

(C) A compound does not retains the physical and chemical properties of its constituent elements.

(D) When two elements react to form two or more than two chemical compounds, the ratio between different masses of one of the elements combining with a fixed mass of the other.

0 0

New answer posted

10 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

If the density of a solution is 3.12 g mL^–1, the mass of 1.5 mL solution in significant figures is _______.

(i) 4.7g

(ii) 4680 * 10^–3g

(iii) 4.680g

(iv) 46.80g

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A)

The density (d ) is given by the formula:

d = $\frac{m}{V}$

Here, m is the mass and v is the volume

On substituting the values in the above equation:

3.12 g mL^-1= $\frac{m}{1.5 m L}$ M = 4.68 g

The mass will be 4.7 upto 2 significant figure.

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Chemistry

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience