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7 months ago

Some Basic Concepts of Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

Chlorophyll extracted form the crushed green leaves was dissolved in water to make 2 L solution of Mg of concentration 48 ppm. The number of atoms of Mg in this solution is x * 10²⁰ atoms. The value of x is _____________. (Nearest Integer)

(Given : Atomic mass of Mg is 24 g mol^-1; N_A = 6.02 * 10²³ mol^-1)

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Vishal Baghel

Contributor-Level 10

Mass of solute = 48 * 2 * 10^-3 gram

Atoms of Mg = 24.08 * 10²⁰

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7 months ago

Some Basic Concepts of Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

A sample of 0.125g of an organic compound when analyzed by Duma's method yields 22.78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mm Hg. The percentage of nitrogen in the given organic compound is__________. (Nearest integer)

[Given: (a) The vapour pressure of water of 280K is 14.2mm Hg

(b) R = 0.082 L atm K^-1 mol^-1]

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Vishal Baghel

Contributor-Level 10

$V_{N_{2}} = 22.78 m l T = 280 K$

PTotal = 759 mm of Hg

$P_{N_{2}} = 759 - 14.2 = 744.8 m m H g$

$n_{N_{2}} = \frac{744.8 * (22.78 / 1000)}{(0.082 * 760) * 280}$

= 0.00097

$% N = \frac{0.02716}{0.125} * 100 = 21.728$

$\approx 22 .$

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7 months ago

Some Basic Concepts of Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

2L of 0.2 M H₂SO₄ is reacted with 2L of 0.1 M NaOH solution, the molarity of the resulting product Na₂SO₄ in the solution is_________ millimolar (Nearest integer)

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Vishal Baghel

Contributor-Level 10

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

Number of mole = 0.2 * 2 2 * 0.1 - -

0.4 = 0.2

No. of moles after reaction = 0.4 – 0.1 0.2 – 0.2 &nbs

...more

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

Number of mole = 0.2 * 2 2 * 0.1 - -

0.4 = 0.2

No. of moles after reaction = 0.4 – 0.1 0.2 – 0.2 0.1

= 0.3 = 0

molarity of Na₂SO₄ = $\frac{n}{V (L)} = \frac{0.1}{4} = 0.025$

$∴$ = 25 milli molar

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7 months ago

Some Basic Concepts of Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

The mass of ammonia in grams produced when $2.8 k g$ of dinitrogen quantitatively reacts with $1 k g$ of dihydrogen is(Mole Concept)

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alok kumar singh

Contributor-Level 10

$N_{2} + 3 H_{2} ? 2 {N H}_{3}$

$\begin{array}{l} \frac{2.8 * 10^{3}}{28} & \frac{1 * 10^{3}}{2} \\ \frac{0.1 * 10^{3} m o l}{L R} & 0.5 * 10^{3} m o l \end{array}$

$Mass of {N H}_{3} produced = 0.2 * 10^{3} * 17 = 3.4 k g = 3400 g$

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7 months ago

Some Basic Concepts of Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

$N_{2 (g)} + 3 H_{2 (g)} ? 2 N H_{3 (g)}$

20g 5g

Consider the above reaction, the limiting reagent of the reaction and number of moles of NH₃ formed respectively are:

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Vishal Baghel

Contributor-Level 10

t = 0 $\frac{20}{28}$ $\frac{5}{2}$ 0

Here, N₂ is the limiting reagent

and No of moles of NH₃ = 2 * No of moles ofN₂ reacted

$2 * \frac{20}{28} = \frac{40}{28} = \frac{1.42}{}$

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7 months ago

Some Basic Concepts of Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

116 g of a substance upon dissociation reaction yields 7.5 g of hydrogen 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12, respectively. The data agrees with how many formulae of the following?

(a) CH₃COOH (b) HCHO (c) CH₃OOCH₃ &n

...more

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Payal Gupta

Contributor-Level 10

Determining empirical formula of the given compound.

$% o f C = \frac{48.5}{116} * 100 % = 41.8 %$

$% o f H = \frac{7.5}{116} * 100 % = 6.5 %$

$% o f O = \frac{60}{116} * 100 % = 51.7 %$

Mass Moles Simplest ratio

C 41.8g $\frac{41.8}{12} = 3.48$ $\frac{3.48}{3.23} \approx 1$ &n

...more

Determining empirical formula of the given compound.

$% o f C = \frac{48.5}{116} * 100 % = 41.8 %$

$% o f H = \frac{7.5}{116} * 100 % = 6.5 %$

$% o f O = \frac{60}{116} * 100 % = 51.7 %$

Mass Moles Simplest ratio

C 41.8g $\frac{41.8}{12} = 3.48$ $\frac{3.48}{3.23} \approx 1$

H 6.5 g $\frac{6.5}{1} = 6.5$ $\frac{6.5}{3.23} \approx 2$

O 51.7 g $\frac{51.7}{16} = 3.23$ $\frac{3.23}{3.23} = 1$

Ratio of moles of C, H and O is 1 : 2 : 1

A. CH₃COOH ⇒ C : H : O = 2 : 4 : 2 = 1 : 2 : 1

B. HCHO ⇒ C : H : O = 1 : 2 : 1

C. CH₃OOCH₃ ⇒ C : H : O = 2 : 6 : 2 = 1 : 3 : 1

D. CH₃CHO ⇒ C : H : O = 2 : 4 : 1

So, CH₃COOH and HCHO formulae agree with the given data

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8 months ago

Some Basic Concepts of Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Assertion (A) : Combustion of 16 g of methane gives 18 g of water.

Reason (R) : In the combustion of methane, water is one of the products.

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

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false.

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alok kumar singh

Contributor-Level 10

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

Option (C) The combustion reaction of methane is given below

CH₄ + O₂ -> CO₂ + H₂O

In this reaction, water and carbon dioxide are formed.

According to the reaction, 1 mole of methane gives 2 moles of water. So, 16 g of methane gives 36 g of water

Thus, X is false but R is true

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8 months ago

Some Basic Concepts of Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Assertion (A) : Significant figures for 0.200 is 3 where as for 200 it is 1.

Reason (R) : Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.

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

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

(iii) A is true but R is false.

(iv) Both A and R are false.

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alok kumar singh

Contributor-Level 10

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

Option (C)

In the decimal portion, only trailing or final zeros are significant. This is in accordance with the rules of significant figures.

The given number, 0.200 has 3 significant figures whereas the number, 200 has only one significant figure because zeros present on the right side or at the end of the decimal number are significant and it is provided that they are present on the right side of the decimal point.

Also, zeros are not significant in numbers without decimals. Thus, X is true but R is false.

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8 months ago

Some Basic Concepts of Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Assertion (A) : One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.

Reason (R) : Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard.

(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.

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alok kumar singh

Contributor-Level 10

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

Option (B)

Atomic mass unit is defined as a mass that is equal to accurately the mass of carbon-12 atom.

When scientists compare the relative atomic masses of the elements to the mass of carbon, they find that they are near to a whole number value.

So, both X and B are true but R is the correct explanation of A.

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8 months ago

Some Basic Concepts of Chemistry Chemistry NCERT Exemplar Solutions Class 11th Chapter One

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): The empirical mass of ethene is half of its molecular mass.

Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound.

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

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false

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alok kumar singh

Contributor-Level 10

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

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

Explanation: The molecular formula gives the actual number of different atoms present in a molecule and the empirical formula gives a simple whole number ratio of different atoms present in a molecule.

Since ethene C₂H₄ can be further simplified, the empirical formula becomes CH₂. The molecular mass of C₂H₄ is 28 g/mol and the empirical mass of CH2 is 14 g/mol.

Hence, the empirical mass of ethene is half of its molecular mass.

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