Ncert Solutions Chemistry Class 11th

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

The neutralization occurs when 10 mL of 0.1 M acid ‘A’ is allowed to react with 30 mL of 0.05 M base M base M(OH)₂. The basicity of the acid ‘A’ is___________.

[M is a metal]

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Acid = M (OH)₂ ® Salt + H₂O

M.E of Acid = me of Base

$∴ 10 \times (0.1 \times n f) = (0.05 \times 2) \times 30$

$∴ n_{f} = 3$

Thus basicity of acid = 3

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

A rigid nitrogen tank stored inside a laboratory has a pressure of 30 atm at 06:00 am when the temperature is $27 ° C$ . At 03:00 pm, when the temperature is $45 ° C$ , the pressure in the tank will be___________atm. [nearest integer]

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$T_{1} = 27 ° C = 300 K$ T₂ = 45° C = 45 + 273 = 318 K

P₁ = 30 atm P₂ =?

V is constant due to rigid tank

$∴ \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

$\frac{30}{300} = \frac{P_{2}}{318} ∴ P_{2} = \frac{1}{10} \times 318 = 31.8 a t m \approx 32 a t m$

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

200mL of 0.01 M HCl is mixed with 400mL of 0.01 M H₂SO₄. The pH of the mixture is___________.

Given: log2 = 0.30, log 3 = 0.48, log5 = 0.70, log7 = 0.84, log11 = 1.04

26 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Total concentration of H+ after the mixing of HCl and H2SO4;

$[H^{+}] = \frac{(200 \times 0.01) + (400 \times 2 \times 0.01)}{600}$

$[H^{+}] = \frac{1}{60}$

$∴ p H = l o g_{10} (\frac{1}{60}) = 1.78$

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

C(s) + O₂(g) → CO₂(g) + 400 kJ

C(s) + $\frac{1}{2} O_{2} (g) \to C O (g) + 100 k J$ 

When coal of purity 60% is allowed to burn in presence of insufficient oxygen, 60% of carbon is converted into ‘CO’ and the remaining is converted into ‘CO₂’.

The heat generated when 0.6 kg of coal is burnt is

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Mass of pure carbon in coal = 0.6 × 1000gm × $\frac{60}{100} = 360 g m$

Mass of carbon converted into CO = 6 $\frac{60}{100} \times 360 = 216 g m \Rightarrow M o l = \frac{216}{12} = 18$ mole of carbon

Mass of carbon converted into CO₂ = 360 – 216 = 144gm Mole = $\frac{144}{12} = 12$ mole of carbon.

C (s) + O₂ (g) CO₂ (g) + 400KJ

Mole – 12 for CO₂ production, Hence total energy produced = 400 × 12 =

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

Consider the reaction

4HNO₃(l) + 3 KCl(s) → Cl₂(g) + NOCl(g) + 2H₂O(g) + 3KNO₃(s)

The amount of HNO₃ required to produce 110.0g of KNO₃ is

(Given: Atomic masses of H, O, N and K are 1, 16, 14 and 39, respectively)

13 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$4 H N O I_{3 (l)} + 3 K C l_{(s)} \to_{}^{} C l_{2 (g)} + N O C l_{(g)} + 2 H_{2} O (g) + 3 K N O_{3} (g)$

$↓$

4 moles of HNO₃ produced 3 mol of KNO₃

Here mole of produced KNO₃ = $\frac{110}{101}$

If 3 mol of KNO₃ produced by 4 moles of HNO₃

$∴$ 1 mole of KNO₃ produced by $\frac{4}{3}$ moles of HNO₃

and $\frac{110}{101}$ mole of KNO₃ produced by $\frac{4 \times 110}{3 \times 101}$ moles of HNO₃ = 1.45 mole of HNO₃

Hence mass of HNO₃= mole × mol.wt = 145 ×

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Periodic Table Elements

The correct order of electron gain enthalpies of Cl, F, Te and Po is

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Element e^- gain enthalpy

F -333

Cl &nbs

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Thermodynamics

At  $25 ° C$ and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are -3268 kJ mol^-1 and -1300 kJ mol^-1, respectively. The change in enthalpy for the reaction 3C₂H₃(g) ® C₆H₆(I), is

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$Δ H = 3 {(Δ H_{c o m b})}_{C_{2} H_{2} (g)} - {(Δ H_{c o m b})}_{C_{6} H_{6} (l)}$

$= - 1300 \times 3 - (- 3268) k J / m o l e$

$= (- 3900 + 3268) k J / m o l e$

$= - 632 k J / m o l e$

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

The molar heat capacity for an ideal gas at constant pressure is 20.785JK^-1 mol^-1. The change in internal energy is 5000 J upon heating it from 300 K to 500 K. The number of moles of the gas at constant volume is __________.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$C_{P, m} = C_{v, m} + R$

$C_{v, m} = 20.785 - 8.314 = 12.471 J K^{- 1} m o l^{- 1}$

$n = \frac{5000}{12.471 \times 200} = \frac{25}{12.471} \approx 2$

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

Amongst the following, the number of oxide(s) which are paramagnetic in nature is Na₂O, KO₂, NO₂, N₂O, ClO₂, NO, SO₂, Cl₂O (Nearest Integer)

16 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$K O_{2}, N O_{2}, C l O_{2}, N O$ are paramagnetic.

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

In base vs, acid titration, at the end point methyl orange is present as

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

For an acid- base titration, Methly orange exist at end point as quinonoid form.

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Ncert Solutions Chemistry Class 11th

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