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Class 11th Chemistry

Consider the following orders given below.

(i) S > Se > Te > O (Electron Affinity)

(ii) $N^{- 3} > O^{2 -} > F^{-} > N a^{+} > M g^{2 +}$ (Ionic Size)

(ii) Fe²⁺ > Fe³⁺ > Fe⁴⁺ (Ionization Enthalpy)

(iv) H < B < P < C < Cl (Electronegativity)

(v) Na⁺ < Mg²⁺ < Al³⁺ (Extent of hydration)

(vi) Li > Na > K > Rb > Cs (Reducing Nature)

(vii) LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)

If x = number of correct orders

y = number of incorrect orders

then the value of $(1 + \frac{y}{x})$ is.

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Vishal Baghel

Contributor-Level 10

Kindly go through the answers

(1.75)

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New answer posted

6 months ago

Class 11th Chemistry

Nitric oxide is formed in the combustion of fuel. An internal combustion engine produces 600 ppm (600 mg L^-1) of NO was oxidized to NO₂. The NO₂ formed was dissolved in 10L of water. The pH of the resulting solution is

(log 2 = 0.30, log 3 = 0.48)

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Vishal Baghel

Contributor-Level 10

Moles of NO = $\frac{600 * 1 0^{- 3} * 100}{30} = 2$

$∴$ moles of NO₂ = 2

3NO₂ + H₂O -> 2HNO₃ + NO

$0.2 M \frac{0.4}{3} M o r p H = - l o g \frac{0.4}{3} = 0.88$

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Class 11th Chemistry

Find out the number of following orders which are INCORRECT against the mentioned properties:

(i) Ortho hydrogen > Para hydrogen (Stability at low temperature)

(ii) Ethanol > Glycerol (Viscosity)

(iii) D₂ < He (Boiling point)

(iv) HF > H₂O (Melting point)

(v) H₃BO₃ > BF₃ (Melting point)

(vi) NH₃ < SbH₃ (Boiling point)

(vii) H₂O₂ > H₂O (Strength of hydrogen bond)

(viii) H₂O < D₂O (Freezing point)

(ix) $H F_{2}^{Θ}$ < HF (Strength of hydrogen bond)

(x) D₂O < H₂O (Bond energy)

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Vishal Baghel

Contributor-Level 10

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(7.00)

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New answer posted

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Class 11th Chemistry

If x = Group number of Europium (atomic number 63)

y = Period number of Americium (atomic number 95)

z = Most stable oxidation state shown by Gadolinium (atomic number 64)

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Vishal Baghel

Contributor-Level 10

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(13.00)

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6 months ago

Class 11th Physics Thermal Properties of Matter

Two rods of length d₁ and d₂ and coefficients of thermal conductivities K₁ and K₂ are kept touching each other. Both have the same area of cross?section, the equivalent thermal conductivity is

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Raj Pandey

Contributor-Level 9

When Two rods are connected in series

$Q = \frac{A (T_{1} - T_{2}) t}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{1}}} = \frac{A (T_{1} - T_{2}) t}{(d_{1} + d_{2}) / K}$

$\frac{d_{1} + d_{2}}{K} = \frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}}$

$K = \frac{(d_{1} + d_{2})}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{1}}}$

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Class 11th Physics

The diagram showing the variation of gravitational potential of earth with distance from the centre of earth is

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Raj Pandey

Contributor-Level 9

$V_{in} = \frac{- G M}{2 R} [3 - {(\frac{r}{R})}^{2}],$

$V_{s u r f a c e} = \frac{- G M}{R}, V_{out} = \frac{- G M}{r}$

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Class 11th Physics

A 20 kg block B is suspended from a cord attached to a 40 kg cart A. Find the ratio of the acceleration of block in cases (i) and (ii) shown in the figure immediately after the system is released from rest. (Neglect friction)

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Raj Pandey

Contributor-Level 9

Case I:

$T - N = 40 a$

and $20 g - T = 20 a$

Also $N = 20 a$

After simplifying, we get

$a = \frac{g}{4}$

Acceleration of block $B, = \sqrt[]{2} a = \frac{g}{2 \sqrt[]{2}} .$

Case II:

$T = 40 a$

and $20 g - T = 20 a$

After simplifying above equation, we get

$a = g / 3$

Ratio $= \frac{g / 2 \sqrt[]{2}}{g / 3} = \frac{3}{2 \sqrt[]{2}} .$

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Class 11th Thermodynamics

The only INCORRECT information is

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Vishal Baghel

Contributor-Level 10

q = 0, w = 0, U = 0 = f (V, T)

As volume changes, temperature must change to maintain constant internal energy.

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Class 11th Chemistry

If Q is a colourless gas with rotten fish smell then select CORRECT statement.

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Vishal Baghel

Contributor-Level 10

Zero acidic hydrogen present in anionic part of (R)

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Class 11th Equilibrium

An equilibrium mixture contains 2.0 moles of A(g) and 4.0 moles of B(g). Now, 2.0 moles of A(g) is added in this equilibrium mixture and the system is allowed to re-achieve equilibrium at constant volume and temperature. Now, the volume of system is doubled at constant temperature. What should be the moles of B(g) at new equilibrium? The reaction involved is

$A (g) \overset{}{?} 2 B (g)$ .

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Vishal Baghel

Contributor-Level 10

$A (g) ? 2 B (g) .$

$E_{m} \frac{2 m o l}{V l} \frac{4 m o l}{V l}$

$N e w {E_{q}}^{m} \frac{4 - x}{V} \frac{4 + 2 x}{V}$

$V * 2 \frac{4 - x}{2 V} \frac{4 - 2 x}{2 V}$

New Eqm $\frac{4 - x - y}{2 V} \frac{4 + 2 x + 2 y}{2 V}$

= $\frac{4 - Z}{2 V} = \frac{4 + 2 Z}{2 V} (Z = x + y)$

Now, KC = $\frac{^{}}{}$ $\frac{{[B]}^{2}}{[A]} =$ constant

$o r, \frac{{(\frac{4}{V})}^{2}}{(\frac{2}{V})} = \frac{(\frac{4 + 2 Z}{2 V})}{(\frac{4 - Z}{2 V})} \Rightarrow Z = 1.29$

mole of B at new Eqm = 4 + 2Z = 6.58

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