Class 12th

Given below are two statements:

Statement I: Ethyl pent -4-yn-oate on reaction with CH₃MgBr gives a 3° -alcohol.

Statement II: In this reaction one mole of ethyl pent-4-yn-oate utilizes two moles of CH₃MgBr.

In the light of the above statements, choose the most appropriate answer from the options given below.

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Class 12th Physics Ncert Solutions Class 12th

Contributor-Level 10

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If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160m, then the maximum range of LOS communication is_______ km.

(Take radius of Earth = 6400km)

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Raj Pandey

Contributor-Level 9

$d M = \sqrt[]{2 R h_{T}} + \sqrt[]{2 R h_{R}}$

$? h_{T} + h_{R} = 160$

[Given]

So, hR = 160 - hT

$d_{M} = \sqrt[]{2 R h_{T}} + \sqrt[]{2 R (160 - h_{T})}$

$∴ d_{M} = \sqrt[]{2 * 6400000 * 80} + \sqrt[]{2 * 6400000 * 80}$

$= 2 * 4 * 8 * 1 0^{3}$

dM = 64 km

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Which one of the following is formed (mainly) when red phosphorus is heated in a sealed tube at 803K?

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Class 12th Coordination Compounds

Contributor-Level 10

When red phosphorus is heated in a sealed tube at 803 K, a - black phosphorus is formed.

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The number of miles of NH₃, that must be added to 2L of 0.80 M AgNO₃ in order to reduce the concentration of Ag⁺ions to 5.0 * 10^-8 M (K_formation for [Ag(NH₃)₂]⁺ = 1.0 * 10⁸) is_________.

(Nearest integer)

[Assume no volume change on adding NH₃]

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Class 12th Physics Electromagnetic Waves

Contributor-Level 10

$A g^{+} (a q) + 2 N H_{3} (a q) ? A g {(N H_{3})}_{2}^{+} (a q)$

t = 00.8 M $(\frac{a}{2}) M$

0 $= \infty$ 5 * 10^-8 M $(\frac{a}{2} - 1.6) M$ 0.8M

$k_{f} = \frac{[A g {(N H_{3})}_{2}^{+}]}{[A g^{+}] {[N H_{3}]}^{2}}$

$1 0^{8} = \frac{0.8}{(5 * 1 0^{- 8}) {(\frac{a}{2} - 1.6)}^{2}}$

$\frac{a}{2} = 2$

Concentration of NH₃ added = $\frac{a}{2} = 2 M$

Volume of solution = 2L

Moles of NH₃ added = 2 * 2 mol = 4 mol.

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The electric field in an electromagnetic wave is given by E = $(50 N C^{- 1}) s i n ω (t - x ? / ? c)$ The energy contains in a cylinder of volume V is 5.5 * 10^-12 J. The value of V is________cm³.

(given by $\in_{0} = 8.8 * 1 0^{- 12} C^{2} N^{- 1} m^{- 2})$

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Raj Pandey

Contributor-Level 9

E = (50 N/s) sin (t – x/c)

Energy 5.5 * 10^-12 J., vol v =?

$u = \frac{1}{2} ε_{0} E^{2} \to$ energy density

$? u = \frac{U}{v} \Rightarrow U = u v$

= $\frac{1}{2} ε_{0} E^{2} . v$

$v = \frac{5.5 * 1 0^{- 12} * 2}{8.55 * 2500 * 1 0^{- 12}} * 1 0^{6} c m^{3}$

= 497

$≅ 500 c m^{3}$

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Class 12th Physics

Match list - I with list – II

List – I List – II

(Name of ore/mineral) &n

...more

Match list - I with list – II

List – I List – II

(Name of ore/mineral) (Chemical formula)

(A) Calamine (i) ZnS

(B) Malachite (ii) FeCO₃

(C) Siderite (iii) ZnCO₃

(D) Sphalerite (iv) $C u C O_{3} \cdot C u {(C O)}_{2}$

Choose the most appropriate answer from the options given below:

less

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Contributor-Level 10

(a) $C a l a m i n e : Z n C O_{3}$

(b) Malachite : $C u C O_{3} . C u {(O H)}_{2}$

(c) Siderite : FeCO₃

(d) Sphalerite : ZnS

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Class 12th Redox Reactions

When 10 mL of an aqueous solution of KMnO₄ was titrated in acidic medium, equal volume of 0.1 M of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of KMnO₄ is grams per litre is________* 10^-2. (Nearest integer)

[Atomic mass of K = 39, Mn = 55, O = 16]

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Contributor-Level 10

Let molarity of KMnO₄ = x

$K M n O_{4} + F e S O_{4} \to F e_{2} {(S O_{4})}_{3} + M n^{2 +}$

n = 5 n = 1 Ferric sulphate

equivalent of KMnO₄ = equivalent of FeSO₄

5 * x * 10 = 1 * 0.1 * 10

x = 0.02 M

Strength = (0.02 * 158) = 3.16g/L

= 316 * 10^-2 g/L

Ans. = 316

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The number of f electrons in the ground state electronic configuration Np (Z = 93) is_________.

(Nearest integer)

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$\underset{(z = 93)}{N p} \Rightarrow 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2}$

$4 d^{10} 5 p^{6} 6 s^{2} 4 f^{14} 5 d^{10} 6 p^{6} 7 s^{2} 5 f^{4} 6 d^{1}$

Total number of f-electrons = 14 + 4 = 18 electrons

Ans. = 18

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The number of moles of CuO, that will be utilized in Dumas method for estimating nitrogen in a sample of 57.5g of N, N-dimethylaminopentane is________ * 10^-2. (Nearest integer).

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Contributor-Level 10

Dumas method,

Moles of N in N, N-dimethylaminopentane (C₇H₁₇N)

$= \frac{57.5}{115} = 0.5 m o l e$

$C_{7} H_{17} N + \frac{45}{2} C u O \to 7 C O_{2} + \frac{17}{2} H_{2} O + \frac{1}{2} N_{2} + \frac{45}{2} C u$

$\frac{n_{C u O}}{(\frac{45}{2})} = \frac{n_{c_{7} H_{17} N}}{1}$

$n_{C u O} = (\frac{45}{2}) * 0.5 m o l$

= 11.25 mol

= 1125 * 10^-2 mol

Ans. = 1125

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The compound/s which will show significant intermolecular H-bonding is/are:

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