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9 months ago

Probability Maths

The sum and product of the main and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is……………

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9 months ago

Haloalkanes and Haloarenes Chemistry

The number of chlorine atoms in bithionol is_____________.

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Payal Gupta

Contributor-Level 10

Structure of Bithinol is;

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9 months ago

Organic Chemistry - Some Basic Principles and Tech Chemistry

The number of stereoisomers formed in a reaction of

$(\pm) P h (C = O) C (O H) (C N) P h w i t h H C N i s____________.$

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Payal Gupta

Contributor-Level 10

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9 months ago

Alcohol Phenol And Ethers Chemistry

A 1.84 mg sample of polyhydric alcoholic compound ‘X’ of molar mass 92.0 g/mol gave 1.344mL of H₂ gas at STP. The number of alcoholic hydrogens present in compound ‘X’ is_________.

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Payal Gupta

Contributor-Level 10

Mole of polyhydric alcohol = $\frac{1.84 \times 1 0^{- 3}}{92} = 2.0 \times 1 0^{- 5} m o l e .$

Mole of H₂ gas produced = $\frac{1.344}{22400} = 6.0 \times 1 0^{- 5} m o l e .$

No of OH gp present = $\frac{6.0 \times 1 0^{- 5}}{2.0 \times 1 0^{- 5}} = 3$

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9 months ago

P Block Elements Chemistry

Consider the following sulphur based oxoacids.

H₂SO₃, H₂SO₄, H₂S₂O₈ and H₂S₂O₇.

Amongst these oxoacids, the number of those with peroxo (O-O) bond is______________.

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Payal Gupta

Contributor-Level 10

Only H₂S₂O₈ has per-oxo bond.

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9 months ago

Chemistry Coordination Compounds Chemistry

Sum of oxidation state (magnitude) and coordination number of cobalt in $N a [C o (b p y) C l_{4}]$ is___________.

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Payal Gupta

Contributor-Level 10

Oxidation state of Co = 3

And co-ordination No = 6

Sum = 3 + 6 = 9

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9 months ago

Chemistry Class 12th

Assuming 1 $μ g$ of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is __________× 10^-1 $μ g$

[Given: In 10 = 2.303; log 2 = 0.30]

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Payal Gupta

Contributor-Level 10

$λ t = l n \frac{N_{o}}{N_{t}}$

$\frac{0.693}{t_{1 / 2}} \times t = l n \frac{1}{N_{t}}$

Or $\frac{0.693 \times 100}{30} = l n \frac{1}{N_{t}}$

$\frac{1}{N_{t}} = 10$

$∴ N_{t} = \frac{1}{10} = 0.1$

Or N_t = 1 × 101 $μ g$

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9 months ago

Nernst Equation Chemistry Electrochemistry

For a cell, $C u (s) | C u^{2 +} (0.001 M) | | A g^{+} (0.01 M) | A g (s)$

The cell potential is found to be 0.43V at 298 K. The magnitude of standard electrode potential for Cu²⁺ / Cu is______________ × 10^-2V.

$[G i v e n : E_{A g^{+} / A g}^{Θ} = 0.80 V a n d \frac{2.303 R T}{F} = 0.06 V]$

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9 months ago

Solutions Chemistry

1.80g of solute A was dissolved in 62.5cm³ of ethanol and freezing point of the solution was found to be 155.1 K . The molar mass of solute A is __________g mol^-1.

[Given: Freezing point of ethanol is 156.0 K. Density of ethanol is 0.80 g cm^-3. Freezing point depression constant of ethanol is 2.00 K kg mol^-1]

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Payal Gupta

Contributor-Level 10

$Δ T_{f} = K_{f} \times m$

$(156 - 155.1) = 2 \times \frac{1.8}{M W} \times \frac{1000}{(62.5 \times 0.8)}$

0.9 = $\frac{2 \times 1.8 \times 1000}{M W \times 50}$

$∴ M W = 80 g m / m o l$

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9 months ago

Hybridization Chemistry Chemical Bonding and Molecular Structure

Consider, PF₅, BrF₅, PCl₃, SF₆, ${[I C l_{4}]}^{-}$ , CIF₃ and IF₅.

Amongst the above molecule(s) / ion(s), the number of molecule(s) / ion(s) having sp³d² hybridisation is_____________.

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Payal Gupta

Contributor-Level 10

Following have the sp³d² hybridisation

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9 months ago

Biyani Girls B.Ed College Colleges in Jaipur

Which courses are offered by Biyani Girls B.Ed. College?

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abhishek gaurav

Contributor-Level 9

Biyani Girls BEd College offers programmes at UG and PG level. The courses include:

B.Ed. + Microsoft Office Specialist
M.Ed. + Microsoft Office Specialist
D.El.Ed. + Microsoft Office Specialist (Training)
B.Ed. – M.Ed. (3 Years Integrated Course)
B.Sc. – B.Ed. (4 Years Integrated Course)

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9 months ago

States of Matter Chemistry

‘x’ g of molecular oxygen(O₂) is mixed with 200g of neon (Ne). The total pressure of the non-reactive mixture of O₂ and Ne in the cylinder is 25 bar. The partial pressure of Ne is 20 bar at the same temperature and volume. The value of ‘x’ is_________.

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Payal Gupta

Contributor-Level 10

Po₂ = P_Total - $P_{N_{2}}$

= 25 – 20 = 5bar

Po₂ = $\frac{n_{O_{2}}}{n_{O_{2}} + n_{N e}} \times P_{T o t a l}$

$5 = (\frac{x / 32}{x / 32 + \frac{200}{20}}) \times 25 \Rightarrow x = 80 g m$

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9 months ago

Biyani Girls B.Ed College Colleges in Rajasthan

Is there any entrance exam at Biyani Girls B.Ed. College?

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abhishek gaurav

Contributor-Level 9

Biyani Girls B.Ed. College accepts entrance exam scores for admission. The college accepts PTET for B.Ed and MPET for M.Ed programme. Admission to B.Ed-M.Ed three-year Integrated course is based on PBMET scores of the candidates.

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9 months ago

Electromagnetic Induction Physics Ncert Solutions Class 12th

A capacitor of capacitance 500 $μ F$ is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. Then maximum current in LC circuit will be_______ A.

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Payal Gupta

Contributor-Level 10

C = 500 $μ F,$ V = 100 v, L = 50 mH

In this LC – oscillation

q = q₀ cos $ω t$

$i = \frac{- d q}{d t} = q_{0} ω s i n ω t ω = \frac{1}{\sqrt[]{2 c}} = \frac{1}{\sqrt[]{50 \times 1 0^{- 3} \times 5 \times 1 0^{- 4}}}$

= $\frac{1000}{5} = 200$

So, i_max = $q_{0} ω = 500 \times 1 0^{6} \times 100 \times 200$

= 10A

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9 months ago

Physics Nuclei Physics

Two radioactive materials A and B have decay constants 25 $λ$ and 16 $λ$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of B to that of A will be “e” after a time $\frac{1}{a λ} .$ The value of a is__________.

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Payal Gupta

Contributor-Level 10

$λ_{A} = 25 λ,$ $λ_{B} = 16 λ$

At t = 0 N_A = N_B = N₀

after t = $\frac{1}{a λ} : - \frac{N_{B}}{N_{A}} = \frac{N_{0} e^{- 16 λ t}}{N_{0} e^{- 25 λ t}} = e^{(25 λ - 16 λ) t}$

e = $e^{(9 λ \frac{1}{a λ})}$

$\Rightarrow \frac{9}{a} = 1 \Rightarrow a = 9$

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9 months ago

Biyani Girls B.Ed College Colleges in Jaipur

What documents are required during admission to Biyani Girls B.Ed. College?

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abhishek gaurav

Contributor-Level 9

At the time of admission in Biyani Girls B.Ed. College the candidates are required to upload their caste certificate (in case of SC/ST), copies of marksheet of Class 10, 12, graduation and a formal passport size photograph with a white background.

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9 months ago

Physics Ncert Solutions Class 12th Semiconductor Electronics: Materials, Devices and

A 8 V Zener diode along with a series resistance R is connected across a 20V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be_______ $Ω$ .

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Payal Gupta

Contributor-Level 10

$i_{(z e n e r - m a x)} = 25 m A$

20 – i_max R – 8 = 0

i_max R = 12

At minimum zener current $(μ A) : -$

$20 - i_{m i n} R - i_{m i n} R_{L} = 0$

$\frac{R}{R_{L}} = \frac{12}{8} = \frac{3}{2}$

$l_{m i n} R = 12$

$i_{m i n} R_{L} = 8$

At maxm zener current –

$20 - i_{m a x} R - 8 = 0$

$i_{L} = O {a s i_{z} m a x^{m} = 25 m A}$

i_maxR = 12v

25 × 103 R = 12

$R = \frac{12 \times 1 0^{3}}{25} = 12 \times 40 = 480 Ω$

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9 months ago

Physics Ncert Solutions Class 12th Semiconductor Electronics: Materials, Devices and

The metallic bob of simple pendulum has the relative density. 5 The time period of this pendulum is 10 s. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt[]{x} s .$ The value of x will be_______.

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Payal Gupta

Contributor-Level 10

g_eff = g – $(\frac{ρ_{w}}{ρ_{b}}) g$

$T = 2 π \sqrt[]{\frac{l}{g}}$

$T' = 2 π \sqrt[]{\frac{l}{g_{e f f}}}$

$\Rightarrow T' = \sqrt[]{\frac{5}{4}} T$

= $\sqrt[]{\frac{5}{4}} \times 10$

= $5 \sqrt[]{5} s e c$

So, x = 5

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9 months ago

Physics Oscillations Physics

The speed of a transverse wave passing through a string of length 60 cm and mass 10g is 60 ms^-1. The area of cross-section of the wire is 2.0 mm² and its Young’s modulus is 1.2 × 10¹¹ Nm^-2. The extension of the wire over its natural length due to its tension will be x × 10^-5 m. The value of x is________.

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Payal Gupta

Contributor-Level 10

m = 10g

$l = 50 c m$

A = 2mm²

Y = 1.2 × 10¹¹N/m²

$Δ x = x \times 1 0^{- 5} m$

As, $\frac{T}{A} = Y \frac{Δ x}{l}$

$Δ x = \frac{T l}{A Y}$

$T l = V^{2} m$

= $\frac{V^{2} m}{A Y}$

$= \frac{3600 \times 10 \times 1 0^{- 3}}{2 \times 1 0^{- 6} \times 1.2 \times 1 0^{11}} = \frac{1800 \times 1 0^{- 3} \times 1 0^{6}}{1.2}$

$= \frac{18}{12} \times 1 0^{- 4} = \frac{3}{2} \times 1 0^{- 4} = 15 \times 1 0^{- 5} m$

So, x = 15

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9 months ago

Physics Mechanical Properties of Solids Physics

A modulating signal 2sin (6.28 × 10⁶)t is added to the carrier signal 4 sin(12.56 ×10⁹)t for amplitude modulation. The combined signal is passed through a non-linear square law device. The output is then passes through a band pass filter. The bandwidth of the output signal of band pass filter will be_________ MHz.

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