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1.26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour could be produced?

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Marri Laxman Reddy Institute of Technology and Management (MLRITM) Marri Laxman Reddy Institute of Technology and Management (MLRITM) Admission Process

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1.26. H₂and O₂ react according to the equation

2H₂ (g) + O₂ (g) ——>2H₂O (g)

Thus, 2 volumes of H₂ react with 1 volume of O₂ to produce 2 volumes of water vapour. Hence, 10 volumes of H₂ will react completely with 5 volumes of O₂ to produce 10 volumes of water vapour.

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11 months ago

When will Marri Laxman Reddy Institute of Technology and Management application forms be out?

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Taru Shukla

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Marri Laxman Reddy Institute of Technology and Management application forms are out for the academic year 2025-26 for all courses. Candidates seeking admission to any courses can apply online through the official website of the college before the last date. The admission to the course is merit and entrance-based. Additionally, applicants must have to full fill out the basic eligibility criteria for their desired courses.

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11 months ago

Physics Ncert Solutions Class 12th Electrostatic Potential and Capacitance

2.21 Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively.

(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?

(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.

(c) How much work is done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

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2.21 Charge –q is located at (0,0,-a) and charge +q is located at (0,0,a). Hence, they form a dipole. Point (0,0,z) is on the axis of this dipole and point (x,y,0) is normal to the axis of the dipole. Hence electrostatic potential at point (x,y,0) is zero. Electrostatic potential at point (0,0,z) is given by,

V = $\frac{1}{4 π ε_{0}}$ ( $\frac{q}{z - a}) + \frac{1}{4 π ε_{0}}$ ( $- \frac{q}{z + a})$ = $\frac{q (z + a - z + a)}{4 π ε_{0} (z^{2} - a^{2})}$ = $\frac{2 q a}{4 π ε_{0} (z^{2} - a^{2})}$ = $\frac{p}{4 π ε_{0} (z^{2} - a^{2})}$

Where $ε_{0}$ = Permittivity of free space

p = Dipole moment of the system of two charges = 2qa

Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to the dist

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Where $ε_{0}$ = Permittivity of free space

p = Dipole moment of the system of two charges = 2qa

Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to the distance. i.e. V $\propto$ $\frac{1}{r^{2}}$

Electrostatic potential ( $V_{1})$ at point (5,0,0) is given by

$V_{1}$ = $\frac{- q}{4 π ε_{0}} \frac{1}{\sqrt{{(5 - 0)}^{2} + {(- a)}^{2}}} + \frac{q}{4 π ε_{0}} \frac{1}{\sqrt{{(5 - 0)}^{2} + {(a)}^{2}}}$

= 0

Electrostatic potential ( $V_{2})$ at point (-7,0,0) is given by

$V_{2}$ = $\frac{- q}{4 π ε_{0}} \frac{1}{\sqrt{{(- 7)}^{2} + {(- a)}^{2}}} + \frac{q}{4 π ε_{0}} \frac{1}{\sqrt{{(- 7)}^{2} + {(a)}^{2}}}$

= 0

Hence, no work is done in moving a small test charge from point (5,0,0) to point (-7,0,0) along the x-axis.

The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

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11 months ago

1.25. How are 0.50 mol Na₂CO₃and 0.50 M Na₂CO₃different?

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1.25. Molar mass of Na₂CO₃= (2 x 23) + 12 + (3 x 16) = 106g mol^-1

0.50 mol Na₂CO₃means 0.50 x 106 g = 53 g of Na₂CO₃

0.50 M Na₂CO₃ means 0.50 mol per volume in litre,

i.e., half of 106 g Na₂CO₃is present in 1 L solution.

i.e.,53 g Na₂CO₃is present in 1 L of the solution

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11 months ago

1.24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:

N₂ (g) + 3H₂(g) —–> 2NH₃ (g)

(i) Calculate the mass of ammonia produced if 2.00 × 10³ g dinitrogen reacts with 1.00 ×10³ g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?

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Physics Ncert Solutions Class 12th Electrostatic Potential and Capacitance

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1.24. According to the given equation, 1 mol of N₂ reacts with 3 mol of H₂.

Or, 28 g of N₂ react with 6 g of H₂.

So, 2000 g of N₂ will react with H₂= 6/28 x 2000 g = 428.6 g of H₂.

(i) 2 mol of N₂ or 28 g of N₂ produce NH₃ = 2 mol = 34 g

So, 2000 g of N₂ will produce NH₃ = 34/28 x 2000 g = 2428.57 g

(ii) Yes, N₂ is the limiting reagent while H₂ is the excess reagent. So, H₂will remain unreacted.

(iii) H₂ will remain unreacted. Mass left unreacted = 1000 g – 428.6 g = 571.4 g

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11 months ago

2.20 Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

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2.20 For Sphere A: We assume, radius = a, Charge on the sphere = $Q_{A}$ , Capacitance = $C_{A}$ , Electric field = $E_{A}$

For Sphere B: We assume, radius = b, Charge on the sphere = $Q_{B}$ , Capacitance = $C_{B}$ , Electric field = $E_{B}$

$E_{A} i s g i v e n b y, E_{A} =$ $\frac{Q_{A}}{4 π ε_{0} a^{2}}$ and $E_{B} i s g i v e n b y, E_{B} =$ $\frac{Q_{B}}{4 π ε_{0} b^{2}}$

$\frac{E_{A}}{E_{B}}$ = $\frac{Q_{A}}{Q_{B}} \times \frac{b^{2}}{a^{2}}$ ……….(1)

We also know Q = CV, hence $Q_{A}$ = $C_{A}$ V and $Q_{B}$ = $C_{B}$ V and $\frac{C_{A}}{C_{B}}$ = $\frac{a}{b}$

Then, $\frac{Q_{A}}{Q_{B}}$ = $\frac{C_{A}}{C_{B}} = \frac{a}{b}$ ……….(2)

Combining equations (1) and (2), we get

$\frac{E_{A}}{E_{B}}$ = $\frac{a}{b} \times \frac{b^{2}}{a^{2}}$ = $\frac{b}{a}$

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11 months ago

1.23. In the reaction,

A + B₂——> AB₂,

identify the limiting reagent, if any, in the following reaction mixtures

(i) 300 atoms of A + 200 molecules ofB

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

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Marri Laxman Reddy Institute of Technology and Management (MLRITM) Marri Laxman Reddy Institute of Technology and Management (MLRITM) Admission Process

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1.23. (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B
Thus, 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be left unreacted. Hence, B is the limiting reagent while A is the excess reagent.

(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B
Thus, 2 mol of A will react with 2 mol of B. Hence, A is the limiting reactant since 1 mol of B is left unreacted.

(iii) No limiting reagent.

(iv) 2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reagent.

(v) 2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reagent

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11 months ago

Can I get direct admission at Marri Laxman Reddy Institute of Technology and Management?

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Mukul Shukla

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Marri Laxman Reddy Institute of Technology and Management offers entrance-based admissions along with counseling for all the courses. The college accepts EAMCET entrance exam scores and counseling for admission to BTech courses and GATE, PGECET, TS ICET exam scores and counseling scores for M.Tech and MBA courses, respectively.

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11 months ago

1.22. If the speed of light is 3.0 × 10⁸ms^–1, calculate the distance covered by light in 2.00 ns.

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Jaipuria Institute of Management, Ghaziabad Justice K S Hegde Institute of Management

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1.22. Speed = Distance / Time

Or, Distance = Speed x Time = 3.0 * 10⁸ms^–1x 2.00 ns = 3.0 * 10⁸ms^–1x 2.00 x 10^-9 s = 6 x 10^-1 m = 0.600 m

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11 months ago

Which is better for MBA: JKSHIM or JIM?

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saurya snehal

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Jaipuria Institute of Management and Justice KS Hegde Institute of Management both offer a two year MBA programme for candidates. However, based on the average package and the career path of the alumnus we can say that JIM is a better option for MBA. Candidates should choose the institute on the basis of their personal requirements, average package still remains to be one of the most important deciding factors and JIM has performed better in the talked about criteria.

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11 months ago

Physics Ncert Solutions Class 12th Electrostatic Potential and Capacitance

2.19 If one of the two electrons of a $H_{2}$ molecule is removed, we get a hydrogen molecular ion $H_{2}^{+}$ . In the ground state of an $H_{2}^{+}$ , the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

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National Institute of Fashion Technology, Delhi Colleges in Delhi

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2.19

Let the charge in Proton 1 be $q_{1}$ = 1.6 $\times 10^{- 19}$ C, in Proton 2, $q_{2}$ = 1.6 $\times 10^{- 19}$ C and the charge in the electron, $q_{3}$ = $-$ 1.6 $\times 10^{- 19}$ C

Distance between Proton 1 & Proton 2, $d_{1}$ = 1.5 Å = 1.5 $\times 10^{- 10}$ m

Distance between Proton 1 and Electron, $d_{2}$ = 1.0 Å = 1.0 $\times 10^{- 10}$ m

Distance between Proton 2 and Electron, $d_{3}$ = 1.0 Å = 1.0 $\times 10^{- 10}$ m

The potential energy of the system,

V = $\frac{q_{1} q_{2}}{4 π ε_{0} d_{1}}$ + $\frac{q_{2} q_{3}}{4 π ε_{0} d_{3}}$ + $\frac{q_{3} q_{1}}{4 π ε_{0} d_{2}}$ ,

Where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

V = $\frac{1}{4 π ε_{0}}$ ( $\frac{q_{1} q_{2}}{d_{1}}$ + $\frac{q_{2} q_{3}}{d_{2}}$ + $\frac{q_{3} q_{1}}{d_{2}}$ )

= $\frac{1}{4 π \times 8.854 \times 10^{- 12}} \times$ ( $\frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{1.5 \times 10^{- 10}}$ + $\frac{1.6 \times 10^{- 19} \times - 1.6 \times 10^{- 19}}{1.0 \times 10^{- 10}}$ +&n

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2.19

Let the charge in Proton 1 be $q_{1}$ = 1.6 $\times 10^{- 19}$ C, in Proton 2, $q_{2}$ = 1.6 $\times 10^{- 19}$ C and the charge in the electron, $q_{3}$ = $-$ 1.6 $\times 10^{- 19}$ C

Distance between Proton 1 & Proton 2, $d_{1}$ = 1.5 Å = 1.5 $\times 10^{- 10}$ m

Distance between Proton 1 and Electron, $d_{2}$ = 1.0 Å = 1.0 $\times 10^{- 10}$ m

Distance between Proton 2 and Electron, $d_{3}$ = 1.0 Å = 1.0 $\times 10^{- 10}$ m

The potential energy of the system,

V = $\frac{q_{1} q_{2}}{4 π ε_{0} d_{1}}$ + $\frac{q_{2} q_{3}}{4 π ε_{0} d_{3}}$ + $\frac{q_{3} q_{1}}{4 π ε_{0} d_{2}}$ ,

Where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

V = $\frac{1}{4 π ε_{0}}$ ( $\frac{q_{1} q_{2}}{d_{1}}$ + $\frac{q_{2} q_{3}}{d_{2}}$ + $\frac{q_{3} q_{1}}{d_{2}}$ )

= 8.988 $\times 10^{9} \times (1.707 \times 10^{- 28}$ $- 2.56 \times 10^{- 28} - 2.56 \times 10^{- 28})$

= $- 3.067 \times 10^{- 18}$ J

= $- 3.067 \times 10^{- 18} \times \frac{1}{1.6 \times 10^{- 19}}$ eV

= - 19.17 eV

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11 months ago

What is the course fees for four-year BDes in NIFT Delhi?

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Virajita Mehra

Contributor-Level 10

The fee structure is inclusive of different components, such as tuition fees, hostel fees, refundable security deposit, registration fee, etc. A few of these components are one-time chargeable, while some have to be paid annually or semester-wise. Once selection procedure is completed, students need to pay the NIFT Delhi BDes fees. The following table showcases the fee bifurcation:

Fee components	Amount (4 years)
Tuition Fees	INR 12.34 Lakh
Hostel Fees	INR 4.19 Lakh
One-time Payment	INR 27,200

NOTE: The aforementioned fee details are as per the official website/ sanctioning body. It is still subject to changes and hence, is indicative.

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1.21. The following data were obtained when dinitrogen and dioxygen react together to form compounds: (Beginner)

Mass of dinitrogen Mass of dioxygen

(i) 14 g 16 g

(ii) 14 g 32 g

(iii) 28 g

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Physics Ncert Solutions Class 12th Electrostatic Potential and Capacitance

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1.21. (a) Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32,64, 32 and 80 g in the given four oxides. Theseare in the ratio 1: 2: 1: 5 which is a simple whole number ratio. Hence, the given data obeys the law of multiple proportions.

(b)

(i) 1 km = 10³ m and 1 m = 10³ mm. So, 1 km = 10³x 10³ mm = 10⁶ mm

Now, 1 pm = 10^-12 m. So, 1 km = 10³ m x 10¹² m = 10¹⁵ pm

Therefore, 1 km = 10⁶ mm = 10¹⁵ pm

(ii) 1 mg = 10^-3 g and 1 g = 10^-3kg. So, 1 mg = 10^-3x 10^-3 kg = 10^-6 kg

Now, 1 mg = 10^-3 g and 1 g =

Therefore, 1 mg = 10^-6 kg = 10⁶ ng

1L = 1000 mL.So 1 mL = 10^-3 L.

Now, 1 mL = 1cm³ and 1dm = 10cm. So, 1 mL = 1

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11 months ago

2.18 In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:

(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separations?

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Marri Laxman Reddy Institute of Technology and Management (MLRITM)

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2.18 The distance between electron – proton of a hydrogen atom, d = 0.53 Å = 0.53 $\times 10^{- 10}$ m

Charge of an electron, $q_{1}$ = $-$ 1.6 $\times 10^{- 19}$ C

Charge of a proton, $q_{2}$ = 1.6 $\times 10^{- 19}$ C

Potential at infinity = 0

Potential energy of the system = Potential energy at infinity – Potential energy at a distance d

= 0 - $\frac{q_{1} q_{2}}{4 π ε_{0} d}$ , where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ == 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

= 0 - $\frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{4 π \times 8.854 \times 10^{- 12} \times 0.53 \times 10^{- 10}}$ = - 4.34 $\times 10^{- 18}$ J = $\frac{4.34 \times 10^{- 18}}{1.6 \times 10^{- 19}}$ = - 27.13 eV

(Since 1 eV = 1.6 $\times 10^{- 19}$ J)

Kinetic energy = $\frac{1}{2}$ of potential energy = $\frac{1}{2}$ $\times -$ 27.13 eV = -13.57 eV

Total energy = - 13.57 – (-27.13) = 13.57 eV

Ther

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2.18 The distance between electron – proton of a hydrogen atom, d = 0.53 Å = 0.53 $\times 10^{- 10}$ m

Charge of an electron, $q_{1}$ = $-$ 1.6 $\times 10^{- 19}$ C

Charge of a proton, $q_{2}$ = 1.6 $\times 10^{- 19}$ C

Potential at infinity = 0

Potential energy of the system = Potential energy at infinity – Potential energy at a distance d

= 0 - $\frac{q_{1} q_{2}}{4 π ε_{0} d}$ , where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ == 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

(Since 1 eV = 1.6 $\times 10^{- 19}$ J)

Kinetic energy = $\frac{1}{2}$ of potential energy = $\frac{1}{2}$ $\times -$ 27.13 eV = -13.57 eV

Total energy = - 13.57 – (-27.13) = 13.57 eV

Therefore, the minimum work required to free the electron is 13.57 eV

In case of zero potential energy, $d_{1}$ = 1.06 Å = 1.06 $\times 10^{- 10}$ m

Potential energy of the system = Potential energy at $d_{1}$ - Potential energy at d

= $\frac{q_{1} q_{2}}{4 π ε_{0} d}$ $\times \frac{1}{1.6 \times 10^{- 19}}$ - 27.13 eV = $\frac{{(1.6 \times 10^{- 19})}^{2}}{4 π \times 8.854 \times 10^{- 12} \times 1.06 \times 10^{- 10}} \times \frac{1}{1.6 \times 10^{- 19}} -$ 27.13 = 13.56 -27.13 eV= -13.56 eV

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11 months ago

What is the fees of Marri Laxman Reddy Institute of Technology and Management?

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Mohit Shekhar

Contributor-Level 9

Marri Laxman Reddy Institute of Technology and Management offers three courses at UG and PG levels. The below-mentioned course fee is as per the official website/sanctioning body. It is subject to changes and hence, is indicative. The total tuition fee for all the courses is given below:

Courses	Tuition Fees
BTech	INR 3.2 Lacs - INR 4 lakh
MTech	INR 1.2 lakh
MBA	INR 60,000

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11 months ago

Invertis University Pharmacy

What is the final step for BPharm admission at Invertis University?

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Himanshu Singh

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The final step for BPharm admission at Invertis University is the counselling round, where selected candidates must participate for document verification and seat allotment. To confirm their admission, students are required to pay the admission fee. As per the official sources, the total tuition fee for the BPharm course is INR 5.6 lakh, while the hostel fee is INR 2.20 lakh.

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11 months ago

Physics Ncert Solutions Class 12th Electrostatic Potential and Capacitance

2.17 A long charged cylinder of linear charged density $λ$ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

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2.17

Let charge density of the long charged cylinder, with length L and radius r be $λ .$

Another cylinder of same length surrounds the previous cylinder with radius R.

Let E be the electric field produced in the space between the two cylinders.

Electric flux through the Gaussian surface is given by Gaussian’s theorem as

$φ = E (2 π$ dL) = $\frac{q}{ε_{0}}$ , where

$q = c h a r g e o n t h e i n n e r s p h e r e o f t h e o u t e r c y l i n d e r$

$ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$

Then, $φ = E (2 π$ dL) = $\frac{q}{ε_{0}}$ = $\frac{λ L}{ε_{0}}$

E = $\frac{λ}{{2 π d ε}_{0}}$

Therefore, the electric field in the space between the two cylinders is $\frac{λ}{{2 π d ε}_{0}} .$

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11 months ago

1.20. Round up the following upto three significant figures:

(i) 34.216

(ii) 10.4107

(iii) 0.04597

(iv) 2808

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1.20. The values after round up with three significant figures are:

(i) 34.2

(ii) 10.4

(iii) 0.0460

(iv) 2810

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11 months ago

1.19. How many significant figures are present in the following?

(i) 0.0025

(ii) 208

(iii) 5005

(iv) 126,000

(v) 500.0

(vi) 2.0034

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