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8 months ago

Class 11th Chemistry

While performing a thermodynamics experiment, a student made the following observation.

HCl + NaOH → NaCl + H₂O $Δ H$ = -57.3 kJ mol^?1

CH₃COOH + NaOH → CH₃COONa + H₂O $Δ H$ = -55.3 kJ mol^-1

The enthalpy of ionization of CH₃COOH as calculated by the student is _________kJ mol^-1. (nearest integer)

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Payal Gupta

Contributor-Level 10

$Δ H_{i o n i z a t i o n} o f C H_{3} C O O H = - 55.3 (- 57.3)$

= 2 KJ/mole

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8 months ago

Class 11th Chemistry

A sealed flask with a capacity of 2 dm³ contains 11g of propane gas. The flask is so weak that it will burst if the pressure becomes 2 MPa. The minimum temperature at which the flask will burst is _________ $° C$ . [Nearest integer]

(Given: R = 8.3 JK^-1 mol^-1. Atomic masses of C and H are 12u and 1u, respectively)

(Assume that propane behaves as an ideal gas.)

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Payal Gupta

Contributor-Level 10

V = 2 $l$ , n = 0.25 Moles, P = 2 * 106 Pa, R = 8.314 J/mol.k

PV = nRT

$2 * 1 0^{6} * 2 * 1 0^{- 3} = 0.25 * 8.314 * T$

T = 1654.7 $\approx$ 1655 K

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8 months ago

Class 11th Chemistry Some Basic Concepts of Chemistry

56.0L of nitrogen gas is mixed with excess of hydrogen gas and it is found that 20L of ammonia gas is produced. The volume of unused nitrogen gas is found to be _______L.

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Payal Gupta

Contributor-Level 10

$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$

t = 0 56 $l$ Excess 0

-10 $l$ +20 $l$

Final 46 $l$ 20 $l$

Vol. of N₂ (g) remained unreacted = 46 $l$

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8 months ago

Class 11th Physics Waves

Two waves executing simple harmonic motions travelling in the same direction with same amplitude and frequency are superimposed. The resultant amplitude is equal to the $\sqrt[]{3}$ time amplitude of individual motions. The phase difference between the two motions is _______(degree).

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Payal Gupta

Contributor-Level 10

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8 months ago

Class 11th Physics Thermodynamics

A Block of ice of mass 120g at temperature $0 ° C$ is put in 300g of water at $25 ° C .$ The xg of ice melts as the temperature of the water reaches $0 ° C .$ The value of x is _________.

[Use specific heat capacity of water = 4200Jkg^-1K^-1, Latent heat of ice = 3.5 * 10⁵Jkg^-1]

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Payal Gupta

Contributor-Level 10

Using concept of calorimeter, we can write

Gain in heat = Loss in heat

$x * L = m c Δ T \Rightarrow x * 3.5 * 1 0^{5} = \frac{300}{1000} * 25 * 4200$

$\Rightarrow x = \frac{300 * 25 * 4200}{1000 * 3.5 * 1 0^{5}} = \frac{3 * 25 * 42}{35 * 1000} = \frac{90}{1000} k g = 90 g m$

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8 months ago

Class 11th Physics

Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $\sqrt[]{x} m .$ The value of x is __________.

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Payal Gupta

Contributor-Level 10

$X_{c} = Y_{c} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{m * 0 + m * 0 + m * 3}{3 m} = 1 m$

$O C = X \sqrt[]{x_{0}^{2} + Y_{c}^{2}} = \sqrt[]{{(1)}^{2} + {(1)}^{2}} = \sqrt[]{2} m$

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8 months ago

Class 11th Physics

A particle is moving in a straight line such that its velocity is increasing at 5 ms^-1 per meter. The acceleration of the particle is _________ms^-2 at a point where its velocity is 20 ms^-1.

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Payal Gupta

Contributor-Level 10

$t a n θ = \frac{d v}{d x} = 5 (m / s) / m$

$\Rightarrow d v = 5 d x \Rightarrow \frac{d v}{d t} = 5 \frac{d x}{d t}$

$\Rightarrow a = 5 v = 5 * 20 = 100 m / s^{2}$

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8 months ago

Class 11th VSEPR Theory

What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.

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Satyendra Dhyani

Beginner-Level 5

Molecules usually form chemical bond through either sharing or through rtransfering the electrons. During covalent bonding the electron pairs shared between atoms to form covalent bond are called shared pair or bond pair. At the same time, the electron pair which is not involved in sharing is called lone pair of electrons.

For example: CH₄ has 4 bond pairs but H₂O has 2 bond pairs and 2 lone pairs.

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Class 11th VSEPR Theory

What is VSEPR theory in class 11th Chemistry?

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Esha Garg

Beginner-Level 5

The Valence Shell Electron Pair Repulsion (VSEPR) Theory is an algorithm developed to predict the molecular geometry of the compounds. The VSEPR theory predict the molecular shape based on the repulsion between electron pairs (bonding and lone) around the central atom. As per the NCERT Textbooks:

“According to this theory, the shape of a molecule depends upon the number of valence shell electron pairs around the central atom. Electron pairs repel each other and try to remain as far apart as possible to minimise repulsion, thus determining the geometry of the molecule.”

You can use this theory in primarily explaining the molecular str

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