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(d) The products of the concerned reaction react each other forming back the reactants. 

XeF₆ +3H₂O →XeO₃ +6HF .

The ccp lattice is formed by the atoms of the element N.

Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element N.

According to the question, the atoms of element M occupy 1/3^rdof the tetrahedral voids.

Therefore, the number of atoms of M is equal to 2x1/3 = 2/3^rd of the number of atoms of N.

Therefore, ratio of the number of atoms of M to that of N is M: N =2/3: 1 = 2 : 3

Thus, the formula of the compound is M₂N₃.

a. Al 

b. Au 

c. Cu 

d. Zn

Avogadro’s number (N) is equal to 6.023 x 10²³ .

Correct option is C) faintest

Answer: Protein-mediated mechanism.

(i) w = –P_ext × ΔV

As expansion taks place into the evacuated bulb, i.e.,

against vacuum, P_ext = 0. Henc, w = 0.

For adiabatic process, q = 0 

∴ ΔU = q + w = 0 + 0 = 0.

(ii) V = V₂ – V₁ = 5 – 1 = 4 litres

P = 1 atm ∴ w = – PV

 = –1 × 4 litre atm = – 4 litres atm

 = – 4 × 101.3 J = – 405.2 J (1 L – atm = 101.3J) 

The negative sign implies that the work is done by the system.

For adiabatic process, ΔU = q + w = 0 – 405.2 J = – 405.2 J.

Aim : C (s) + 2H₂ (g) + 1/2 O₂ (g)→ CH₃ OH,Δ _fH°(l)

Eqn. (ii) + 2 × Eqn. (iii) – Eqn. (i) gives the required eqn. 

with ΔH= – 393 + 2 (–286) – (– 726) kJ mol^–1 =

(c) NH₄OH is neutralised by CH₃ COOH 

(d) ΔH <ΔE only if the number of moles of products is less than the number of moles of the reactants.

(c) graphite is stabler than diamond

Remember, enthalpy of combustion is always negative we are given

(i) C (graphite) + O₂ (g) → CO₂ (g) ; Δ_c H° = – 393.5 kJ mol^–1

(ii) C (diamond) + O₂ (g) → CO₂ (g) ; Δ_c H°  = – 395.4 kJ mol^–1

We aim at C(graphite) → C(diamond), Δ_trans H°= ?

Subtracting eqn. (ii) from eqn. (i), we get

 C(graphite – C(diamond) → 0;

Δ_r H° = – 393.5 – (–395.4) = + 1.9 kJ 

or C(graphite) → (diamond) ; Δ_trans H = + 1.9 kJ

Given:

The LCM of two numbers is 6 times their HCF.

One of them is 45 and the sum of HCF and LCM is 315.

Formula used:

LCM × HCF = N₁ × N₂

LCM = least common factor, HCF = Heighest common factor

N₁ = first number, N₂ = second number

Calculation:

According to the question,

N₁ = 45

LCM = 6 HCF

LCM + HCF = 315

⇒ 6 × HCF + HCF = 315

⇒ 7 HCF = 315

⇒ HCF = 315/7

⇒ HCF = 45

⇒ LCM = 6 × HCF 

⇒ LCM = 6 × 45 = 270

Then,

LCM × HCF = N₁ × N₂

⇒ 270 × 45 = 45 × N₂

⇒ N₂ = 270

∴ The other number is 270.

w = – 2.303 n RT log V₂/V₁

– 4754 = – 2.303 × n × 8.314 × 300 log 20/10.

This given n = 2.75 moles.

The Correct option is (b) 10 kJ 

L: Conformers,

M: Regulators

i. To regulate the body temperature – M/Regulators 

ii. To keep their body temperature constant by behavioural response for coping with variations in environment – L/Conformers

Correct option is (1) ΔH = ΔU - ΔPV

ΔH = ΔU + ΔPV

(C) (Fallopian tubes - Amnion)

Fossil fuels have the following, environmental effects:
(i) Air pollution: Burning of fossil fuels releases oxides and sulphides in the air and many other harmful gases like carbon monoxide, sulphur dioxide, etc. These cause various health problems and also lead to; acid rain which further affects water and soil resources.
(ii) Greenhouse effect: On burning fossil fuels, a large amount carbon dioxide is released into the atmosphere. This is a greenhouse gas and does not allow the sun rays reflected from the earth surface to escape into the atmosphere. Thus, increasing the temperature of the atmosphere. This is called greenhouse effect and results in global, warming. 

Following steps can be taken to minimize pollution:
(i) Use of smokeless appliances.
(ii) Use of refined technology to increase the efficiency of the combustion process and to reduce the escape of harmful gases into the atmosphere.
(iii) Judicious use of energy.

Relationship between CP and CV for an Ideal Gas

From the equation q = n C ∆T, we can say:

At constant pressure P, we have qP = n CP∆T

This value is equal to the change in enthalpy, that is, q_P = n CP∆T = ∆H

Similarly, at constant volume V, we have q

V = n CV∆T

This value is equal to the change internal energy, that is, q

V = n CV∆T= ∆U

We know that for one mole (n=1) of ideal gas,

∆H = ∆U + ∆(pV )

= ∆U + ∆(RT )

= ∆U + R∆T

Therefore, ∆H = ∆U + R ∆T

Substituting the values of ∆H and ∆U from above in the former equation,

CP∆T = CV∆T + R ∆T

Or CP = CV + R

Or CP – CV= R

A. False It is formed in choroid plexuses by active and passive processes. 

B. False Most of the brain’s nutrition comes from the blood. 

C. True This is its main function and it does so through cushioning and buoyancy. 

D. False Its higher pressure allows drainage by filtration to the dural venous sinuses via the arachnoid villi. 

E. True This is about four times its volume. 

A. True This ‘resting membrane potential’ is about 90 mV. 

B. True These are released on excitation. 

C. False A single neurone supplies a group of muscle fibres. 

D. True It becomes more excitable as its membrane potential approaches the firing threshold (about 70 mV). 

E. False Decreasing extracellular Ca²⁺ increases excitability and may lead to spontaneous contractions (tetany), possibly by increasing sodium permeability.

A. True Both have highly organized actin and myosin filaments. 

B. True The iris is an example of multiunit smooth muscle. 

C. True Isolated hearts and gut segments show spontaneous activity in the organ bath. 

D. True Only the pace maker cells in the heart have unstable membrane potentials. 

E. True The Frank–Starling relationship describes this with respect to cardiac muscle.

A. False Skeletal muscle has no pacemaker cells and shows no spontaneous activity. 

B. False Calcium is released from intracellular stores when it contracts. 

C. False The filaments do not shorten but slide together over one another. 

D. True There is greater overlap of the actin and myosin fibrils. 

E. True Moderate stretch increases contraction strength as in the heart.

A. True These paralyse muscle by blocking transmission at neuromuscular junctions. 

B. True Lower motor neurones also paralyse skeletal muscle. 

C. False Loss of supraspinal influences results in spasticity of the affected muscles. 

D. True The cerebellum helps to maintain normal muscle tone. 

E. False These increase spindle sensitivity to stretch and hence muscle tone.

A. True The intrinsic ‘myogenic’ response in smooth muscle opposes stretch; skeletal muscle requires a nerve reflex arc for this type of response. 

B. True It continues to contract due to local pacemakers. 

C. True They show spontaneous depolarization between contractions. 

D. False In smooth muscle, actin and myosin filaments occur but are less obvious on microscopy. 

E. True Smooth muscle has a less well-developed sarcoplasmic reticulum.

A. True The more precise the movement required, the fewer the fibres supplied by one motor neurone. 

B. False Somatic motor neurones conduct at 60–120 m/sec; autonomic at about 1 m/sec. 

C. False Fast-conducting fibres are large and myelinated. 

D. False Impulses carried by somatic motor neurones are excitatory to skeletal muscle. 

E. False Some carry impulses to inhibitory (Renshaw) cells in the anterior horn. 

A. False The synapse may amplify or attenuate the signal. 

B. True The uptake of Ca²⁺ by the nerve ending facilitates release of transmitter. 

C. True The neurotransmitter contained in the vesicles is released by exocytosis. 

D. False The impulses may be inhibitory; even if they are excitatory, the post-synaptic neurone may be strongly inhibited by inputs from other pre-ganglionic neurones. 

E. True Neurotransmitter is thought to be synthesized and packaged in the Golgi apparatus in the neurone cell body before travelling down the axon to the nerve terminals.

A. False No such potentials have been recorded here. 

B. True The hyperpolarization of the anterior horn cell reduces the likelihood of the cell firing action potentials by moving the membrane potential further from the firing threshold. 

C. False Its amplitude is about 5 mV and its duration about 5 msec. 

D. True It may be produced by increased permeability to potassium or to chloride ions. 

E. True In this way the post-synaptic cell integrates the various signals it receives.

d. It is used to make bhakri and khakhras

It is used to make bhakri and khakhras is incorrect statement for pulses.

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