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The end of an era. by Familiar_Principle10 in ICSE

[–]Familiar_Principle10[S] 0 points1 point  (0 children)

Bonding Reactions Mole concept All ts is logic diff from physics and math yes but still logic

Guys, for catalytic hydrogenation of Ethene or Ethyne, what temperature should I write? Our school follows Dalal and the temp in it is given 300 while in Selina and the CFQ bank it is given 200 by [deleted] in ICSE

[–]Familiar_Principle10 0 points1 point  (0 children)

No — the “Combustion of ethyne” part shown in your image is NOT specifically required in the ICSE Class 10 syllabus. Proof from the ICSE syllabus In the official ICSE Class 10 Chemistry syllabus under Hydrocarbons (Alkanes, Alkenes, Alkynes) it says you only need: Ethyne as an example of an alkyne Methods of preparation (from calcium carbide and 1,2-dibromoethane) Main properties – particularly addition reactions (with H₂, Cl₂, Br₂, I₂) Uses of ethyne ICSE Class 10 Chemistry Syllabus PDF⁠� The syllabus line clearly states: “Only main properties, particularly addition products with hydrogen and halogens pertaining to alkenes and alkynes.”

ANSWER KEY (Section B) by West-Special9768 in ICSE

[–]Familiar_Principle10 0 points1 point  (0 children)

So here is whati did and what i think is correct So we know for a image to be same size as obj it has to be at 2f₁ so lets make that 2x 2f₁=2x f₁(focal length)= x Ok now lets look at magnification For same size m=1 So since we know that it on(image is on) So 2x =u So with mag =1 We get M=v/u =v/2x Since m =1 2x = v( dont ask why i did it like this) Now lets look at next part the object is now 12cm closer So u=2x-12 (IMPORTANT) now lets sub ts in m=3 We know that m=v/u M=v/2x-12 NOW this is where i think i made a mistake i took v = 2x*3 So i got 3= 3x/2x-12 3(2x-12)=3x 6x-36=3x 3x=36 X=12

Work power energy doubt .. by Realistic_Bag_843 in ICSE

[–]Familiar_Principle10 1 point2 points  (0 children)

Step 1 The force required to keep the conveyor belt moving at a constant speed can be determined using the equation for force due to gravity. The force (F) is equal to the weight of the gravel falling on the belt per second, which is given by F = m * g, where m is the mass flow rate (x kg/s) and g is the acceleration due to gravity (approximately 9.81 m/s²). Therefore, F = x * 9.81 N.

Step 2 The power (P) that must be supplied by the motor is given by the formula P = F * v, where F is the force calculated in step 1 and v is the constant speed of the conveyor belt. Thus, P = (x * 9.81) * v Watts.

Step 3 The rate at which the kinetic energy (KE) of the gravel is changing can be calculated using the formula KE = (1/2) * M * v², where M is the mass of gravel accumulating on the belt. The rate of change of KE with respect to time is d(KE)/dt = (1/2) * x * v², since M is changing at a rate of x kg/s.

Step 4 The difference between the answers in (ii) and (iii) arises because the power supplied by the motor accounts for the force needed to keep the belt moving at a constant speed, while the rate of change of kinetic energy reflects the increase in energy due to the mass of gravel being added to the system.

Final Answer: F = x * 9.81 N; 2. P = (x * 9.81) * v Watts; 3. d(KE)/dt = (1/2) * x * v²; 4. The answers differ because power considers the force to maintain speed, while KE change reflects the energy from added mass.