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All posts related to Science

Tazo Mazo
Shared from BITSAT Mock - 8 on May 21, 2019 10:10 PM

The chemical added to leaded petrol to prevent the deposition of lead in the combustion chamber is

 
iso-octane
ethylene dibromide
tetraethyl lead
mercaptan
n-heptane
Please type your answer before submitting.
Tazo Mazo

Somebody plij batauuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

Kalpataru Dash
Shared from BITSAT Mock - 8 on May 21, 2019 9:46 PM

During acetylation of amines, what is replaced by acetyl group?

 
Hydrogen atom attached to nitrogen atom
One or more hydrogen atoms attached to carbon atom
One or more hydrogen atoms attached to nitrogen atom
Hydrogen atoms attached to either carbon atom or nitrogen atom
Please type your answer before submitting.
Kalpataru Dash

How A wrong

Sai Ganesh
Shared from BITSAT Mock - 3 on May 21, 2019 8:40 PM

The voltage 7.25 sin (300t) is applied to a series RLC circuit with R = 120 ohms, L = 0.140 H and C = 1.45 F. What is the impedance Z and the phase angle ?

 
120 ohms, -10.2o
306 ohms, -68.1o
2340 ohms, -25.7o
2260 ohms, -87o
Please type your answer before submitting.
Sai Ganesh

THE CALCULATIONS IN QUESTIONS LIKE THESE ARE TOO LENGTHY AND COMPLEX. CANT SOLVE IT IN TIME AND APPROXIMATIONS DON'T WORK EITHER. HMPH.

Mahaashri K
Shared from JIPMER Mock - 2 on May 21, 2019 8:06 PM

The condition for a uniform spherical mass m of radius r to be a black hole is:

[G = gravitational constant and g = acceleration due to gravity]

 
Please type your answer before submitting.
Mahaashri K

I did not understand this question. please explain it clearly

Kwopac Tm
Shared from BITSAT Mock - 4 on May 19, 2019 9:07 PM

The heat of reaction for C10H8 (s) + 12O2 (g) 10 CO2 (g) + 4H2O () at constant volume is -1228.2 kcal at 25 oC. The heat of reaction at constant pressure and the same temperature is

 
-1228.2 kcal
-1229.4 kcal
-1232.9 kcal
-1242.6 kcal
Please type your answer before submitting.
Kwopac Tm

C10H8 (s) 12O2 (g) 10 CO2 (g) 4H2O () We know, = Heat at constant pressure = Heat at constant volume = -1228.2 x 103 cal = Change in gaseous moles = 10 - 12 = -2 T = 298 K and R = 2 cal = -1228.2 x 103 (-2)(2)(298) = -1229392 cal = -1229.392 kcal