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Answer:Wb = 12.4nB= 62g/molWA = 100gkf for water = 1.86 K kg/molkb for water = 0.512 K kg/mol 
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Find CBSE, ICSE State Boards Textbook Solutions of Class 8 to 12 of all subjects. Read NCERT Textbooks Questions and Answers & Find Previous year solved papers.
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At a particular spot of an object made of iron oxidation takes place and that spot behaves as an anode.At anode: Electrons released at anode spot moves throug
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The cell isMg(s) | Mg2+(aq) (0.130 M) || Ag+(aq) (0.0001 M) | Ag(s)The Nernst equation isE(cell) = E°cell-RT2FlnMg2+Ag+Ecell = 3.1
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(i) BaCl2(aq) + Na2SO4(aq) →BaSO4(s) + 2NaCl(aq)(ii) NaOH(aq) + HCl (aq) → NaCl(aq) + H2O(l)
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इस कहानी में लेखक ने गधे की सरलता और सहनशीलता की ओर हमारा ध्यान खींचा है। गधे को स्वभाव के कारण मूर्खता का पर्याय समझा जाता है। आमतौर पर हम गधे के लिए मूर्ख शब्द का प्रयोग करते हैं परन्तु उसके स्वभाव में सरलता और सहनशीलता भी देखने को मिलती है। गधा ही एक एक मात्र ऐसा प्राणी है जो
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Let AC = y and BC = x and x + y = k (k is a constant)Suppose θ be the angle between BC and AC.Let A be the area of a triangle. Then,
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A (4, 5, 10), B (2, 3, 4), C (1, 2, – 1) are three vertices of the parallelogram ABCD.
Direction-ratios of AB are 2 – 4, 3 – 5, 4 – 10 i.e.. – 2, –
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Given : Two concentric circles C1 and C2 of radii 13 cm and 8 cm respectively. AB is a diameter of the bigger circle (C1) and BD is a tangent to the smaller
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Electric field intensity at a point on the broad side-on position equatorial line:

Consider an electric dipole consisting of two equal but opposite charges -q and +q separated by a vector distance of 2l.

Let P be a point at a distance r from the center of the dipole O.

The electric intensity at P due to the dipole is the vector sum of the field due to the charge -q at A and +q at B.

The resultant intensity is the vector sum of EA and EB.

EA and EB can be resolved into two components.

The y-components of the field cancel each other because,

EA sin straight theta = EB sin s