Class 11th
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New answer posted
a year agoContributor-Level 10
(i) Energy of photon (E) = hν, where h= Plank's const, ν= Frequency
h= 6.626 * 10-34 J s ; ν = 3 * 1015 Hz = 3 * 1015 s-1
∴ E = (6.626 * 10-34 J s) * (3 * 1015 s-1) = 1.986 * 1018 J
(ii) Energy of photon (E) = hν = hc/λ, where λ= wavelength
h = 6.626 * 10-34 J s ; c = 3 * 108 m s-1 λ= 0.50 Å = 0.5 * 10-10 m.
∴ E = (6.626 * 10-34 J s) * (3 * 108 ms-1) / 0.5 * 10-10 m = 3.98 x 10-15 J.
New question posted
a year ago
New question posted
a year agoNew answer posted
a year agoContributor-Level 10
(i) One mole of methane (CH4) has = 6.022 * 1023 molecules
No. of electrons present in one molecule of CH4 = 6 + (4 x 1) = 10
No. of electrons present in 6.022 * 1023 molecules of CH4 = 6.022 * 1023 * 10
= 6.022 * 1024 electrons
(ii) (a)
Calculation of total number of carbon atoms
Gram atomic mass of carbon (C-14) = 14 g = 14 * 103 mg
14 g or 14 * 103 mg of carbon (C-14) has 6.022 * 1023 ato
New answer posted
a year agoContributor-Level 10
43. A process which can take place of its own or initiate under some conditions. For example: Common salt dissolves in water of its own.
New answer posted
a year agoContributor-Level 10
42. ?G = ?H – T?S
Where ?G = free energy change.
?H = enthalpy change.
?S = entropy change.
New answer posted
a year agoContributor-Level 10
(i) Mass of an electron = 9.10939*10−31 kg = 9.1 * 10-28 g
⇒ No. of electrons in 9.1 * 10-28 g = 1 electron
Therefore, No. of electrons in 1 g = 1/9.1 * 10-28 = 1.098*1027
(ii) One mole of electrons = 6.022 * 1023 electrons
Mass of 1 electron = 9.1 * 10-31 kg
Mass of 6.022 * 1023 electrons = (9.1 * 10-31kg) * (6.022 * 1023) = 5.48 * 10-7 kg
Charge on one electron = 1.602 * 10-19 C
Charge on one mole electrons = 1.602 * 10-19 * 6.022 * 1023 = 9
New answer posted
a year agoContributor-Level 10
41. Amount of heat required to vapourize one mole of a liquid at constant temperature and under standard pressure (1bar) is called its standard enthalpy of vapourization or molar enthalpy of vapourization, ΔvapH?.
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