216 - Chemistry Questions Answers

M of H₂SO₄ = 1/200

so [H⁺] = 2*1/200 = 1/100 = 10-2

now calculate pH

no of hydroxide ion in pure water = 10^-7

so according to the given condition no. of hydroxide ions in solution = 2*10^-7

so pOH = -log2*10^-7

so pH = 14 - pOH 

HCN ------------------->  H⁺ + CN^-

0.1                                 0          0

0.1-x                              x           x

pH = 5.2

so -log[H⁺] = 5.2 

solve for [H⁺], it will give x

then calculate K_a

2 heat of atomization of C + 4 heat of atomization of H - 4 bond energy of C-H - 1 bond energy of C=C = heat of formation of C₂H₄

2*171 + 4*52.1 - 4*99.3 - x = 12.5

solve

n is the no. of electrons in anode or cathode reaction.

If no. of electrons are different then make these no. common

for ex. if reactions are 

             Al ---------------> Al⁺⁺⁺ + 3e^-

and     Cu⁺⁺ + 2e^- ------------> Cu

then multiply (1) to 2 and (2) to 3

after that electrons will be 6 in both reactions so n = 6

b depends on size so on the basis of size order will be AlCl₃>Cl₂>H₂O

This answer is given by Sarika and is correct answer

I₂ is non polar molecule so it will be ion-induced dipole interactions 

N₂, SO₂, ClF3, K₂O, LiF

N₂, SO₂, ClF3, K₂O, LiF

This order is due to difference of electronegativity of atoms in a molecule

more difference more will be the ionic character

dipole moment = charge * bond length

now solve

f/c = R[ 1/2² - 1/3² ]

R and f are Rydberg constant and frequency