67 - General Chemistry Questions Answers

NaNO₃  −−−−>    NaNO₂ + 1/2 O₂

molar mass of sodium nitrate = 85

so 85 gm will give 11200 ml O₂

so x gm will give 11200*x/85 = 2240x/17

on comparing with 44.8

we get 2240x/17 = 44.8

so x = 44.8*17/2240 = 17/50 = 34/100 = 0.34 gm

means 0.34 gm sodium nitrate is pure in 0.5 gm. sample so % purity = 0.34*100/0.50 = 68%

question is not complete, you should give the information about other elements present.

T α r/v    α is symbol for proportional

so f α v/r

v α 1/n

r α n²

now solve

O.N. change = 1-2 = -1

2 A are present so net change = 2

so e.w. = M/2

electronegativity of H = 2

and for all other elements in this question electronegativity is less than 2 

least for Bi 

so diff. for Bi is max so max polar

formula is {3RT/2}/N = 3kT/2

N and k are gas and Boltzmann constant

Consider about the amount of energy required. It can gain from its surrounding

0.75 mole water contains ...................................... 0.25 mole C₂H₅OH

so 0.75*18 gm contains ....................................... 0.25 mole

now calculate moles of alcohol in 1000 gm water. It will be molality

similarly

0.75 mole water contains ...................................... 0.25 mole C₂H₅OH

so 0.75*18 gm contains ....................................... 0.25*46 gm

so 0.75*18+0.25*46 gm solution contains ........................... 0.25 mole alcohol

so (0.75*18+0.25*46)*0.967 ml solution contains ............................. 0.25 mole alcohol

now calculate moles of alcohol in 1000 ml solution. It will be molarity

matter is one of the fundamental constituent of the universe

A hydrogen like system has ionisation enthalpy 11808KJ/mol

on converting it into eV we get [11808*1000]/[6.023*10²³*1.6*10^-19]  = 122.53 eV

now formula for this energy is [-Z²/n²]13.6 eV

for n = 1, Z will be 3