- Organic Chemistry
- Aldehydes and Ketones
- Alkyl Halides, Alcohols and ethers
- Amines and other nitrogen compounds
- Aromatic Chemistry
- Carbohydrates, Amino acids, protein, Vitamin and Fat
- Carboxylic acids and its derivatives
- Chemistry in daily life
- General Mechanism in organic compounds
- Hydrocarbons
- Nomenclature and isomerism
67 - General Chemistry Questions Answers
HOW MUCH WATER IS TO BE ADDED TO DILUTE 10ml OF 10N HCl TO MAKE IT DECINORMAL ?
ANS 100ml
N1V1 = N2V2
so 10*10 = 0.1*v
or v = 1000 ml
so extra water added will be 990 ml
the normality of 10% (w/v) acetic acid is ?
1.7N
10 % w/v acetic acid means 10 gm. in 100 ml. soln.
so 1000 ml soln. contains 100 gm acetic acid
so 1 litre contains 100/60 moles of acetic acid
so M = 100/60 = 1.67
so N = M * basicity = 1.67*1 = 1.67
the ques which i have already submitted i.e. d ques in which we have to find min transitional energy in terms ofRH , IS THE ANS IS RH. PLS AGAIN TELL THAT SOLn so that i can understand more easily
for max transitional energy we will use
E = hcRH {1/12 - 1/∞2}
so E will be proportional to RH
oxide of E react with KOH & give compd like KMnO4 WHAT IS THE FORMULA OF OXIDE OF E
EO
E2O3
E2O7 OR ANY OTHER ANS (& IF ANY OTHER ONE THEN GIVE THAT ANS WITH SOLn)
- 2 EO2 + 4 KOH + O2 → 2 K2EO4 + 2 H2O
- and
- 3 K2EO4 + 2 CO2 → 2 KEO4 + 2 K2CO3 + EO2
- so oxide will be EO2
Q ONE LITRE MIXTURE OF CO &CO2 IS PASSED THRU RED HOT CHARCOAL IN A TUBE THE NEW VOL . BECOMES 1.2L WHAT % OF CO IN MIXTURE INITIALLY?
Q AN Aq SOLn CONTAINING 100 g OF DISSOLVED MgSO4 IS FED TO A CRYSTALLIZER WHERE 80% SALT CRY. OUT AS MgSO4.6H2O CRYSTAL HOW MANY gm OF HEXAHYDRATE SALT CRY. ARE OBTAINED FROM CRYSTALLIZER=?
ELECTROCHEM
Fe3+ + e- -----> Fe2+ = 0.77V
Fe --------.> Fe3+ + 3e- E = +0.04V
WHAT IS VALUE OF E* for Fe2+ +2e--------> Fe ?
CO2 + C (charcoal) -------> 2CO
let us consider that initially volume of CO2 is x litre and that of CO is (1-x) litre
so after the above written process net CO in final = 2x+1-x = 1.2 given
on solving x = 0.2 and 1-x = 0.8 now calculate %
MgSO4 --------------> MgSO4 .6H2O
it means 120 gm (mw of compound) gives 228 gm MgSO4 . 6H2O
so 80 gm will give 228*80/120 solve it
here 80 is used for 80% of 100 gm.
Fe+++ + e- -----> Fe++ E= +0.77 V
Fe -----> Fe+++ + 3e- E = +0.04 V
on adding we get
Fe -----> Fe++ + 2e-
so E for the new reaction = [1F(0.77) + 3F(0.04)]/2F = 0.89/2 = 0.445
A decapeptide (Mol.wt 796) on complete hydrolysis gives glycine(Mol.wt 75) , alanine and phenylalanine. Glycine contributes 47 % to the total weight of the hydrolysed products . The number of glycine units present in the decapeptide is
47 % of (796+162) = 958*47/100 = 450 approx
so no of glycene units = 450/75 = 6 approx
here 162 is taken for the weight of 9 water molecules required for breaking the decapeptides.
A compound contains 3.2% of oxygen . the minimum molecular weight of the compound is ?
atomic wt. of oxygen = 16
so m.w./100 = 16/3.2 or m.w. = 500