- 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
217 - Chemistry Questions Answers
show that (du/dv)s=T-sdp
question is incorrect
because (du/dv)s represents pressure and T is temperature
Which of the following hydrides has the highest M-H bond polarity
(a) AsH3 (b) BiH3 (c) SbH3 (d) None of these
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
The average kinetic energy of an ideal gas, per molecule in S.I. units, at 25 °C will be ..........
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
q. = Mass of solution of 1 molal glucose solution to get 0.2 molal glucose is.
a.200
b.300
c.236
d.108
I think question is incomplete, correct it first
acetic acid in a solvent undergoes 20% dissociation and 70% association, find vant hoff factor.
CH3COOH = CH3COO- + H+
1 0 0
0.8 0.2 0.2
CH3COO- + H+ = CH3COOH
0.2 0.2 0.8
0.06 0.06 0.8+0.14
0.06 0.06 0.94 sum = 1.06
so i = 1.06/1 = 1.06
calculate the molecular weight of cellulose acetate if its 0.2%wt/vol solution in acetone(sp. gravity 0.8) shows osmotic rise of 23.1mm against pure acetone at 27C.
By formula πV = nRT
π = hdg, n=w/M
given
0.2%wt/vol solution in acetone means w= 0.2 gm, V = 100 ml= 100/1000 litre = 0.1*10-3 m3 , h = 23.1/1000 m, d = 800 kg/m3 ,
now solve
BI DENTATE means having 2 pair of electrons to donate
With acids Al(OH)3 gives Al3+ and with bases it gives Al(OH)4- , but stable valency for Tl is 1 so it treats only with acids to form Tl+
Please check it is Ti or Tl, I think it is Tl belonging the family of Al
what is the ph of the solution formed by mixing 20 ml of 0.05 M H2SO4 with 5 ml of 0.45 M NaOH at 298 k ?
mili eq of acid = VN = 20*0.1 = 2
mili eq of base = VN = 5*0.45 = 2.25
total milieq = 2.25-2 = 0.25 of base in 25 ml
so N = 0.25/25 = 10-2
so pOH = 2 so pH = 12