217 - Chemistry Questions Answers

THE STRUCTURE OF HYDROGEN PEROXIDE IS NON-PLANER.HOW?

Asked By: SHIVANI MALIK
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Joshi sir comment

The structure of HYDROGEN PEROXIDE is like an open book. O-O bond is present in the joint and both H atom on the opposite side. Hybridisation will be sp3 for all bonds 

A mixture of FeO and Fe304 when heated in air to constant weight gains 5% in its weight. The % of FeO in the sample

Asked By: AMIT DAS
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Joshi sir comment

let FeO = x gm and FeO+Fe2O3= y gm so FeO = [72/(72+160)]*y and Fe2O= [160/(160+72)]*y

2FeO + 1/2O2  ---------->  Fe2O3

so amount of Fe2Oformed by x gm FeO  = 160x/144

similarly amount of Fe2Oformed by  [72/(72+160)]*y gm FeO = 160 [72/(72+160)]*y/144

so total Fe2O3 = 160x/144 + 160 [72/(72+160)]*y/144 +  [160/(160+72)]*y = 160x/144 + [160*72/232]y/144 + 160y/232

now according to the given condition {[160x/144 + [160*72/232]y/144 + 160y/232}/{x+y) = 105/100

now solve

Bond angle is maximum in which of the following - PCl3 , NCl3 , BCl3 & AsCl3 and why

Is it true that bond angle decreases down the group?

Asked By: AVANTIKA PURI
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Joshi sir comment

out of PCl3, NCl3, AsCl3 second has maximum bond angle of 107 degree whereas angle for BClis 120 degree

0.25 g lyophilic colloid is added to 100 ml gold solution to prevent the coagulation on adding 1ml 10% NaCl soln . What will be gold number of lyophilic colloid??

Asked By: SARIKA SHARMA
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Joshi sir comment

 “weight of the dried protective agent in milligrams, which when added to 10 ml of a standard gold sol (0.0053 to 0.0058%) is just sufficient to prevent a colour change from red to blue on the addition of 1 ml of 10 % sodium chloride solution, is equal to the gold number of that protective colloid.”

 

0.25 g lyophilic colloid is added to 100 ml gold solution

so 250 mg lyophilic colloid is added to 100 ml gold solution

so 25 mg lyophilic colloid is added to 10 ml gold solution

so answer will be 25

Most effective coagulant for a colloid soln of arsenic sulphide in water is

1) 0.1 M sodium phosphate

2) 0.1M zinc sulphate

3) 0.1 M zinc nitrate

4) 0.1 M aluminium chloride

Plzz annex explanation also

Asked By: SARIKA 19 Day ago
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Asked By: SARIKA SHARMA
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Joshi sir comment

its 4th option 

because As2Sis usually a negative colloid and positive ion is responsible for its coagulation and according to Hardy Schulze priciple more positive ion will be more effective

which of the following have maximum value of enthalpy of physisorption?

(a) H2

(b) N2

(c) H2O

(d) CH4

Asked By: HIMANSHU MITTAL
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(d) CH4 based on the value of energy which is 6 Kcal/mole for 0.1 milimole/gram and is greatest among the given 4

volume of gases H2 , CH4 , CO2 , and NH3 absorbed by 1g of charcoal at 288K are in the order

Asked By: HIMANSHU MITTAL
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Joshi sir comment

H2        4.7 cc per gm

CH4   16.2 cc per gm

CO   48 cc per gm

NH3   181 cc per gm

the radiation from a naturally ocuring radioactive substance as seen after deflection by a magnetic field in one direction are

(a) α rays

(b) β rays

(c) both (a) & (b)

(d) either α or β-rays

Asked By: HIMANSHU MITTAL
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Joshi sir comment

(d) because one will move opposite to the other

the amount of U-235 required per day to run a power house of capacity 50 MW (efficiency of nuclear reactor 75%. Assume energy liberated by fission of 1 U-235 atom is 200 Mev) is

(a) 15.8gm

(b) 28.1gm

(c)21.1gm

(d) none of these

Asked By: HIMANSHU MITTAL
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Joshi sir comment

let x gm is required

then according to the given condition 

(x/235)*6.023*1023*200*106*1.6*10-19 = 50*106*86400

solve

there are two radio nuclei A and B. A is α-emitter and B is a β-emitter, their decay constant are in the ratio 1:2. what should be the no. of atoms of A&B at time t=0, so that probability of getting α and β-particles are same at time t=0

(a) 2:1

(b) 1;2

(c) 1:4

(d) 4:1

Asked By: HIMANSHU MITTAL
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Joshi sir comment

its (b)

decay constant ratio = 1:2

so half life ratio = 2:1

so for getting probability of finding alpha and beta same ratio will be 1:2

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