217 - Chemistry Questions Answers

a carbon sample from the frame of a picture gives 7 counts of C-14 per minute per gram of carbon . if freshly cut wood gives 15.3 counts of C-14 per minute. calculate the age of frame (t½ of C-14 = 5570 yrs)

(a) 6286 yrs

(b) 5527 yrs

(c) 5570 yrs

(d) 4570 yrs

Asked By: HIMANSHU MITTAL
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Joshi sir comment

use t = 2.303/k * log(15.3/7)

k = 0.6932/t1/2

the binding energy per nucleon for C-12 is 7.68 MeV and that for C-13 is 7.47 MeV. calculate the energy required to remove a neutron from C-13

(a) 4.5MeV

(b) 45MeV

(c) 0.45MeV

(d) 0.045MeV

Asked By: HIMANSHU MITTAL
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Joshi sir comment

7.47*13-7.68*12 =  97.11-92.16 = 4.5 Mev approx

1gm of  94Pu-239 is an α-emitter with a half life of 24400 yrs. given 1g sample of this plutonium how many α-particles will it emit per second?

(a) 2.27*10^9

(b) 2.27*10^10

(c) 2.27*10^11

(d) 2.27*10^12

Asked By: HIMANSHU MITTAL
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Joshi sir comment

dn/dt = kn

so dn = (0.6932/24400*3.1*107)*(1/239)*6.023*1023*1

solve 

a radioactive substance (M= 100g/mol) have t½ of 5 days. if today 0.125mg is left over what was its activity 20 days earlier?

(a) 522.8 Ci

(b) 1.93*10^7 Rd

(c) 1.93*10^13 Bq

(d) all of these

Asked By: HIMANSHU MITTAL
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Joshi sir comment

20 day means 4 half lifes so today 0.125mg means 0.125*24 = 2gm earlier 

so activity dn/dt = kn

solve and convert in required units

a uranium mineral contains U-238 and Pb-206 in the ratio 4:1 by weight. Calculate the age of mineral (t½ for U-238 = 4.5 *10^9 years)

(a) 1.45*10^9 yrs

(b) 16.6*10^9 yrs

(c) 16.6*10^10 yrs

Asked By: HIMANSHU MITTAL
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Joshi sir comment

U:Pb = 4:1

taking these ratio as mass

we get the initial mass of U = 4+(1*235/208)

and present mass of U = 4

now use the first order equation of radioactivity

for an endothermic rx, higher value of energy of activation will be

(a) less than enthalpy of rx.

(b) 0

(c) more than enthalpy of rx.

(d) equal to enthalpy of rx.

Asked By: HIMANSHU MITTAL
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Joshi sir comment

(c) numerically it will be (c)

What is the volume of a gas at NTP and STP ?wink

Asked By: AVANTIKA PURI
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Joshi sir comment

question is irrelevant, amount of gas is not given

how can we find log and antilogs without the help of log and antilogs tablessad

Asked By: AVANTIKA PURI
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Joshi sir comment

there are methods but these are not good for all cases

16 kg oxygen gas expands at STP isobarically to occupy double of its original volume. The work done during the process is 244 kcal. HOW????????

Asked By: AVANTIKA PURI
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Joshi sir comment

W = PdV = nRdT = (16*1000/32)*(8.31/4.18)*273 

now solve

What is the difference b/w metalloids & semimetals.

Asked By: SARIKA
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Joshi sir comment

A metalloid is a chemical element  that has properties that are in between or a mixture of those of metals and nonmetals and is consequently difficult to classify unambiguously as either a metal or a nonmetal. There is no standard definition of a metalloid, nor is there agreement as to which elements are appropriately classified as such. Despite this lack of specificity the term remains in use in chemistry literature.The six elements commonly recognised as metalloids are B, Si, Ge, As, Sb, Te

 

 A semimetal is a material with a very small overlap between the bottom of the conduction band and the top of the valence band. A semimetal thus has no band gap and a negligible density of states at the fermi level. A metal, by contrast, has an appreciable density of states at the Fermi level because the conduction band is partially filled.

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