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Grade: 12
        

please give detailed answer with full solution otherwise leave it for some good brains.


1.600ml of a mixture of O3 and O2 weighs 1g at NTP.calculate the volume of ozone in the mixture?


2.how many litres of liquid CCl4 (density=1.5g/cc) must be measured out to contain 1*10^25(or 10 raised to the power 25) CCl4 molecules?


3.which would be larger :an atomic mass unit based on the current standard or one based on the mass of Be - 9 atom  set at exactly 9 amu?how?


4.the enzyme carbonic anhydrase catalysis the hydrogen of CO2. this reaction : CO2 + H2O ----}  H2CO3 is involved in the transfer of CO2 from tissues to the lungs via the bloodstream . one enzyme molecule hydrates 10^6 molecules of CO2 per second.how many kg of CO2 are hydrated in one hour in one litre by 1*10^-6M enzyme?


5.22.4 litres of water vapour at NTP ,when condensed to water ,occupies an approximate volume of-?how?


6.is 1 g of water =1ml of water?give detail?


7.what is detonating gas?give the formula?

8 years ago

Answers : (1)

ankitesh gupta
63 Points
							

1.                ASSUME A MIXTURE OF O2 AND O3 THE TOTAL VOLUME IS 0.6LITRES AND TOTAL WT IS 1gm AT NTP

                   LET US CONSIDER ''a'' LITRES AND ''b'' gm OF O3 IN THE MIXTURE THEREFORE THE VOLUME AND WT OF O3 PRESENT IN MIXTURE WOULD BE ''0.6-a'' litres and '' 1-b '' gm AT NTP

                NOW MOLECULAR WT OF O2 AND O3 ARE 32 AND 48

    AT STP AND NTP MOLES=VOLUME(LITRES)/22.4

    ALSO MOLES = MASS/MOLECULAR MASS THEREFORE FOR O2 b/32=a/22.4...........................(1)

                     simillarly for O3   (1-b)/48=(0.6-a)/32.................(2)

                     SOLVING EQUATION (1) AND  (2) GET THE VALUE OF ''a'' and ''b'' AND THEN GET YOUR REQUIRED ANSWER

.............................................................................................................................................................................................................................................................................................................................

 

2.  NO. OF MOLECULES=NO. OF MOLES X AVAGADROS NO.                      AVAGADRO NO. IS 6.022 X 10^23

     NO. OF MOLES = 10^25/6.022 X 10^23...........................(1)

     NO. OF MOLES ALSO EQUAL TO= MASS / MOLECULAR MASS...............................(2)  MOLECULAR MASS OF CCL4 IS 12+4 X 35.5=142+12=154

   FROM USING (1) AND (2) GET THE VALUE OF MASS NOW DENSITY = 1.5gm/ml

        DENSITY = MASS/VOLUME...................(3) YOU KNOW THE VALUE OF MASS AND DENSITY SUBSTITUTE THOSE VALUES IN EQUATION (3) AND GET YOUR ANSWER WHICH WILL BE ''ml''

..............................................................................................................................................................................................................................................................

3. 

Be would be lighter on this scale than it is on the C-12 scale

however
nature would not be changed by this new definition
the ratio of masses between any two given atoms of elements would be the same
so
the measured masses of all atoms would be slightly less by this new definition
an atom twice as heavy as Be would now weigh 18 as opposed to 18.02 by C-12 standard

Of course, the mass of the atoms have not changed in an absolute sense
we have just redefined the standard of measurement
 

.............................................................................................................................................................................................................................................................................................................................

5. SINCE DENSITY OF WATER IS 1000KG/M^3   SO 22.4 L OF H2O GAS AT NTP MOLES=V(LITRES)/22.4  THE NO. OF MOLES @NTP IS 1 ALSO MOLES =WT/MOLECULAR MASS

MOLECULAR MASS OF H20 IS 18 THEREFORE WEIGHT OF H20 GAS WOULD BE 18 gm .SINCE H20 GAS IS CONVERTED TO H2O LIQUID THE MASS WOULD BE CONSERVED THEREFORE FORM DENSITY=MASS / VOLUME WE CAN CALCULATE VOLUME REMEMBER KEEP WT IN KG THE YOU WILL GET YOUR ANSWER IN M^3

..............................................................................................................................................................................................................................................................

 6.  SINCE WE KNOW DENSITY OF WATER IS 1000KG/M^3   

1gm=(1/1000)KG AND 1ML=(1/1000000)M^3

BY SUBSTITUTING THE VALUES OF MASS AND VOLUME THE DENSITY COMES OUT TO BE 1000KG/M^3 AND HENCE 1gm=1ml

................................................................................................................................................................................................................................................................

 

4TH ANSWER IS TOO BIG AND HARD TO EXPLAIN ON P.C AND THE LAST ONE I DONT KNOW SORRY 

 AND IF YOU ARE SATISFIED PLZZZZZZZZZZZZ APPROVE THE ANSWER BY CLICKING YES BUTTON ....................Smile

 

 

 

 

 

 

 

 


6 years ago
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