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Wed July 30, 2014  

Ans: Given: P 1 = 800 mm of Hg
V 1 = 10000 L
T 1 = -3 o C = 273 - 3 = 270 K

P 2 = 400 mm of Hg
V 2 = Volume required for filling gases with New temp and pressure condition= ?
T 2 = 273 K

Let us apply combined gas law, We get,

P 1 V 1 /T 1 = P 2 V 2 /T 2 

V 2 = P 1 V 1 T 2 /T 1 P 2  

V 2 =(800 10000 × 273)/(270 × 400) = 20222 L of gas is to be filled

Thus no. of cylinders required = 20222/10                                           = 2022 (approx.) ......Ans.

Wed July 30, 2014  

Ans: Isotonic solutions:  If there are two solutions and the concentrations of solutes in both the solutions are same, then they are isotonic solution. Hypertonic solutions and Hypotonic solutions:  Consider, there are two solutions A and B. When solution A has higher concentration of solute than solution B, then solution A is called hypertonic solution. When solution A has lower concentration of solutes than solution B, then solution A is called hypotonic solution.  

Wed July 30, 2014  

Ans: The distance-time graph of a train standing on a platform will be a straight line on the time axis. Since, the train is stationary it does not cover any distance.  

Tue July 29, 2014  

Ans: At high altitude, atmospheric pressure is lower than that at sea level. At high altitude, atmospheric temperature decreases as pressure decreases.

Tue July 29, 2014  

Ans: F r o m space t h e space b a s i c space r e l a t i o n space b e t w e e n space i n t e g r a t i o n space a n d space d i f f e r e n c i a t i o n integral f to the power of apostrophe open parentheses x close parentheses space d x space equals space f open parentheses x close parentheses plus C  S o space i f space f to the power of apostrophe open parentheses x close parentheses equals 0 comma space w h i c h space m e a n s space w e space a r e space d i f f e r e n t i a t i n g space e i t h e r space a space c o n s tan t space t e r m space o r space z e r o space w i t h space r e s p c t space t o space x space w h i c h space i s space g i v i n g space u s space z e r o. space  rightwards double arrow A s space p e r space a b o v e space d e f i n i t i o n comma space w e space g e t space f left parenthesis x right parenthesis space a s space e i t h e r space z e r o space o r space s o m e space c o n s tan t space f u n c t i o n. space

Tue July 29, 2014  

Ans: A strong acid will react with a weak base to form an acidic (pH < 7) solution.

Tue July 29, 2014  

Ans: I n space t h e space g i v e n space t r i a n g l e comma space i f space w e space j o i n space A comma space B space a n d space C space w i t h space O space a n d space c o n s i d e r space t r i a n g l e s increment O C E comma space O C squared equals O E squared plus C E squared increment O B D comma space O B squared equals O D squared plus B D squared increment O A F comma space O A squared equals O F squared plus A F squared  A d d space a l l space t h e space a b o v e space t h r e e space e q u a t i o n s comma space w e space g e t rightwards double arrow O C squared plus O B squared plus O A squared equals O E squared plus C E squared plus O D squared plus B D squared plus O F squared plus A F squared rightwards double arrow O C squared plus O B squared plus O A squared minus O D squared minus O E squared minus O F squared equals A F squared plus B D squared plus C E squared  C o n s i d e r space f o l l o w i n g space t r i a n g l e s comma space increment O C D comma space O C squared equals O D squared plus C D squared increment O B F comma space O B squared equals O F squared plus B F squared increment O A E comma space O A squared equals O E squared plus A E squared  A d d space a l l space t h e space a b o v e space t h r e e space e q u a t i o n s comma space w e space g e t rightwards double arrow O C squared plus O B squared plus O A squared equals O D squared plus C D squared plus O F squared plus B F squared plus O E squared plus A E squared rightwards double arrow O C squared plus O B squared plus O A squared minus O D squared minus O E squared minus O F squared equals A E squared plus C D squared plus B F squared

Tue July 29, 2014  

Ans: Earth's gravity attracts all objects in it. All objects are pulled towards the Earth by the same amount of force. However, gravity is a very weak force. Objects can overcome gravity with the help of various external factors. Example: 1) A feather falls slowly as air resists its fall.              2) A book held in hand cannot fall down as the hand helps in overcoming gravity.   When water evaporates, the water molecules gains kinetic energy and the temperature rises. This causes the air containing vapour to become lighter than the colder air above. Hence, the cold air from above falls down and the air containing vapour rises up. Also, the negligible mass of vapour  makes it easier for air to rise up and overcome gravity.

Tue July 29, 2014  

Ans: Photosynthesis is a physiological process by which plant cells containing chlorophyll produce food in the form of carbohydrates by using carbon dioxide, water and light energy. Oxygen is released as a by-product.   Chemical equation for photosynthesis is as follows:  

Tue July 29, 2014  

Ans: Dear nand030599@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
In case of multiple questions within a query, please post each question individually and let us know where you are getting stuck so that we would be able to explain things better.   however, answer to your first query is,   A polar molecule has a net dipole as a result of the opposing charges (i.e. having partial positive and partial negative charges) from polar bonds arranged asymmetrically.  Water (H 2 O) is an example of a polar molecule since it has a slight positive charge on one side and a slight negative charge on the other. The dipoles do not cancel out resulting in a net dipole. Due to the polar nature of the water molecule itself, polar molecules are generally able to dissolve in water.  A molecule may be  non-polar  either when there is an equal sharing of electrons between the two atoms of a diatomic molecule or because of the symmetrical arrangement of polar bonds in a more complex molecule. Most non-polar molecules are water-insoluble (hydrophobic) at room temperature. However, many non-polar organic solvents, such as turpentine, are able to dissolve polar substances.      Regards

Topperlearning Team.

Tue July 29, 2014  

Ans: Pure, crystalline solids have a characteristic melting point . It is the temperature at which the solid melts to become a liquid.

Tue July 29, 2014  

Ans: Froth Floatation Process Principle : The principle of froth floatation is that sulphide ores are preferentially wetted by pine oil, whereas the gangue particles are wetted by water. Collectors are added to enhance the non-wettability of the mineral particles. Examples of collectors are pine oil, fatty acids and xanthates. Froth stabilisers are added to stabilise the froth. Examples of froth stabilisers are cresols, aniline. If two sulphide ores are present, then it is possible to separate the two sulphide ores by adjusting the proportion of oil to water or by adding depressants. Example: For an ore containing ZnS and PbS, the depressant used is NaCN. It selectively prevents ZnS from coming to froth but allows PbS to come with the froth. Method: This method employs a mixture of water and pine oil which is made to froth in a tank to separate sulphide ores. The differences in the wetting properties of the ore and gangue particles separate them. A mixture of water, pine oil, detergent and powdered ore is first taken in a tank. A blast of compressed air is blown through the pipe of a rotating agitator to produce froth. The sulphide ore particles are wetted and coated by pine oil and rise up along with the froth (froth being lighter). The gangue particles wetted by water sink to the bottom of the tank (water being heavier). Sulphide being more electronegative attracts the covalent oil molecules. The gangue being less electronegative is attracted by the water. The froth containing the sulphide ore is transferred to another container, washed, and dried.

Tue July 29, 2014  

Ans: We know that a current carrying wire produces magnetic field around it.When a current is passed through a conducting wire a magnetic field is produced around it. On mapping the magnetic field by means of a compass needle or iron filings we find that the magnetic field lines forms concentric circles around the wire with their plane perpendicular to the straight wire and with their centre lying on the wire.And it is found that the magnetic field lines at points near the wire are more and lie closer to each other while at points far away from the wire the magnetic field lines are lesser. Thus we can conclude that the magnetic field lines are inversely proportional to distance,as we move away from the current carrying wire the concentric circles around it representing magnetic field lines becomes larger and larger indicating the decreasing strength of magnetic filed.

Tue July 29, 2014  

Ans: मित्रवर आपके प्रश्न के लिए धन्यवाद। "विशेषज्ञ से पूछो" ("Ask the Expert" ) की सेवा छात्रों के संदेहों को स्पष्ट करने के लिए है न कि लंबे उत्तर प्रदान करने के लिए । दुर्भाग्य से आपके सवाल का जवाब हमारी प्रदान सेवाओं के दायरे से बाहर हो जाता है । यदि आपने इस विषय पर जवाब तैयार किया है तो एक विशेष मामले के रूप में हम उसकी समीक्षा अवश्य करेंगे । इस प्रश्न के स्थान पर आप एक और प्रश्न पूछ सकते हैं । अध्ययन के लिए हमारी हार्दिक शुभकामनाएँ ।

Tue July 29, 2014  

Ans: b) Graphite is an element which is composed of only carbon atoms. Al other are compounds.

Tue July 29, 2014  

Ans: Birds sitting on a wire (with one leg or both legs) do not feel any current through their body as there is no potential difference on the body as it is not connected to ground. Hence, nothing will happen to the birds.

Tue July 29, 2014  

Ans: Colloids are a type of solution, in which the size of the solute particles is intermediate between solutions and suspensions. When observed through a high power microscope the particles in a colloid appear to be floating over the surface of the liquid. This shows that colloids are heterogeneous in nature though they appear to be homogeneous.

Tue July 29, 2014  

Ans: C o n s i d e r space t h e space g i v e n space e x p r e s s i o n space 2 cross times 7 n plus 3 cross times 5 n minus 5 A s s u m e space t h a t space space n equals 1 2 cross times 7 n plus 3 cross times 5 n minus 5 equals 2 cross times 7 cross times 1 plus 3 cross times 5 cross times 1 minus 5 equals 14 plus 15 minus 5 equals 24 T h u s comma space t h e space g i v e n space e x p r e s s i o n space i s space t r u e space f o r space n equals 1 N o w space l e t space u s space a s s u m e space t h a t space t h e space g i v e n space e x p r e s s i o n space i s space d i v i s i b l e space b y space 24 space f o r space n equals k W e space n e e d space t o space p r o v e space t h a t space t h e space g i v e n space e x p r e s s i o n space i s space d i v i s i b l e space b y space 24 space f o r space n equals k plus 1 C o n s i d e r space t h e space g i v e n space e x p r e s s i o n space f o r space n equals k plus 1 2 cross times 7 open parentheses k plus 1 close parentheses plus 3 cross times 5 open parentheses k plus 1 close parentheses minus 5 equals 14 k plus 14 plus 15 k plus 15 minus 5 equals 29 k plus 24 I t space i s space c l e a r space f r o m space t h e space a b o v e space e x p r e s s i o n space t h a t space i t space i s space n o t space d i v i s i b l e space b y space 24.

Tue July 29, 2014  

Ans: Gold and silver are the least reactive metals in the rectivity series, hence can not bond with oxygen easily. Therefore they both can not get corroded easily.

Tue July 29, 2014  

Ans: 1 amu = 1.66 x 10 -24 grams 126 amu = 126 x 1.66 x 10 -24               = 209.16 x 10 -24 grams 1 mole HNO 3 = 63.01 g Hence 209.16 x 10 -24 grams = 3.32 x 10 -24 moles   1 mole = 6.023 x 10 23 molecules hence 3.32 x 10 -24 moles = 19.99 x 10 23  molecules   1 molecule of HNO 3 contains 5 atoms. Hence, 19.99 x 10 23  molecules = 99.96 x 10 23 atoms   Therefore, 126 amu of HNO3 contains 99.96 x 10 23 atoms.
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