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  • 1 answers

Gaurav Seth 5 years, 4 months ago

Answer

(a) Molar mass of KI = 39 + 127 = 166 g mol - 1

20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.

That is,

20 g of KI is present in (100 - 20) g of water = 80 g of water

Therefore, molality of the solution = Moles of KI Mass of water in kg

= 20/166 0.08 m

= 1.506 m

= 1.51 m (approximately)

 

(b) It is given that the density of the solution = 1.202 g mL - 1

∴Volume of 100 g solution = Mass /  Density

= 100g 1.202g mL-1

= 83.19 mL

= 83.19 × 10 - 3 L

Therefore, molarity of the solution = 20/166 mol 83.19 × 10 - 3 L

= 1.45 M

 

(c) Moles of KI = 20/166 = 0.12 mol

Moles of water = 80/18 = 4.44 mol

Therefore, mole fraction of KI = Moles of KI (Moles of KI + Moles of water)

= 0.12 / (0.12+4.44)

= 0.0263

  • 1 answers

Yogita Ingle 5 years, 4 months ago

  • An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species. An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.
  • The formation of hydrogen fluoride is an example of a redox reaction. We can break the reaction down to analyze the oxidation and reduction of reactants. The hydrogen is oxidized and loses two electrons, so each hydrogen becomes positive.

H2​+2F→2HF

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Vansh Prajapati 5 years, 4 months ago

 Biochemical production of natural rubber

Vansh Prajapati 5 years, 4 months ago

Natural rubber consists almost entirely of the cis-1,4 structure and hence is, chemically, cis-1,4 polyisoprene. When the chain units in a polymer consists of the same isomer it is said to be stereoregular.N
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Hiral Solanki 5 years, 3 months ago

6.022*10^23

Vansh Prajapati 5 years, 4 months ago

6.022 * 10'''''''‹«'23
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Devil ? 5 years, 4 months ago

Hexagonal closed packing

Vansh Prajapati 5 years, 4 months ago

Hexagonal closed packing

Ashutosh Bind 5 years, 4 months ago

hcp is Hexagonal closed packing

Nikhil Badaliya 5 years, 4 months ago

Hexagonal closed packing

Aakansha Chhimwal 5 years, 4 months ago

Hexagonal closed packing
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Mishti ???? 5 years, 4 months ago

In easy language because they have the tendency to flow very slowly.. Regards. ??

Gaurav Seth 5 years, 4 months ago

The amorphous solids are called as super cooled liquid which allows molecules in the material to continue to move but in very very less large time and as it does not form a crystalline structure , or we can say they have a tendency to flow like liquids that's why they are called as pseudo solids or supercooled liquids.

Yogita Ingle 5 years, 4 months ago

The amorphous solids are called as super cooled liquid which allows molecules in the material to continue to move but in very very less large time and as it does not form a crystalline structure , or we can say they have a tendency to flow like liquids that's why they are called as pseudo solids or supercooled liquids.

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Gaurav Seth 5 years, 4 months ago

The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. In order to obtain the reaction order, the rate expression (or the rate equation) of the reaction in question must be obtained.

In order to determine the reaction order, the power-law form of the rate equation is generally used. The expression of this form of the rate law is given by r = k[A]x[B]y.

In the expression described above, ‘r’ refers to the rate of reaction, ‘k’ is the rate constant of the reaction, [A] and [B] are the concentrations of the reactants. The exponents of the reactant concentrations x and y are referred to as partial orders of the reaction. Therefore, the sum of all the partial orders of the reaction yields the overall order of the reaction.

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Yogita Ingle 5 years, 4 months ago

Methods of Preparation of haloalkanes

  • By halogenations of alkanes in presence of light, catalyst or heat:

  • By halogenations of alkenes with HX :

 

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? Yang ? 5 years, 4 months ago

There are haloalkanes that can undergo elimination in two different ways resulting in two different products. Alkenes with less number of hydrogens on the double-bonded carbon atoms are the preferred product. This process is known as Saytzeff’s rule. Hope It Helps You?✌
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Yogita Ingle 5 years, 4 months ago

Markovnikov proposed a rule called Markovnikov rule for the prediction of major product in the electrophilic addition of unsymmetrical alkenes. According to Markovnikov rule, the negative part of the adding molecule gets attached to that carbon atom which possesses a lesser number of hydrogen atoms.

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Yogita Ingle 5 years, 4 months ago

A volatile solute produces vapour at the boiling point of the solution. At the same temperature, they have higher vapour pressure than non-volatile solutes.

Examples: Alcohol, ether, mercury, and gasoline.

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Mamta Yadav 5 years, 4 months ago

The BCC unit cell consists of a net total of two atoms, the one in the center and eight eighths from the corners. In the FCC arrangement, again there are eight atoms at corners of the unit cell and one atom centered in each of the faces.
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Anshika Sharma 5 years, 4 months ago

When catalyst is added to the reaction then it lowers the activation energy of reactants and hence it increases the rate of reaction increases. Since on adding catalyst, enthalpy, entropy and temperature remain same therefore there is no effect on Gibbs free
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Ragini Yadav??? 5 years, 4 months ago

Thanks

Gaurav Seth 5 years, 4 months ago

Henry law states whenever solid is dissolved in liquid or gas is dissolved in liquid its solubility is determined by pressure and temperature. it states that partial pressure is directly proportional to mole fraction. ie . p=khx where kh is Henry law constant. Raoult law states that when two volatile liquid are mixed their vapour pressure is directly proportional to mole fraction ie p1 = p*1 x

  • 2 answers

? Yang ? 5 years, 4 months ago

1. Calculate the mass percentage of benzene () and carbon tetrachloride () if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Ans. Mass percentage of   = 15.28% Mass percentage of =  =84.72% Alternatively, Mass percentage of = (100 - 15.28) % = 84.72% 2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. Ans. Let the total mass of the solution be 100 g and the mass of benzene be 30 g. ∴Mass of carbon tetrachloride = (100 - 30) g = 70 g Molar mass of benzene () = (612 + 6)  = 78  ∴Number of moles of  = 0.3846 mol Molar mass of carbon tetrachloride ( ) = 112 + 4355 = 154  ∴Number of moles of  = 0.4545 mol Thus, the mole fraction of is given as:  = 0.458 3. Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2 .6H2O in 4.3 L of solution (b) 30 mL of 0.5 M diluted to 500 mL. Ans. Molarity is given by:  (a) Molar mass of Co(NO3)2 = 59 + 2 (14 + 3 16) + 618 = 291  Therefore, Moles of Co(NO3)2 = 30/291 mol = 0.103 mol Therefore, molarity  = 0.023 M (b) Number of moles present in 1000 mL of 0.5 M  ∴Number of moles present in 30 mL of 0.5 M  = 0.015 mol Therefore, molarity = 0.03 M 4. Calculate the mass of urea () required in making 2.5 kg of 0.25 molal aqueous solution. Ans. Molar mass of urea  = 60  0.25 molar aqueous solution of urea means: 1000 g of water contains 0.25 mol =  of urea = 15 g of urea That is, (1000 + 15) g of solution contains 15 g of urea Therefore, 2.5 kg (2500 g) of solution contains = = 36.95 g = 37 g of urea (approximately) Hence, mass of urea required = 37 g 5. Calculate (a)molality (b)molarity and (c)mole fraction of KI if the density of 20% (mass/mass) aqueous KI is. Ans. (a) Molar mass of KI = 39 + 127 = 166  20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution. That is, 20 g of KI is present in (100 - 20) g of water = 80 g of water Therefore, molality of the solution =  = 1.506 m = 1.51 m (approximately) (b) It is given that the density of the solution =  Therefore, Volume of 100 g solution =  = = 83.19 mL =  Therefore, molarity of the solution =  = 1.45 M (c) Moles of KI =  Moles of water =  Therefore, mole fraction of KI =  =  = 0.0263 6. , a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of  in water at STP is 0.195 m, calculate Henry's law constant. Ans. It is given that the solubility of in water at STP is 0.195 m, i.e., 0.195 mol of is dissolved in 1000 g of water. Moles of water =  = 55.56 mol ∴Mole fraction of , x =  =  = 0.0035 At STP, pressure (p) = 0.987 bar According to Henry's law: p= KHx  = 282 bar 7. Henry's law constant for  in water is 1.67 x 108 Pa at 298 K. Calculate the quantity of in 500 mL of soda water when packed under 2.5 atm pressure at 298 K. Ans. It is given that: KH = 1.67 x 108 Pa = 2.5 atm = 2.5 x 1.01325 x 105 = 2.533 x 105 Pa According to Henry's law: PCO2 = KHx x=  = 0.00152 We can write,  [Since, is negligible as compared to] In 500 mL of soda water, the volume of water = 500 mL [Neglecting the amount of soda present] We can write: 500 mL of water = 500 g of water = mol of water = 27.78 mol of water Now,  Hence, quantity of in 500 mL of soda water =  = 1.848 g 8. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. Ans. It is given that:  = 450 mm of Hg = 700 mm of Hg P total= 600 mm of Hg From Raoult's law, we have:  Therefore, total pressure,   600= (450-700)xA + 700 -100 = -250 xA xA = 100/250 = 0.4 Therefore,  = 1 - 0.4 = 0.6 Now,  =  = 180 mm of Hg  =  = 420 mm of Hg Now, in the vapour phase: Mole fraction of liquid A =   = 0.30 And, mole fraction of liquid B = 1 - 0.30 = 0.70 9. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea () is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering. Ans. It is given that vapour pressure of water, of Hg Weight of water taken,  Weight of urea taken,  Molecular weight of water,  Molecular weight of urea,  Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1. Now, from Raoult's law, we have:    Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173. 10. Boiling point of water at 750 mm Hg is . How much sucrose is to be added to 500 g of water such that it boils at . Molal elevation constant for water is. Ans. Here, elevation of boiling point = (100 + 273) – (99.63 + 273) = 0.37 K Mass of water,  Molar mass of sucrose ( ),  =  Molal elevation constant, Kb =  We know that:  = 121.67 g (approximately) Hence, 121.67 g of sucrose is to be added. 11. Calculate the mass of ascorbic acid (Vitamin C,) to be dissolved in 75 g of acetic acid to lower its melting point by.. Ans. Mass of acetic acid,  Molar mass of ascorbic acid  =  Lowering of melting point,  We know that:   = 5.08 g (approx) Hence, 5.08 g of ascorbic acid is needed to be dissolved. 12. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at. Ans. It is given that: Volume of water, V= 450 mL = 0.45 L Temperature, T = (37 + 273) K = 310 K Number of moles of the polymer,  We know that: Osmotic pressure,   = 30.98 Pa = 31 Pa (approximately)

Sanjana Dua?????❣️ 5 years, 4 months ago

Of which chapter and subject
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Anjali Kumawat 5 years, 4 months ago

Molar conductivity at infinite dilution of a strong electrolyte can be determined by extrapolation as linear graph is obtained. → For weak electrolyte molar conductivity at infinite dilution cannot be determined experimentally and by extrapolation because graph is not linear, maximum value of molar conductivity at infinite dilution cannot be obtained.
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Yash Bhadauria 5 years, 4 months ago

Melting point from nitrogen to arsenic increases & from arsenic it decreases upto bismuth because down the group as the size of the element increases the tendency of elements to form three covalent bonds increases

Yash Bhadauria 5 years, 4 months ago

Hi pooja

Yogita Ingle 5 years, 4 months ago

Melting point from nitrogen to arsenic increases & from arsenic it decreases upto bismuth because down the group as the size of the element increases the tendency of elements to form three covalent bonds increases (inert pair effect).Thus, there are weak forces of attraction between their atoms in solid state leading to lowering in melting point.

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Anjali Kumawat 5 years, 4 months ago

The best chemistry channel in you tube is Dr. Virendra Soni.
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Ragini Yadav??? 5 years, 4 months ago

Which obey Raoult law is called ideal solution or which not obey Raoult law called non ideal solution
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Ragini Yadav??? 5 years, 4 months ago

crystalline solid is anisotropic in nature but amorphous solids are isotropic in nature

Sushmita Yadav 5 years, 4 months ago

Crystalline solid are true solid.Amorphous solid are psuedo solid (super cooled liquid)

Sushmita Yadav 5 years, 4 months ago

Crystalline solid

Sushmita Yadav 5 years, 4 months ago

Hi

Royal Only Panda Ji 5 years, 4 months ago

I
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Rishu Singh 5 years, 4 months ago

Aman Singh class 12 C biology

Gaurav Verma 5 years, 4 months ago

Sorption and absorption is called adsorption

Yogita Ingle 5 years, 4 months ago

Adsorption is a process which involves the accumulation of a substance in molecular species in higher concentration on the surface. If we look at Hydrogen, Nitrogen and Oxygen, these gases adsorb on activated charcoal. Meanwhile, we have to note that adsorption is different from absorption. The two processes involve totally different mechanisms.

For the adsorption process, two components are required,

  • Adsorbate: Substance which is deposited on the surface of another substance. For example, H2, N2 and O2 gases.
  • Adsorbent: Surface of a substance on which adsorbate adsorbs. For example, Charcoal, Silica gel, Alumina.
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Yogita Ingle 5 years, 4 months ago

Crystalline solids can be classified into different categories depending upon the type of constituent particles and the nature of attractive forces operating between them.

Atomic Solid: In these solids the constituent particles are atoms. These closely packed atoms are held up by London dispersion forces. Some examples are crystals of noble gases. Such solids are very soft, possess very low melting points and poor conductors of heat and electricity.
Molecular Solids: In these solids, the constituent particles which pack up together are molecules of the substance. These molecules may be non – polar (dipole moment = 0) such as etc. or they may be polar (dipole moment > 0) like etc.  

Ionic Solids: In ionic solids, the constituent particles are ions of opposite charges. Each ion is surrounded by a definite number of ions of opposite charge.
Covalent Solids: In these types of solids the constituent particles are atoms of same or different elements connected to each other by covalent bond network.

Metallic Solids: In these type of solids, the constituent particles are metal atoms. The interparticle forces in these solids are metallic bonds. In the metallic crystals the metal atoms occupy the fixed positions but their valence electrons are mobile.

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Chandu B 5 years, 4 months ago

It is standard oxidation potential It is electrode potential of electrode where oxidation takes place measured at standard conditions
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? Yang ? 5 years, 4 months ago

Retention of Configuration: A process in which the relative configuration of an atom is retained. If the atom in question is a stereocenter, retention of configuration usually (but not always) changes R absolute configuration into R, and S into S. Hope It Helps You?
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Tannu Rao 5 years, 4 months ago

It is a type of adsorption in which accumulation of gas occurs on solid surface due to weak van der waal forces

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