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The balanced reaction is 4Al + 3O2= 2Al2O3

​3moles of O2requires 4 moles of Al0.5 moles of O2requires = 4 X 0.5/3 = 2/3 moles of Al

Mass of 2/3 mol of Al = 2/3 mol X 27g/mol= 18 g

answer

when l = 2 , the orbital is D and no..of electrons for d orbital is = 2(2l+1) = 2(5) = 10

1) n = 1 , l = 0 ====> 1s 2) n= 2 and l = 1 ====>>> 2P.

On addition towaterthe Na+ section of NaCl is attracted to the oxygen side of thewater molecules, while the Cl- side is attracted to the hydrogens' side of thewatermolecule. This causes the sodium chloride to split inwater, and the NaCldissolvesinto separate Na+ and Cl- atoms.

The common feature is that they all are transition elements and they all have 7 valenceelectrons in their outermost shell.

option d is the correct answer

option D

Acid - Base indicators (also known as pH indicators) aresubstanceswhich change colour with pH. They are usually weak acids or bases, which when dissolved in water dissociate slightly and form ions.

option D is the correct answer

The minimum energy required for molecules to enter into chemical reaction is called (a)Threshold energy (b) Activation energy

The minimum energy required for molecules to enter intochemical reactionis called a)threshold energy b)activation energy

Answer :iii) AgNO3Explanation : AgNO3 will finally trun blue as silver nitrate react with copper to form hairlike crystal of silver metal and blue solution if copper nitrate will be formed 2AgNO3+Cu-->Cu(NO3)2+2Ag

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Basically the request is for deriving the expression for Kinetic Energy.

It is given that mass 'm' at rest whose velocity changes to 'v' in 't' seconds .

It means there is acceleration and hence a force F is acting on the body.

First we will compute the acceleration = a from the relation V= U+at

In our case U the initial velocity = 0 ∴ V = at ⇒ a = V/t

Force = m*a = m*V/t

Distance moved is (1/2)*a*t² = (1/2)*(V/t)*t² = Vt/2

Work done on the body :

(which is stored as Kinetic energy in it) = KE = F*s = (m*a)*s = m*(V/t)* (Vt/2)

= (1/2)*m*V²

Acetone is highly sublimable. it converts into vapour at a fast rate. this increases the liquid molecules around it in vapour phase thus our palm feel cold.

acetone,petrol or perfume has very low boiling temperature. when we put some acetone on our palm the particles of it gain energy from our palm and evaporate quickly which causes cooling .

Calculate the volume of 5.0 M sodium hydroxide needed to prepare the dilution.

Mreagent × Vreagent = Mdilution × Vdilution 5.0 M × Vreagent = 0.25 M × 0.500 L Vreagent = 0.025 L = 25 mL

Slowly add 25 mL of 5.0 M sodium hydroxide to a 500-mL volumetric flask half-filled with distilled or deionized water and swirl the flask to mix. Once the solution is at room temperature, dilute to the mark with water and invert the flask several times to mix.

Magnesium hydroxide contains two hydroxyl groups. One-half mole of magnesium hydroxide, therefore, accepts one mole of protons. To prepare a 1.0 N solution of magnesium hydroxide, slowly add 29 g magnesium hydroxide to a 500-mL volumetric flask half-filled with distilled or deionized water and swirl the flask to mix. When all the solid is dissolved and the solution is at room temperature, dilute to the mark and invert the flask several times to mix.

Calculate the mass of impure potassium chloride needed.

mass of impure potassium chloride = M pure × V pure × gram formula weight / percent purity = 1.0 M × 0.500 L × 74.56 g/mol0.930

= 40 gSlowly add 40 g of 93% potassium chloride to a 500-mL volumetric flask half-filled with distilled or deionized water and swirl the flask to mix. When all the solid is dissolved and the solution is at room temperature, dilute to the mark and invert the flask several times to mix.

Calculate the mass of impure potassium chloride needed.

mass of impure potassium chloride = M pure × V pure × gram formula weight / percent purity = 1.0 M × 0.500 L × 74.56 g/mol0.930

= 40 gSlowly add 40 g of 93% potassium chloride to a 500-mL volumetric flask half-filled with distilled or deionized water and swirl the flask to mix. When all the solid is dissolved and the solution is at room temperature, dilute to the mark and invert the flask several times to mix.

Tocalculatethe mass needed multiply the number of moles by the molecular weight in grams. massKClneeded = volume x molarity x molecular weight. To get exactly 0.100 Mol solution you need to use 7.455 g ofKclto 1 liter of water.

with the help of own

Since it is very dilute acidic solution, so H+concentrations from acid and water are comparable, and the concentration of H+from water cannot be neglected.

Therefore,[H+]total= [H+]acid+ [H+]waterSince HCl is a strong acid and is completely ionized[H+]HCl= 1.0 x 10-8The concentration of H+from ionization is equal to the [OH-] from water,[H+]H2O= [OH-]H2O= x (assume)[H+]total= 1.0 x 10-8+ xBut[H+] [OH-] = 1.0 x 10-14(1.0 x 10-8+ x) (x) = 1.0 x 10-14X2+ 10-8x – 10-14= 0Solving for x, we get x = 9.5 x 10-8Therefore,[H+] = 1.0 x 10-8+ 9.5 x 10-8= 10.5 x 10-8= 1.05 x 10-7pH = – log [H+]

= – log (1.05 x 10-7)

= 6.98

No. of moles of atoms = weight / atomic weight.100 gms of N₂ = 100/2 x 14 moles = 100/28 molesNumber of molecules = 100 / 28 x 6.022 x 10²³Molar mass of N₂ = 2 x molar mass of monoatomic NMolar mass of N₂ = 2 x 14.0067 = 28 moles.Number of molecules = 100/28 x 6.022 x 10²³No. of atoms = 2 x 100/28 x 6.022 x 10²³ = 43.01 x 10²³For NH₃ :-

100 gm of NH₃ = 100/17 molesNumber of molecules = 100/17 x 6.022 x 10²³ moleculesNo. of atoms in NH₃ = (1 + 3) = 4 x 100/17 x 6.022 x 10²³ =141.69 x 10²³ atoms.Therefore, NH₃ has more atoms than N₂.

No. of moles of atoms = weight / atomic weight.

100 gms of N₂ = 100/2 x 14 moles = 100/28 moles

Number of molecules = 100 / 28 x 6.022 x 10²³

Molar mass of N₂ = 2 x molar mass of monoatomic N

Molar mass of N₂ = 2 x 14.0067 = 28 moles.

Number of molecules = 100/28 x 6.022 x 10²³

No. of atoms = 2 x 100/28 x 6.022 x 10²³ = 43.01 x 10²³

For NH₃ :-

100 gm of NH₃ = 100/17 moles

Number of molecules = 100/17 x 6.022 x 10²³ molecules

No. of atoms in NH₃ = (1 + 3) = 4 x 100/17 x 6.022 x 10²³ =141.69 x 10²³ atoms.

Therefore, NH₃ has more atoms than N₂.

It means that the liquidscanflow only on one plane or the container in which itiskept.

Ingases, the particles vibrate inanydirection.

The molecules in agas havevery large intermolecular spacesand high kinetic energy as compared tosolidsand liquids. They donot experience intermolecular forces.

According to law of conservation of mass, the total mass in a reaction remains conserved.In other words,the total mass of reactants is equal to the total marks of products formed in a chemical reaction. Mass can be converted into other forms but the overall mass involved in reaction remains the same.

molarity and normality will be same as V.f of HCl is = 1

1N HCL - add 36.46 gm of HCL in 1 litre of water (Molar mass of HCL = 36.46 g/mole)

36.46 g of HCL is equivalent to 30.67 ml of HCL ( Volume = Mass/Density = 36.46/1.19 = 30.67)

Density of HCL is = 1.19 g/ml

Percent Concentration of Most Conc HCL is = 37.5 %

30.67 ml of HCL is equal to = (30.67)*(100/37.5) = 81.8 ml HCL

Hence to prepare 1N HCL - add 81.8 ml of HCL in 1000 ml of Water.

HCl + AgNO3…............>AgCl+ HNO310ml …...excess .... ….............00ml …..excess.. …..................0.1435g...................................................=(0.1435/143.5) mol=0.001mol AgCl made by 10 ml of HClFrom reaction it is clear that,1 mole of AgCl would be obtained by1 mole HClSo number of moles of HCl in 10ml would be=0.001SO number of equivalents of HCl in 10ml = 0.001*1 …..(as basicity of HCl =1)Hence normality of HCl=(number of equivalents of HCl/volume of HCl in ml)*1000normality= (0.001/10)*1000=0.1N

HCl + AgNO3 …............>AgCl+ HNO310ml …...excess .... ….............00ml …..excess.. …..................0.1435g...................................................=(0.1435/143.5) mol=0.001mol AgCl made by 10 ml of HClFrom reaction it is clear that,1 mole of AgCl would be obtained by1 mole HClSo number of moles of HCl in 10ml would be=0.001SO number of equivalents of HCl in 10ml = 0.001*1 …..(as basicity of HCl =1)Hence normality of HCl=(number of equivalents of HCl/volume of HCl in ml)*1000normality= (0.001/10)*1000=0.1N

The law ofconstantcomposition says that, in any particular chemical compound, all samples of that compound will be made up of the same elements in the sameproportion or ratio. Forexample, any water molecule is always made up of two hydrogen atoms and one oxygen atom in a ratio

Phenolpthalein :

A solution of 250 g ofphthalic anhydridein 200g of concentratedsulfuric acidis prepared withcareful heating. It is cooled to 115° C, and treated with500 g ofphenol, and the mixture is then heated at115-120° Cfor 10-12 hours, care being taken that the temperaturedoes not exceed 120° C. The reaction mixtureis then poured intoboiling water, and the excess ofphenolremoved by steamdistillation. The granular yellow solid is extracted withdilutecaustic soda, which dissolves out the phenolphthalein,leaving behind a by-product, phthalein anhydride. Aftercooling, the liquid is filtered, acidified withacetic acid,mixed with a few drops of hydrochloric acid, and left tostand for 24 hours. The crude phenolphthalein, a sandy yellowpowder, is then filtered off and dried. The further purification ofphenolphthalein is performed bycrystallization. 10 gof the air-dried crude phenolphthalein are boiled under reflux for90 minuteswith 60 gof absolute alcohol and 5 gofactivated charcoalto two-thirds theiroriginal volume, and treated with water. The milky liquiddeposits, on standing, crystals of phenolphthalein mixedwith a gummy impurity. To remove this, the alcoholicsolution is added to water, in the proportion of 40 grams to320 mlthe mixture is vigorously shaken, and after standingfor a short time separated quickly from the precipitatedresin. The solution is then heated, when the milkinessdisappears, a white crystalline powder being precipitated.This is filtered off, washed with water, and dried at a lowtemperature. A further quantity is obtained by removalby distillation of the alcohol from the filtrate. The yield ofphenolphthalein is75 %, calculated from the weight of phthalicanhydride employed.Phenolphthaleinforms a white powder, almost insolublein water; readily soluble in alcohol. It melts at250-253° C and dissolvescompletely in sodium or potassium hydroxide solution.

Phenolphthaleinis oftenusedas an indicator in acid–base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions. ...Phenolphthaleinis slightly soluble in water and usually is dissolved in alcohols forusein experiments.

Methyl Orange :

The 4-diazobenzenesulfonic acid (4-sulfobenzene-1-diazonium) is dissolved in the smallestpossible quantity of dilutesodium hydroxide,kept cold with ice, and the liquid is pouredinto a solution of the calculated amount ofdimethylanilinemixed withacetic acid. The sodiumsalt of the methyl orangethereupon separates in shining,reddish-yellow, laminar crystals.These crystals are filtered off and recrystallizedfrom hot water. If a warm solution ofthe salt is treated with dilute hydrochloricacid in moderate excess, the free base of methyl orange is precipitatedas a shining, reddish violet, crystallinepowder.

For the preparation of the methyl orange without the isolation the diazo compound – 4-diazobenzenesulfonic acid is used and the reaction is carried by the following procedure.

1 molesulfanilic acidis dissolvedin exactly 1 mole dilutesodium hydroxide, treatedwith 1 molesodium nitrite, and then 1 mole hydrochloric acid is added in the cold. Thissolution is treated straight away with 1 moledimethylanilinein a little hydrochloric acid,and sodium hyroxide is again added. The sodiumsalt of the methyl orange soon separates out. The precipitationmay be rendered more complete byadding common salt.

Methyl orangeis a pH indicator frequentlyusedin titrations because of its clear and distinct colour change. Because it changes colour at the pH of a midstrength acid, it is usuallyusedin titrations for acids.

Normality of the given HCl solution can be determined by the following method:

1. First we took 10 mL or 20 mL of the given unknown normality HCl solution in the conical flask and mark this volume as V₁.

2. Fill the bulb with the NaOH solution of the known normality and mark the normality of NaOH as N₂

3. Add 2-3 drops of the phenolphthalein indicator in the conical flask.

4. Then titrate the HCl with the NaOH as colour of the solution becomes pink, noted the volume of NaOH used as V₂.

Now normality of HCl can be calculated as follows:

N_{1}V_{1}=N_{2}V_{2}

N₁, is the unknown normality of the HCl solution.

So now after having values of N₁, V₁, V₂ normality of the HCl solution can be calculated.

Here as normality of NaOH is given

So the equation for the calculation of normality of HCl will become:

\frac{1\times V_{2}}{V_{1}\times 20}

J.J Thomson was the first scientist who measured charge to mass ratio (e/m) of an electron.When a narrow beam of charged particles are projected at constant speed (v) across a magnetic field in a direction perpendicular to the field, the beam of particles experiences a force, which makes them move in a circular path.

Mendeleev's Periodic law " states that the physical and chemical properties of an element are the Periodic function of their atomic masses.

◆ Achievements -1. It could classify all the elements discovered at that time.2. It helped in discovery of new elements .3. It helped in the correction of atomic mass of some of the elements .

1atom can neither created nor destroyed

2.atom is smallest particle of matter that take part in chemical reactions

3.atom of same elements are similar in all respect

4.atom is extremely small particle

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Atoms combine in the ratio of small whole numbers to form compounds.

In an acid, it is reddish in colour.

a) Unchanged.b) Bluec) Redd) Red

Ans :- Drawbacks/Demerits of Mendeleev's periodic table :-

1) Hydrogen's positionHydrogen resembled the properties of both alkali metals(like lithium) and also that of halogens(like iodine). Hence, the position of hydrogen(whether hydrogen is to be placed with halogens or alkali metals) was not specified.

2) Position for IsotopesIsotopes are those with same atomic no: and different mass no:s.Mendeleev'speriodic table was based on arranging elements in increasing order of atomic masses. But isotopes were not included in his periodic table.

3) Certain elements were arranged in reverse order. Elements having higher atomic mass were placed in front(or before) the elements with less atomic mass. Example - Cobalt and NickelCobalt being more heavier than Nickel ,was placed before Nickel , in the Mendeleev's periodic table.

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solution X:

◆ solution of X gives orange colour when a drop of universal indicator is added to it.◆which tells us that it is acidic In nature. ◆this means the pH would be less than 7.

solution Y:

◆solution Y gives bluish green colour when a drop of universal indicator is added to it .◆ which tells us that it would be little basic in nature.◆ this means that the pH would be more than 7.

{ 7 being neutral }

all 9th class maths solutions graphs in telangana

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. Position of hydrogen:-hydrogen resembles both, alkali metals (IA) and the halogens (VIIA) in properties so Mendeleev could not decide where to place it.

2. Position of Isotopes:- As atomic weight of Isotopes differ, they should have been placed in different position in Mendeleev's periodic table. But there was no such places for Isotopes in his table.

3. Cause of periodicity:- Why physical and chemical properties repeated in a group.

4. Anamolous pairs of elements:- There were some pair of elements which did not follow increasing order of atomic weights.

Example: Ar&K, Te&I, Co&Ni, Th&Pa

Ammonia that is NH3has greater affinity for protons than phosphine that is PH3.Reason:Nitrogen atom has the smallest size in the hydrides of group 15. Therefore, the lone pair is concentrated on a small region and electron density on it is maximum. Consequently this electron density attracts the protons with greater strength. Moving on to the phosphorus atom which has bigger size, the electron density is dispersed, thus have low affinity for protons.