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Halogensareusedin the chemical, water and sanitation, plastics, pharmaceutical, pulp and paper, textile, military and oil industries. Bromine, chlorine, fluorine and iodine are chemical intermediates, bleaching agents and disinfectants. ... Bromine is also a component of military gas and fire-extinguishing fluids.

They arehardbecause of strong electrostatic forces between atoms. They arebrittlebecause of low mobility of defects called dislocations present inionic solids.

b) 6.02 into 10 to the power 23 22400

1 mol of C6H12O6​= 180 g = 6.02*10^23molecules​ 9 g = ?

Number of molecules in 9 g of C6H12O6​ = 9*6.02*10^23/ 180 = 3.01*10^22molecules

Number of C atoms in 9 g of glucose = 6*3.01*10^22​​​​= 18.06 * 10^22​atoms

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A sample of a pure compound contains 2.04g of sodium, 2.65 X 1022atoms of carbons and 0.132 mol of oxygen atoms. Find he empirical formula.

Moles of Na in sample = Given mass/Molar mass of Na = 2.04 g/23 g mol-1​ = 0. 088Mole of carbon1 mole C = 6. 022 x 1023atoms of C​Therefore , 2.65 x 1022atoms = 2.65​x 1022/6. 022 x 1023 = 0. 044Moles of oxygen = 0.132Molar ratio of Na , C and O = 0.088 : 0.044 : 0.132The simple whole number ration = 2 : 1 : 3Empirical formula of compound = Na2CO3

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The mass of glucose can be found out by M=V*Density. = 1*0.8 = 0.8 gms

but the molecular mass of glucose is 180gm.

so no. of moles = 0.8/180 = 4.4 × 10^{-3} and in 1 mole , no. of atoms = 6.02*10^{23} so in 4.4 × 10^{-3} , no. of atoms is = 4.4 × 10^{-3} x 6.02*10^{23}^{21} = 2.68x10^{21}.

The answer is Option B.

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molecularmass of H2SO4=98g98g contain avagadro molecules49g contain 6.022÷2=3.011no.of atoms present inH2S04 =(2+1+4)=7so..no of atoms in 49g=7×3.011×10^23(please like me✔✔🔱

1 -32-43-54-15-2

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1-sugar and water makes a = solution2-solid melts into = liquid3-liquid freeze into = solid4- on heating, liquid change into= gas5-the smaller particle that can exist independently= molecule

Organic chemistry is a chemistry subdiscipline involving the scientific study of the structure, properties, and reactions of organic compounds and organic materials, i.e., matter in its various forms that contain carbon atoms.

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Let's take the example of th methyl carbocation CH3+,In CH3+:two electrons in a lone pair, and 3 bonds, each to a hydrogen.

So the formal charge on H3C: is(Valence Electrons) - (Electrons in Lone Pairs) - (# of bonds)= 4 - 2 - 3= -1

C2H4 is unsaturated hydrocarbon a simplest alkene called ethylene

C3H8 Propane is a three-carbon alkane with the molecular formulaA colorless, easily liquefied, gaseous hydrocarbon (compound of carbon and hydrogen), the third member of the paraffin series followingmethaneandethane. The chemical formula forpropaneis C3H8

a. Ethylene is a hydrocarbon which has the formula C ₂H ₄ or H₂C=CH₂. It is a colorless flammable gas with a faint "sweet and musky" odour when pure. It is the simplest alkene. It contains a double bond, it called an unsaturatedhydrocarbon.

2. propane (C3H8),

Answer is option A(2.189×100000000)

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The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding cloud of electrons

The valency of an element is the number of hydrogen atoms that can combine with or replace (either directly or indirectly) one atom of the element. ... For example, oxygen has six valence electrons, but its valency is 2. Some elements may have more than one combining power (or valency), while others have just one

Decrease downward and increase from left to right in groups & periods of PT respectively.

(D) La > Ac

Atomic size increases as we move down a group and decreases as we move from left to right in a period.

K and Ba are in the same group. K < BaAl and Ga are in the same group. Al < GaIn and Tl are in the same group. In < TlLa and Ac are in the same group. But, the nuclear radii of the d-block elements inside a given series decrease with increment in the nuclear number. This is because of the expansion in the atomic charge that attracts the electron cloud inwards bringing about a decrease in size.

Carbon is a chemical element having the symbol C and the atomic number as 6. Exactly 12 grams of pure carbon-12 is known as one mole of carbon. The number of atoms of carbon-12 present in the one mole sample is - 6.022 136 7 x 10`23

12g of carbon is the overall weight of carbon expressed in the form of gram atomic mass, whereas, 12u of carbon is weight of carbon expressed as relative mass or is the mass of one carbon - 12 atom, where u is the atomic mass whose actual weight is -

1 u = 1.66×10⁻²⁴g

12u c means the mass of 12 electron's c but 12 g c means 1 mole c atom's mass 1u =1.6605×10^-27kg

12u c means the mass of 12 electron's c but 12g c means 1 mole c atom's mass 1u=1.6605×10^-27kg

d) is the correct answer

d.is the right answer

A) is the correct answer

option, d law of gaseous volume is the right answer

d. is the right answer

a i s the correct answer

as the branching increase boiling point decrease and as the mass increase boiling point increase. (iv) is correct answer

d is the correct answer

c>b>d>a is the correct answer

this is due to the emission spectrum

Emission of excess energy absorbed as a radiation in the visible region

Nature of the elements:alkali metals are soft in nature when compared to alkaline earth metals.

Oxidation state:alkali earth metals exhibit an oxidation state of +1. Alkaline earth metals which have two electrons in the outermost s-orbital exhibit an oxidation state of +2.

Reactivity with nitrogen:alkali metals do not react with nitrogen directly. Alkaline earth metals combine with nitrogen directly and form nitrides.

Reaction with ammonia:alkali metals react with ammonia forming amides. Alkaline earth metals react with ammonia at low temperature forming hexamines M (NH3)6

Stability of the carbonates:carbonates formed by alkali metals are very stable compared to those formed by alkaline earth metals. Carbonates of alkaline earth metals decompose easily upon heating.

Decomposition of nitrates:nitrates formed by alkali metals decompose evolving oxygen. NO2is liberated by decomposition of nitrates formed by alkaline earth metals.

2KNO3→ 2KNO2+ O2

2Ba (NO3)2→ 2BaO + 4NO2+ O2

Solubility of sulphates:Sulphates formed by alkali metals are highly soluble and form alums very easily. Sulphates of alkaline earth metals are sparingly soluble and do not form alums.

Solubility of bicarbonates:bicarbonates of alkali metals are less soluble than their corresponding carbonates. But, the bicarbonates of alkaline earth metals are more soluble than their corresponding carbonates.

Solubility of fluorides, carbonates, phosphates and oxalates:fluorides, carbonates, phosphates and oxalates of alkali metals are soluble. While those of alkaline earth metals are insoluble.

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Equi. Wt of Element = 32 gEqui. Wt of Oxygen= 8 gOne Equivalent of Oxide = 32 +8 = 40gPercentage of oxygen in oxide = 8/40 * 100 = 20%

2KClO3 ---> 2KCl + 3O2Therefore, 2 mol of KClO3 gives 3 mol of O2122.5 g of KClO3 gives 22.4X 3 L of O210 g of KClO3 = 0.081 mol (approx.)2.24 L of O2 = 0.1 molif 3 mol of O2 is obtained from 2 mol of KClO30.1 mol of O2 is obtained from 2/3 X 0.1 = 0.06 mol of KClO3Therefore, amount of pure potassium chlorate in sample = 0.06 mol% age purity = 0.06 / 0.081 X 100 = 74.07%

MgCO₃ ⇒ MgO + CO₂84g 40gy g 8 g (y is variable used)

y=(84 x 8)/40 = 16.8 g

Percentage purity of MgCO₃ = (16.8 x 100)/20=84%

option 2 )During this process, it generates hydrogen, which has the potential to behave as an explosive gas. ・It is highly hygroscopic, and absorbs the moisture,carbon dioxide, or sulfur dioxide in theair. It is also highly deliquescent and absorbs moisture to form an aqueous solution.

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Sodium (Na) Z= 11electronic configuration 1s2 2s2 2p6 3s1the differentiating electron is in 3s

the four quantum number is

n - 3l - 0m - 0s - 1/2

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Lithium/Electron configuration

[He] 2s1

Chlorine/Electron configuration

[Ne] 3s2 3p5

Potassium/Electron configuration

[Ar] 4s1

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I) Electronic configuration of sodium and Na+

II) electronic configuration of chlorine and Cl-

The answer coming is 8%. See the solution given below.

Lets say the salt is MSO3, M is any metal

The n factor of this salt will be obviously 2 .

Now its given 20 ml of this requires 10 ml of 0.02M acidified KMno4 ( whose n factor will be 5 as its acidic medium).

So we can write 10* 0.02 *5 = 0.5*1000/Ewhere E is equivalent weight,E is coming 500 g, so the molar wt. of the salt is coming as 1000 g.

% SO3 will be = (80/1000) *100 = 8 %

N1V1=N2V2hence N=gram equivalent/volume0.5=gram equivalent/1hence amount =0.5gm as n factor for HCl is 1

Antimony-SbIron-FeSilver-AgGold- AuMercury-HgLead-PbSodium -Na

antimony=SBiron=Fegold=ausilver=agmercury=hglead=pbsodium=na

Sb = antimonyFe = ironAu = goldAg = silverHg = mercuryPb = leadNa = sodium

Entropy is the extend of disorder in a system.

= In option A, Four moles of reactants give rise to two moles of products. This reduces the disorder in the system.= In option B, 4 moles or reactants give rise to 5 moles of products. Disorder of the system increases.= In option C, one mole of reactant gives rise to two moles of product. again disorder increases.=. In option D, reactants and products have same number of moles. No disorder in system.In option A, as disorder decreases, it shows negative entropy.

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Both acids and bases have a few common characteristics. Some of them are:

# They have the capability of conducting electricity.

# They become neutralized when they are disassociated in water or when they release their ions into water.

# They can change the color of litmus paper although the color differs for each of them.