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Suryansh Goyal 5 years, 4 months ago
Meghna Thapar 5 years, 4 months ago
Covalent compounds have bonds where electrons are shared between atoms. Due to the sharing of electrons, they exhibit characteristic physical properties that include lower melting points and electrical conductivity compared to ionic compounds. The melting point is low due to weak van der Waal's forces of attractions between the covalent molecules. The force of attraction between the molecules of a covalent compound is very weak.
Posted by Ashi Isha 5 years, 4 months ago
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Ashi Isha 5 years, 4 months ago
Gaurav Seth 5 years, 4 months ago
Structure of Water Molecule
In water molecule, the oxygen is sp3 hybridized and hence has 4 sp3 hybridized orbitals. Two of these sp3 orbitals are half filled and hence overlap with 1s orbital of hydrogen to form two sp3 -s , O-H , σ bond while the other two contain a lone pair of electrons each.
The geometry of water molecule should be tetrahedral and the bond angle ∠HOH should be 109.5° but experimentally it has been determined that the actual bond angle is 104.5°.
Reason:
In water, the oxygen atom is surrounded by two shared pair ,and two lone pair of electrons. According to valence shell electron pair repulsion theory, lone pair- lone pair repulsion are stronger than Bond pair – Bond pair repulsion. As a result the bond angle in water is slightly smaller than the regular tetrahedral angle of 109.5°.The actual bond angle is 104.5° and each O-H bond has a bond length of 95.7pm.
<center> </center>In the gaseous state, water exist as discrete molecules. It is a bent molecule with bond angle of 104.5° and bond length of 95.7 pm.
The electronegativity of oxygen is much higher than that of hydrogen and hence the shared pair of electrons in O-H bonds are attracted slightly more towards the oxygen atom. As a result oxygen carries a partial negative charge and hydrogen atoms carrier a partial positive charge. Since these two dipoles are inclined to each other at an angle of 104.5° therefore water is a highly polar molecule its actual dipole moment is 1.84 D.
Posted by Divyanshi Goswami 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
Suppose volume of 10 M HCl required to prepare 1 L of 6 M HCl = x litre
Volume of 3 M HCl required = (1-x) litre
Applying molarity equation:
Therefore Vol. of 10 M HCl required = 0.428 L = 428 mL.
Vol. of 3 M HCl required = 1-0.428 = 0.527 L = 572 mL
Posted by Dark ..? 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
On July 7, HRD Minister Ramesh Pokhriyal announced a major CBSE syllabus reduction with 30% of the syllabus slashed for the year 2020-21 for classes 9 to 12 because of the reduction in classroom teaching time due to the Covid-19 pandemic and lockdown.
CBSE has rationalized the syllabus with the help of suggestions from NCERT and the same has been notified by a new CBSE notification as well.
Click on the links:
Revised Syllabus: <a href="http://cbseacademic.nic.in/web_material/CurriculumMain21/revisedsyllabi/SrSecondary/REVISEDChemistry_2020-21.pdf">http://cbseacademic.nic.in/web_material/CurriculumMain21/revisedsyllabi/SrSecondary/REVISEDChemistry_2020-21.pdf</a>
Deleted portion:
<a href="http://cbseacademic.nic.in/web_material/CurriculumMain21/revisedsyllabi/Deduction/DELETEDChemistry_2020-21.pdf">http://cbseacademic.nic.in/web_material/CurriculumMain21/revisedsyllabi/Deduction/DELETEDChemistry_2020-21.pdf</a>
Posted by Divyanshi Goswami 5 years, 4 months ago
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Posted by Boparai Boparai 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
Answer :
- Molarity of the solution = 0.4 M.
- Normality of the solution = 0.8 N.
Step-by-step explanation :
Given that,
- Mass of
dissolved = 9.8 g.
- Volume of the solution = 250 cm³ = 0.250 L.
✒ Calculation of molarity :
Molar mass of
∴ No. of moles of
Now,
✒ Calculation of normality :
Eq. mass of
∴ No. of g equivalent of
Now,
Posted by Omkar_ . 5 years, 4 months ago
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Harman Singh 5 years, 4 months ago
Naveen Kumar Das 5 years, 4 months ago
Posted by Aditya K 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
The present set up of the Long Form of the Periodic Table can accommodate at the maximum 118 elements. After this, according to Aufbau principle, 8.y-orbital should be filled. Therefore, the outer electronic configuration of element having atomic number 119 will be 8s1. Since it has one electron in the outermost s-orbital, its valency will be 1 and it should belong to group 1 along with alkali metals. The general formula of its oxide will be M20, where M represents the element.
Posted by Boparai Boparai 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
A homologous series is defined as a family or group of structurally similar organic compounds all the members of which contain the same functional group, Show a gradation in physical and similarity in chemical properties and any two adjacent members of which differ by a -CH2 group. The individual members of such a series are called homologues and the phenomena is called homology.
Alkanes constitute a homologous series.
CH4 Methane
C2H6 Ethane
C3H8 Propane
C4H10 Butane
Posted by Prasant Yadav 5 years, 4 months ago
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Harman Singh 5 years, 4 months ago
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Yogita Ingle 5 years, 4 months ago
A mole is defined as that amount of substance which contains Avogadro's number of atoms if the substance is atomic or Avogadro's number of molecules if the substance is molecular. 1 mole of carbon atoms = 6.022 ×1023 atoms of carbon.
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Gaurav Seth 5 years, 4 months ago
Limiting Reagent
- In a chemical reaction, reactant which is present in the lesser amount gets consumed after sometime and after that no further reaction takes place whatever be the amount of the other reactant present. Hence, the reactant which gets consumed, limits the amount of product formed and is, therefore, called the limiting reagent.
Problem:-
50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g).
Calculate the NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation.
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Yogita Ingle 5 years, 4 months ago
Osmosis is the process in which solvent molecules moves through a semi-permeable membrane from a region of higher concentration to a region of lower concentration until the amount of fluid is equalised on both sides of the semipermeable membrane.
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