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Basic Concept To Chemistry

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Published in: Chemistry
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Mole concept

Parteek S / Jaipur

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  1. Basic Concept of Physical Chemistry By CPS
  2. What we will learn in this module ' Mole concept ' Balancing reactions ' Atomic no and mass no
  3. What is 1 mole? ' It is defined as the amount of any chemical substance that contains as many elementary entities, e.g., atoms, molecules, ions, or electrons, as there are atoms in 12 grams of pure carbon-12 (QC), the isotope of carbon with relative atomic mass 12 by definition. This number is expressed by the Avogadro constant, which has a value of
  4. Some fundamental formulas for calculating moles Moles = Mass(g) / Molecular mass (use the periodic table to find this) Moles= No of molecules/ Avogadro's No Moles = volume of gas in liters / 22.41iters at STP
  5. Calculate the number of moles present in 14g of carbon monoxide 28g of CO = 1 mole 14g of CO = ? = 14 x 1/28 = 1/2 = mole.
  6. Calculate the number of atoms in 0.5 moles of carbon Number of atoms = 0.5 X 6.022x1023 = 3.0115 atoms
  7. Find mole of 4.48 liters 02 normal conditions. Solution: under Under normal conditions, 1 mol gas is 22.4 liters We use following formula to find moles of gas under normal conditions; n=V/22.4 mol
  8. So in every mole computing question these or any one of these formulas will be used
  9. Limiting reagent The limiting reagent (or limiting reactant) in a chemical reaction is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent since the reaction cannot proceed further without it.
  10. A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? First, we need to create a balanced equation for the reaction: 4 NH +50 O 3(g) 2(g)
  11. Next we can use stoichiometry to calculate how much product is produced by each reactant. NOTE: It does not matter which product is chosen, but the same product must be used for both reactants so that the amounts can be compared.
  12. 30.OgNO = 3.53 g NO 30.OgNO = 3.OOgNO The reactant that produces the lesser amount of product in this case is the oxygen, which is thus the "limiting reactant."Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant (ammonia) actually did react with the limiting reactant (oxygen).
  13. 4m 3 3 3 We're not finished yet though. 1.70 g is the amount of ammonia that reacted, not what is left over. To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample. 2.00 g NH3 (original sample) —1.70 g (reacted) = 0.30 g NH3 remaining
  14. Balance the following reaction Step 1: Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions: Cu+(aq)+e-+Cu(s) The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. This yields: Cu+(aq)+e-+Cu(s)
  15. Step 2: Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying the entire Cu+(aq)+e--->Cu(s) half- reaction by 3 and leaving the other half reaction as it is. This gives:
  16. Step 3: Adding the equations give: The electrons cancel out and the balanced equation is left.
  17. The number of protons in the nucleus Of an an elernentls atomic number. Number of Neutrons Number of Protons Mass Number
  18. Standard notation 2 Number or Proton Number(Z) Elcmontll Symbol Atomic Mass in 4.003 amu

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