7 The Mole

[object Object],[object Object],How many moles are present in 0.35g of ammonium hydroxide? ,[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],[object Object],c) 1 mole of Na 2 CO 3 .10H 2 O = (2x23) + 12 + (3x16) + (10x18) = 286g  0.24 moles = 0.24x286 = 68.6g

[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],The abbreviation for moles is mol

[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],1g of helium gas contains 1 mol = 0.25 mol 4  1g helium gas contains 0.25x 6.022x1023atoms = 1.51x10 23 atoms 1g beryllium contains 1 mol = 0.11mol 9  1g beryllium contains 0.11x 6.022x10 23 atoms = 6.69x 10 22 atoms  1g hydrogen gas contains the greatest number of the stated particles

[object Object],[object Object],How many moles are there in 20ml of a solution of concentration 0.5mol/L? 1000ml contains 0.5mol  1ml contains 0.5 mol 1000 And 20ml contains 0.5 x 20 mol = 0.01mol 1000

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Percentage Composition ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],[object Object],Find the percentage oxygen in the CUSO 4 .5H 2 O RAM O = 16  total RM O = 16 x 9 = 144  %O = 144 x 100% = 57.7% 249.5

[object Object],%Cu = 63.5 x 100 = 25.45% 249.5 %S = 32 x 100 = 12.82% 249.5

Calculations based on Chemical Formulas ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

The amounts of substances undergoing reaction, as given by the balanced equation, are called the stoichiometric amounts Stoichiometry is the relationship between the amounts of reactants and products in a chemical reaction ,[object Object]

[object Object],[object Object],[object Object],[object Object],Method 1) find the RMM of potassium chlorate 2) find the moles of chlorate heated 3) find the moles of potassium chloride formed 4) find the mass of potassium chloride formed.

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Moles of oxygen formed = 3 x 0.08 = 0.12 2 Mass Oxygen formed = 0.12 x 32 = 3.84g

[object Object],[object Object],The equation tells us that 1 mole of KIO 3 gives 3 moles of I 2.  1/6 mole of KIO 3 gives 1/6 x 3 moles of I 2 = ½ mole I 2

[object Object],On heating the mixture the following reaction occurs 2NaHCO 3(s)  Na 2 CO 3(s) + CO 2(g) + H 2 O (g) . RMM NaHCO 3 = 84  100g = 100 = 1.190mols 84 1.190 mols NaHCO 3 will give 1.190/2 mols Na 2 CO 3 = 0.590mols RMM Na 2 CO 3 = 106  106 x 0.600g = 63.1g

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7 The Mole

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7 The Mole