Notes: Normal conditions In the case of Gases That's right.: a mol claimed 22.4 L
This means that Applicable: n = 1, V = 22.4 L
From Density=m/V and n=m/M following:: D=n·M/V

1. Density = m/V = n·N/V
       Dichte(CO₂)= 1mol·M(CO₂)/22.4 L = 44/22.4= 1.96 g/L
       Dichte(O₂) = 32/22.4 = 1.43 g/L
       Dichte(N₂) = 28/22.4 = 1.25 g/L
2. Air of which: to 80% from N₂ and 20% O₂
       Components N₂: 80% One of them Manufacture in which all the materials used are:: 0.8 mol
       Components O₂: 20% One of them Manufacture in which all the materials used are:: 0.2 mol
       Dichte(Luft) = (0.8·M(N₂)+0.2·M(O₂))/22.4 = (0.8·28+0.2·32)/22.4 = 1.29 g/L
3. The Density of CO₂ is 1.96 g/L, of The Air 1.29 g/L. This means that is CO₂ 'Heavier' How The Air and displaced This means that This 'lighter' O₂ (Density 1.43 g/L), The Flame suffocated, There you go. not Oxygen to the Candle flame Infiltration can.

Calculate the following: Density:

1. Calculate The Density the following: Gases: O₂, N₂ and CO₂ In the case of Normal conditions
2. Calculate The Density The Air
3. Explanation, Why? a Fire with a CO₂-Löscher deleted will be can.

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1. Density = m/V
2. d = 120 g / 110 Milligrams = 1.09 g/ml

One of them. Lthe island has One Weight of 120 Grams. You . Volumes is 110 Milligrams. Calculate their Density.

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1. c = n/V Responsibility. n = c·V
       n(Lösung) = c·V = 0.3 mol/l·0.15 l = 0.045 mol
       new Volumes: 150 Milligrams + 200 Milligrams = 350 Milligrams Responsibility. 0.35 L
       c = n/V = 0.045 mol / 0.35 L = 0.129 mol/L
2. c = n/V Responsibility. n = c·V
       n₁(Lösung) = c₁·V₁ = 0.3 mol/l·0.3 l = 0.09 mol
       n₂(Lösung) = c₂·V₂ = 0.5 mol/l·0.2 l = 0.1 mol
       new n: n₁ + n₂ = 0.19 mol
       new Volumes: 300 Milligrams + 200 Milligrams = 500 Milligrams Responsibility. 0.55 L
       c = n/V = 0.19 mol / 0.5 L = 0.38 mol/L
3. Additional Volumes: 2.0 L – 1.5 L = 0.5 L
       n₁(1.5 L Lösung) = c₁·V₁ = 0.15 mol/L·1.5L = 0.225 mol
       n₂(2.0 L Lösung) = c₂·V₂ = 0.3 mol/L·2L = 0.6 mol
       with n=m/M Responsibility. m(NaCl) =n·M(NaCl)
       M(NaCl) = 58.5 g/mol
       m₁(NaCl) =0.225 mol · 58.5 g/mol = 13.16 g
       m₂(NaCl) =0.6 mol · 58.5 g/mol = 35.1 g
       Additional Quantity NaCl = 35.1-13.16 = about. 22 Grams

Different calculations round to The Concentration.

1. To the 150 Milligrams One of them Lthe island with c = 0.3 mol/l will be 200 Milligrams Water given. What ? Concentration
2. To the 300 Milligrams One of them Lthe island with c = 0.3 mol/l will be 200 Milligrams One of them (equal to) Lthe island with c= 0.5 mol/L given. What ? Concentration has The Lthe island Now .?
3. Given be 1.5 L One of them Cooking salt solution with c = 0.15 mol/L.
   Well, you know what? to be But 2.0 L One of them Cooking salt solution with c = 0.3 mol/L Manufacture will be. How many Grams Fresh or chilled and Water must to the (c) Cooking salt solution Not yet Additionally Added will be.

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1. c = n/V = m/M/V
       M(C₁₂H₂₂O₁₁) = 342 g/mol
       c = 7.7g/342g/mol / 0.15 L = 0.15 mol/L
2. Additional task:
3. n = m/M
       M(C₈H₁₀N₄O₂) = 194 g/mol
       · n = 0.08/194 = 0.000412 mol = 4.12·10^-4 mol
       · 1 mol = 6.022·10²³
       · 4.12·10^-4 mol → 2.48·10²⁰

1. The idea: about. 60% The Land surface with Water Covered, Radius of the earth about. 6000 km, Average Depth of water about. 3 km
2. all These Assumptions Delivery a Volumes of 0.814·10⁹ km³
3. This is Wikipedia. has almost This equal to Volumes:
       V = 1.338·10⁹ km³ = 1.338·10²¹ Litre
       In the the following will be with to the The value of Wikipedia Continued calculated
4. 1 g Au → n(Au)=1g/197g/mol= 1/197 mol
5. 1/197 mol ≙ 3.057·10²¹ Au-Atoms
6. 3.057·10²¹ / 1.338·10²¹ = 2.28 Atoms for each Litre

Calculate The Concentration of Household sugar (C₁₂H₂₂O₁₁), in One of them Cup Coffee (150 Milligrams), where: in it. a WOther sugar (m = 7.7 g) dissolved will be.
Additionally: How many molecules Caffeine (C₈H₁₀N₄O₂) are The Commission after a Coffee In the KManufacture, where: in One of them Cup Coffee 80 mg of which: dissolved were?

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This Volumes must in Litre indicated will be.

1. c = m(NaCl)/M(NaCl)/V = 5g/58.5g/mol/1.0 L = 0.085 mol/l
2. c = m(C₆H₁₂O₆)/M(C₆H₁₂O₆)/V = 18g/180g/mol/0.12L = 0.833 mol/l
3. c = n/V = m/M/V → m = c·V·M
       m = 0.02 mol/l · 0.01 l · 176.13 g/mol = 0.035 g
4. c = n/V = m/M/V → m = c·V·M
       M(MgBr₂) = 184.3 g/mol
       m = 0.02 mol/L · 0.1 L · 184.3 g/mol = 0.3686 g
       Es must be So what do you mean? 0.3686 Grams MgBr₂ weighted will be and So much Water Added will be, that This Final volume 100 Milligrams is.
5. Adopted: 1 Litre Water be 1.0 kg, The Remains So what do you mean? Fresh or chilled: 340 Grams
       n=m(NaCl)/M(NaCl) = 340g/58.5 g/mol = 5.81 mol
       c = n/V = 5.81 mol / 1.0 L = 5.81 mol/L