Concentration in Chemistry Sciences.pptx

Concentration

1. Quantitative continued

2. Week 1 Monday – Quantitative review Tuesday – Quantitative practice Wednesday – Electricity and energy review + electricity practice Thursday – Atomic structure (Physics and Chemistry) Friday – Test from the week (open book) Homework- forces   Half term - Complete revision booklet + exam Qs   Week 2 Monday - Biology extended responses (cardio, photosynthesis practical, stem cells) Tuesday – Chemistry core practical and bonding review Wednesday – Wave demos and write up Thursday – Bonding and electrolysis Friday – Test from the week (open book)  Homework - forces  

3. Conservation of mass Mg + 2HCl  MgCl 2 + H 2 Mass of reactants = Mass of products M r reactants = M r products  A student tests the law of conservation of mass. She measures the mass of the beaker before the reaction and then after the reaction. This is the equation for the reaction. Na 2 CO 3 ( aq ) + 2HCI( aq )  ⟶  2NaCl( aq ) + CO 2 (g) + H 2 O(I) Explain why this student’s results would not appear to support the law of conservation of mass . A gas was released so the mass decreased What should the student have done to prove that mass was conserved?

4. Using moles to balance equations Reactants Products Mg = 24g MgO = 40g O 2 = 16g Mg O 2 MgO Mass 24 16 40 M r 24 16 x 2 = 32 24 + 16 = 40 Moles n =m/ M r n = 24/24 n = 1 n=m/ M r n = 16/32 n = 0.5 n=m/ M r n = 40/40 n = 1 Simplest ratio 2 1 2 2 Mg + O 2  2 MgO

5. Reactants Products Sb = 488g SbCl 3 = 914g Cl 2 = 426g Sb Cl 2 SbCl 3 Mass 488 426 914 M r 122 (35.5 x 2) = 71 122 + (35.5 x 3) =228.5 Moles n = m/ M r n = 288/122 n=4 n= m/ M r n = 426/71 n = 6 n= m/ M r n = 914/228.5 n = 4 Simplest ratio 2 3 2 2 Sb + 3 Cl 2  2 SbCl 3

6. Limiting reagents In a chemical reaction involving two reactants: One of the reactants is often used in excess The other reactant which is completely used up is called the limiting reactant No more products can be made since one of the reactants is used up

7. Quantity Conversion Dm 3 1L 200mL 300cm 3 4000mL 5cm 3 6000cm 3 12cm 3 75000cm 3 dm 3 cm 3 dm 3 cm 3 X 1000 ➗1000 X 1000 ➗1000

8. Find the concentration of the following in g/dm 3 : 1) 160g of CuSO 4 in 1 litre of water 2)  40g of NaOH in 1 dm 3 of water 3) 80g of NaOH in 2 dm 3 of water 4) 196g of Cu(OH) 2 in 200 cm 3 of water 5) 49g of Cu(OH) 2 in 1 litre of water 6) 20g of NaCl in 2 dm 3 of water 7) 58.5 kg of NaCl in 10,000 cm 3 8) 10g of H 2 CO 3 in 200ml of water

9. Answers 1. 160g/dm 3 7. 5850g/dm 3 2. 40g/dm 3 8. 50g/dm 3 3. 40g/dm 3 4. 980g/dm 3 5. 49g/dm 3 6. 10/dm 3