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When you are busy with these calculations, you will need to remember the following:

1dm 3 = 1 litre = 1000ml = 1000cm 3 , therefore dividing cm 3 by 1000 will give you an answer in dm 3 .

Some other terms and equations which will be useful to remember are shown below:

  • Molarity is a term used to describe the concentration of a solution, and is measured in mol.dm - 3 . The symbol for molarity is M. Refer to chapter [link] for more information on molarity.
  • Moles = molarity (mol.dm - 3 ) x volume (dm 3 )
  • Molarity (mol.dm - 3 ) = m o l e s v o l u m e

Given the equation:

NaOH + HCl NaCl + H 2 O

25cm 3 of a sodium hydroxide solution was pipetted into a conical flask and titrated with 0.2M hydrochloric acid. Using a suitable indicator, it was found that 15cm 3 of acid was needed to neutralise the alkali. Calculate the molarity of the sodium hydroxide.

  1. NaOH: V = 25 cm 3

    HCl: V = 15 cm 3 ; C = 0.2 M

    The equation is already balanced.

  2. M = n V

    Therefore, n(HCl) = M × V (make sure that all the units are correct!)

    M = 0.2mol.dm - 3

    V = 15cm 3 = 0.015dm 3

    Therefore

    n ( H C l ) = 0 . 2 × 0 . 015 = 0 . 003

    There are 0.003 moles of HCl that react

  3. Look at the equation for the reaction. For every mole of HCl there is one mole of NaOH that is involved in the reaction. Therefore, if 0.003 moles of HCl react, we can conclude that the same quantity of NaOH is needed for the reaction. The number of moles of NaOH in the reaction is 0.003.

  4. First convert the volume into dm 3 . V = 0.025 dm 3 . Then continue with the calculation.

    M = n V = 0 . 003 0 . 025 = 0 . 12

    The molarity of the NaOH solution is 0.12 mol.dm 3 or 0.12 M

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4.9 g of sulfuric acid is dissolved in water and the final solution has a volume of 220 cm 3 . Using titration, it was found that 20 cm 3 of this solution was able to completely neutralise 10 cm 3 of a sodium hydroxide solution. Calculate the concentration of the sodium hydroxide in mol.dm - 3 .

  1. H 2 SO 4 + 2 NaOH Na 2 SO 4 + 2 H 2 O

  2. M = n/V

    V = 220 cm 3 = 0.22 dm 3

    n = m M = 4 . 9 g 98 g . m o l - 1 = 0 . 05 m o l s

    Therefore,

    M = 0 . 05 0 . 22 = 0 . 23 m o l . d m - 3
  3. Remember that only 20 cm 3 of the sulfuric acid solution is used.

    M = n/V, therefore n = M × V

    n = 0 . 23 × 0 . 02 = 0 . 0046 m o l
  4. According to the balanced chemical equation, the mole ratio of H 2 SO 4 to NaOH is 1:2. Therefore, the number of moles of NaOH that are neutralised is 0.0046 × 2 = 0.0092 mols.

  5. M = n V = 0 . 0092 0 . 01 = 0 . 92 M
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Acid-carbonate reactions

Demonstration : the reaction of acids with carbonates

Apparatus and materials:

Small amounts of sodium carbonate and calcium carbonate (both in powder form); hydrochloric acid and sulfuric acid; retort stand; two test tubes; two rubber stoppers for the test tubes; a delivery tube; lime water. The demonstration should be set up as shown below.

Questions & Answers

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Kagiso Reply
You can use simultaneous equations or T=mg+ma
sibahle
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to calculate acceleration you use formula Vf -Vi divided by the time (t)
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I don't think there's a formula u just have to use all information in that type of situation
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the sum of all the forces acting on an object
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kinetic friction × Normal force
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A force is a pull or push action that affects the shape, motion and direction of an object
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the different types of friction forces
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different types of frictional forces are static frictional force(the object is stationery)and kinetic frictional force(the object is moving)
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static
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a vector is a physical quantity that have both magnitude and direction
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Newton's second law of motion states that the acceleration of an object is equal to the net force exerted on the objectdivided by the object's mass. The direction of acceleration is the same as the direction of thenet force.
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and the mass is inversely proportional
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is the forces that act in the molecules
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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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