Calculations for Veterinary Nurses by Margaret C. Moore, Norman G. Palmer

By Margaret C. Moore, Norman G. Palmer

This convenient notebook may also help veterinary nurses with every kind of calculations. a number of labored examples are incorporated to strengthen the reader's self belief in engaging in the strategies concerned. each one kind of calculation has its personal separate part within the booklet and the authors have used the best attainable technique in explaining every one. The booklet is established in this sort of means that the reader can development from an easy rationalization of the mathematics rules concerned, to the appliance of those ideas to crucial veterinary calculations.

Qualified veterinary nurses and scholars alike will locate this e-book a useful reference resource, even if appearing suitable veterinary calculations or learning for pro examinations.

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Example text

Alternatively the answer could be calculated thus: weight  100 becomes: volume % solution  volume weight ˆ 100 20  450 ˆ 90 g weight ˆ 100 % solution ˆ Manipulating the formula (see also Chapter 2, Basic Principles) The standard equation for a percentage solution is: % solution ˆ weight in g/volume in ml  100 weight in g  100 iXeX % solution ˆ volume in ml It is possible to manipulate this standard % solution formula to ¢nd the `missing part'. For example weight in g  100 (standard formula) volume in ml weight in g  100 (manipulated formula) volume in ml ˆ % solution volume in ml  % solution weight in g ˆ 100 (manipulated formula) % solution ˆ It may be easier to remember the above manipulation if it is thought of as a triangle: 40 Chapter 3 weight in g Â100 Ä volume in ml Ä Â % To apply the triangle: .

Compare the answers and deduce what has to be done to convert one solution into the other. Calculate the % solution from the facts given in the question. e. 2X5% ˆ  100 (equation 1) 100 ml expressed as weight in mg in 5 ml (which is the volume 125 mg (divide top and of the original solution) gives 5 ml bottom of equation 1 by 20). 5% solution then 115 mg (125 mg À 10 mg) of solute must be added to rectify the situation. 5% solution (x) In each case, the weight of solute in the solution left by the students must be calculated.

Answer ˆ 500 ml À 2  25 ml samples ˆ 450 ml Changing the Concentration of a Solution 39 (iii) What weight of solute would be in the remaining volume? Weight of solute left in this 450 ml of 20% solution? Original solution contained 100 g and 2 samples have been taken out containing 5 g. Therefore, there must be 100 g À (2 samples  5 g) ˆ 90 g left. Alternatively the answer could be calculated thus: weight  100 becomes: volume % solution  volume weight ˆ 100 20  450 ˆ 90 g weight ˆ 100 % solution ˆ Manipulating the formula (see also Chapter 2, Basic Principles) The standard equation for a percentage solution is: % solution ˆ weight in g/volume in ml  100 weight in g  100 iXeX % solution ˆ volume in ml It is possible to manipulate this standard % solution formula to ¢nd the `missing part'.

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