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

By Margaret C. Moore, Norman G. Palmer

This useful notebook might help veterinary nurses with every kind of calculations. a number of labored examples are incorporated to boost the reader's self assurance in engaging in the strategies concerned. each one kind of calculation has its personal separate part within the ebook and the authors have used the best attainable technique in explaining every one. The booklet is dependent in one of these method that the reader can development from an easy rationalization of the mathematics rules concerned, to the appliance of those rules to crucial veterinary calculations.

Qualified veterinary nurses and scholars alike will locate this booklet a useful reference resource, no matter if acting proper veterinary calculations or learning for pro examinations.

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Extra resources for Calculations for Veterinary Nurses

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5% Note: only the weight changes, not the volume. (vi) The technique for answering this question is exactly the same as those above. Use the standard formula to express each of the solutions by weight and volume. Thus a 10% standard solution is: 10 g  100 100 ml 5g  100 5% solution ˆ 100 ml 10% solution ˆ Changing the Concentration of a Solution 51 10 g by multiplying top 200 ml and bottom of the equation by 2. or, it can also be expressed as This enables direct comparison of the 10% and the 5% solutions as both have the same weight but are dissolved in di¡erent volumes.

E. 100 g in 500 ml. 5% solution. 5 g of solute in 100 ml of solvent. e. 5 g in 200 ml. Solution 2 requires 500 ml of a 1% solution. A standard 1% solution is expressed as 1 g in 100 ml of solvent. e. 5 g in 500 ml. Calculate how many ml of the more concentrated 20% solution contains 5 g of the solute. e. e. g. e. g. e. 5 g in 500 ml ˆ  100% 500 ml ˆ 1% solution (ii) What volume would remain of the original 20% solution? 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?

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.

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