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Extra info for Practical physical chemistry
Sample text
Now interchange the weights, placing the 5o-gm. weight on the right DETERMINATIONS OF WEIGHT AND VOLUME 25 and the other weights on the left, and determine what weight must now be added to the left or right, in order to obtain exact In this way we obtain the two equations counterpoise. = 50 + 20 20 4- From 10' 4- 10" 4- these two weighings 50 + i" + + ^gm. = 10' 4- 10" 4- 5 4- 2 4- i = 204- we 1 i'" 4- r gm. 50 obtain 10' 4- 10" 4- . + \(r 4- I) T> we know however, If, the ratio of the balance arms, 1 , then Li only one weighing levers, a weight weight wX w necessary, because, from the theory of placed on the right pan is equivalent to a is "D p ratio v- = on the y- Thus, suppose we have found the left.
To calibrate a flask, the latter is first cleaned and thoroughly dried; it is then counterpoised on a balance, and distilled water, having a temperature of 15 to 18, is run into the flask until the lower edge of the meniscus stands at volume mark on the neck. Any water which may have got on the neck above the mark should be removed by means of filter paper. The weight of the water is then determined. c. ) Having determined the weight of water contained in the up to the mark, the volume can be obtained from the table given above.
1 In the above cases, we have assumed that the value of a magnitude depends only on the measurement of one quantity. e. the relative latter will depend on the accuracy of the Thus, the determination of the mole- error) of the final result will several determinations. by the cryoscopic method, involves measurements of weight and of temperature ; the determination of the velocity of a reaction involves quantity and time. y, dx ) is results = can be readily obtained by differentiation. y, dx = k 2ydy the relative error in .