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Having obtanied the voulme of the flask, the student emptied the flask, dried it, and filled it with an unknown whose density she wished to dtermine. The mass of the stoppered flask when completely filled with liquid was 56.796g. Find the density of the liquid.
Mass of the Liquid=____g-______g=_______g

2006-08-30 11:09:47 · 4 answers · asked by C.T. 2 in Science & Mathematics Chemistry

Volume of flask is 29.21 cm cubed

2006-08-30 11:25:48 · update #1

4 answers

Take the weight of the flask with the liquid in it. (That is the first line.) The subtract the weight of the empy flask. (That is the second line, the result is the third.) Then divide that result by the volume of the liquid in the flask.

2006-08-30 11:18:27 · answer #1 · answered by Cadair360 3 · 1 0

What is the VOLUME of the flask...........?

You can't determine the density from mass alone. Density is mass per unit volume.

Oh yeah, we'll also need the mass of the empty, dried flask and the stopper. Subtracting that weight from the total gives the 'net weight', the mass of the contents. Divide the net weight by the volume and you will have the density of the contents.

2006-08-30 11:15:49 · answer #2 · answered by poorcocoboiboi 6 · 0 0

To solve this one needs to know the volume of the flask and the weight of the bottle, none of which is given here.
With the info 'she' got above, she would have the additional weight over that when the bottle was filled with (I guess) water.

That extra weight as a percentage of the weight of the water only will make the density obvious.

2006-08-30 11:14:33 · answer #3 · answered by Anonymous · 1 0

Density of the unknown liquid = d volume of block = V, so volume of liquid displaced =V Mass of liquid displaced through block = density of liquid x volume of liquid For 1st liquid glaring lack of weight = weight of liquid displaced. 15 = 4 x V V = 15/4 litres For 2d liquid glaring lack of weight = weight of liquid displaced. 27 = d x V = d x 15/4 d = 27x4/15 = 7.2 kg/litre

2016-12-06 00:19:34 · answer #4 · answered by Anonymous · 0 0

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