page 51 61 teChapter Section 36 s3. The same volun ment pictured the right). Exp

Question

page 51

61

teChapter Section 36 s3. The same volun ment pictured the right). Exp different value ement and Chemical Caleulations Significant Figures tate the volume of lia in the figure below te numbeto be e num upon the appropriate y ofines arenb of significant figures markings are calibrated to explain how you decided mber o uid in each graduated cylinder ure below and be read at the and the numbers tom of the curved surface. represent milliliters 10 28 mL 54. W ri . How long is the object measured with the rulers shown? The ruler is calibrated in inches. Explain why vou

Explanation / Answer

Ans-51: In this graduated cylinder each sub-division is of 0.2mL therefore for the first picture the volume is up to 3.6mL because its 3 mL and 3 sub-division that is 3 + 0.6 which gives us 3.6 ( we see lower meniscus to measure the volume).
Similarly for 2nd picture it is 1mL plus 2 subdivision and then then lower meniscus is little bit above 2nd sub division therefore it must be 1.44mL.

Ans-60: mass of Sodium Chloride =2.86gm
               mass of Ammonium Sulfate =3.9gm
               mass of Potassium Iodide =0.896gm
               mass of water                     =246gm
    Total mass will be 253.656gm
It will be 253.7gm when expressed with significant figure.

Ans-61: Initial volume = 22.93mL
               Final volume = 19.4mL
therefore volume leaked is 22.93-19.4mL = 3.53mL.

Ans-62: Mass of the beaker =94.33gm
              Mass after adding chemical to the beaker = 101.209gm
therefore mass of the chemical added = 101.209-94.33gm = 6.879gm.

Ans-63: Mass of one mole of sugar is = 342.3gm
              therefore in half mole of sugar mass will be = ½(342.3gm) = 171.15gm
              similarly in0.764 mole the mass will be = 342.3*0.764 = 261.5172gm.

Ans-64: In one liter of solution there is 31.4gm of substance, so in 2 liter the mass will be doubled that is 31.4*2 =62.8gm. Similarly in 7.37 liter the mass will be= 7.37*31.4 = 231.418.
It will be 231.4gm when expressed with significant figure.

Ans-65: Mass of the empty beaker = 42.3gm
               Mass of the compound and beaker together = 62.87gm
Therefore the mass of the liquid compound is = 62.87-42.3gm = 20.57gm.
Volume of the liquid compound is = 19mL
we know the formula for Density,
     Density = Mass/volume
                   = 20.57/19 =1.08gm/mL

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