During an experiment, a strip of magnesium (mg) was measured for its length. the scientists completed four trials of measurements. the first scientist measured trials 1 and 2 with a ruler that measures with mm markings. the second scientist measured trials 3 and 4 with a ruler that measures with inches. markings and then converted the data into mm. which options most accurately explain why the scientists’ data were imprecise? (select all that apply) - according to the scenario, the scientists used a piece of lab equipment that is not precise enough to measure the strip of magnesium. - when doing calculations like converting units, the number of significant figures is altered. this can change the level of uncertainty among the trial measurements. - during the recording of data, the first scientist could have converted the trial results incorrectly and caused the data to appear imprecise. - during a lab experiment, having different team members collect trial results could alter the precision of the data based on different individuals’ improper or biased reading of the equipment.

The only option that fits is the last one — improper and biased measurements.

A scientist should be able to estimate a measurement to a tenth of the smallest scale division. The millimetre ruler would give an uncertainty of ±0.1 mm. The inch ruler would give an uncertainty of ±0.1 in. Since 1 in = 25.4 mm, 0.1 in = 2.54 mm. Thus, the millimetre ruler is more precise than the inch ruler.

Converting units does not change the degree of precision of a measurement.

The first scientist did not convert the units (the second scientist did!).

BUT, having different team members make the observations could introduce uncertainties caused by improper or biased readings of the equipment.

warm front forms when a moist, warm air mass slides up and over a cold air mass. as the warm air mass rises, it condenses into a broad area of clouds. a warm front brings gentle rain or light snow, followed by warmer, milder weather.

hope i !

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole