a) 9cm+2.8cm=11.8cm~ 12 cm
b) 0.135atm+0.6atm=0.735atm~ . 7 atm
a) rounded to the ones place and 2 significant figures. As we have 9 cm which have no decimal so we have put our answer without any decimal
b) rounded to tenths place and 1 significant figure.we have minimum decimal upto one place. 6 atm so answer should be reported to one decimal
To what decimal place should each answer be rounded? How many significant figures does the rounded...
To how many significant figures should each answer be rounded? (6.626 x 10-34 J·s) (2.9979 x 10 m/s) = 4.880610663391 x 10-19 J(unrounded) 4.070 x 10-'m equation A: After rounding, the answer to equation A should have 5 significant figures. 2 significant figures. O 1 significant figure. O 3 significant figures 4 significant figures. equation B: (6.022 x 1023 atoms/mol) (0.935 g) 20.18 g/mol = 2.790 x 10” atoms (unrounded) After rounding, the answer to equation B should have O...
To how many significant figures should each answer be rounded? Equation A: (6.626× 10−34 J⋅s)(2.9979× 108 m/s)4.290×10−7 m=4.630322937063×10−19 J(unrounded) After rounding, the answer to equation A should have *2 significant figures. *1 significant figure. *4 significant figures. *3 significant figures. *5 significant figures. Equation B: (6.022× 1023 atoms/mol)(0.795 g)20.18 g/mol=2.372×1022 atoms(unrounded) After rounding, the answer to equation B should have *1 significant figure. *3 significant figures. *5 significant figures. *2 significant figures. *4 significant figures.
To how many significant figures should cach ans wer be rounded? (6626 x 10-34 J) (2.9979 x 10 m/s) equation A: #4.290299222462 x 109 J (unroun 4.630 x 107 m rounding, the answer to equation A should have O 3 significant figures O 2 significant figures O5 significant figures. O4 significant figures. O1 significant figure t helpe hp ome 7 tachspacs G K ente pause N alt ctri V Σ (6.022 x 1023 atoms/mol) (0.737 g) equation B 2199 x...
The "best" result is always associated with the measurement with the smallest uncertainty. Measurement B has half the standard uncertainty of measurement A. Therefor e our 68 % coverage probability is associated with a smaller interval (83.44 g o 83.56 g) for measurement B than measurement A (83.38 g to 83.62 g). In other words we have better knowledge about the value of the measurand from measurement B, since we have the same coveroge probability associated with a norrower interval....