Calculate Unknowns with Sig Figs

[Quincy]

Homework Statement

Calculate the unknown variables to the appropriate significant figures.

The Attempt at a Solution

mass of metal: 6.31 g - 3 sig figs
volume of water: 20.00 mL - 2 sig figs
volume of water + metal: 20.50 mL - 3 sig figs

volume of metal: 20.50 - 20.00 = 0.50 mL - 2 sig figs
density of metal: 13 g/mL - so the answer should have 2 sig figs correct?

 

[HallsofIvy]

Homework Statement

Calculate the unknown variables to the appropriate significant figures.

The Attempt at a Solution

mass of metal: 6.31 g - 3 sig figs
volume of water: 20.00 mL - 2 sig figs

No, that's 4 significant figures. That indicates that the measurement might be as much as 20.005 or as little as 19.995 but rounds off to 20.00. If only 2 significant figures were intended, if it might be as high as 20.5 or as low as 19.5, then it would be written as "20. mL". If 3 significant figures were intended (as high as 20.05 or as low as 19.95), it would be written 20.0.

volume of water + metal: 20.50 mL - 3 sig fig

Again, that is 4 significant figures

volume of metal: 20.50 - 20.00 = 0.50 mL - 2 sig figs

Because it is determined by subtracting two figures of 4 significant figures, the result should be to 4 significant figures. It would be better written as 5.000 x 10^-1.

[/quote]density of metal: 13 g/mL - so the answer should have 2 sig figs correct?[/QUOTE]
Since you got that by dividing 6.31 g (3 significant figures) by 5.000 x 10^-1 mL (4 significant figures) the result should be of the lesser of those: 3 significant figures. Which would be better written 13.0 g/mL or 1.30 x 10^{-1 g/L}

 

[Quincy]

0.50 - is this 3 significant figures btw?

 

[mgb_phys]

0.50 - is this 3 significant figures btw?

No it's two significant figures.
It is one that catches people out because 1.50 WOULD be 3 figures.
The trick is that you could write it as "5.0 divided by 10" and so using only two figures give exactly the same accuracy.