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If the last digit is 5 or greater you will round up. If the last digit is 4 or less, round down. In the case of .4566, if you needed 3 sig figs it would be .457 and for .4565 it would also be .457 because the last digit in both is 5 or greater.
Professor Lavelle also has a link on his website (https://lavelle.chem.ucla.edu/wp-conten ... OUT_SF.pdf) that is really helpful for figuring out sig figs :)
The link that Lavelle posted on his website for sig figs states the rule. If the last digit is 5 and no number follows it, then you have to round to the nearest even number. Thus, if you need three sig figs and you have 0.4565, you have to round to the closest even number which is 6. Thus, it becomes 0.456. I noticed one of the problems in the solution book followed this rule.
For rounding numbers that end in 5 in chemistry you want to try to round it to the nearest even number. 0.55 would round up to 0.6, 0.65 would round down to 0.6, etc. Usually when you have many numbers like this, rounding in this manner can somewhat "average" everything out. You would not round like that in any other math or science class though. So in this case, .4566 would round to .457, and .4565 would round to .456
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