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Travka [436]
2 years ago
7

can someone help me with these? I just started chemistry this semester, and since it's involving math I don't see myself doing g

ood.

Chemistry
1 answer:
V125BC [204]2 years ago
4 0

Any non zero digits are significant. (120 has 2 sig figs)

Zeroes between significant digits are significant. (103 has 3 sig figs)

Zeroes only count if there is a decimal point and if they are following a significant digit. (.600 has 3 sig figs; .067 has 2 sig figs)

1) <u>1278.50</u> = 6 sig figs

2) <u>12</u>0000 = 2 sig figs

3) <u>90027.00</u> = 7 sig figs

4) 0.00<u>53567</u> = 5 sig figs

5) <u>67</u>0 = 2 sig figs

6) 0.00<u>730</u> = 3 sig figs

That's all I'm gonna do so yeah.

Now for the rounding, it's basically rounding but it must have the x amount of sig figs after rounding. It's pretty hard to explain.

19, I'm not really sure. Might be 120 with a decimal point at the end.

20) 5.457, alright so here, since we are rounding to three significant digits, we look at the third sig fig. The third sig fig here would be 5. Then we just round. Since the number next to 5 is 7, we round 5 up to 6.

So 5.457 rounded to 3 sig figs would be 5.46

21) 0.0008769, the third sig fig here would be 6. You round 6 up to 7 because the number next to it is 9.

So 0.0008769 rounded to 3 sig figs would be 0.000877

So I hope you get it, I'm gonna let you do the rest.

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