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MEASUREMENTCounting significant digits when measurements are MLMultiply or divide the following measurements_ Be sure each20.94 mL90. mLmol 78.0850. Lmol791.9 mol60...

Question

MEASUREMENTCounting significant digits when measurements are MLMultiply or divide the following measurements_ Be sure each20.94 mL90. mLmol 78.0850. Lmol791.9 mol60.330 Lmoj

MEASUREMENT Counting significant digits when measurements are ML Multiply or divide the following measurements_ Be sure each 20.94 mL 90. mL mol 78.08 50. L mol 791.9 mol 60.330 L moj



Answers

Determine the number of significant figures in each measurement.
a. 0.049450 $\mathrm{s}$ $\quad$ c. $3.1587 \times 10^{-4} \mathrm{g}$
b. 0.000482 $\mathrm{mL}$ $\quad$ d. 0.0084 $\mathrm{mL}$

Change the 36 with six balls figures during the Kirk number, So pain the correct longer in 56 threes being is 36. Seen is five g is too.

We're given three problems, all of which asked the final answer to be rounded to the proper number of significant figures. In part, A were asked to take 24.567 grams and add to that 0.4478 grams start units match, so there's no confusion there. When this is punched into a calculator, you will get a return of 24.61178 grams. The rule for additional subtraction with significant figures is that your answer cannot have any more decimal places than the fewest. Seeing the problem. Our first part had three decimal places where, as our final part had five decimal places. Therefore, my answer must be rounded to three decimal places. So my final answer for part A will be 24.61 two grams because the seven will round up with one in Part B were instead given a division problem, we want to take 4.67 for two grams and divide that by 0.371 leaders. When this is punched into a calculator, you get a return of 1259 point 892 183 grams per liter. The rule for division and multiplication is that your answer cannot have any more significant figures in the fewest scene in the problem here we were given five significant figures since they were all non zeros. Here we follow the rule that leading zeros on a decimal number or not significant. So this only had three. So my answer must conform to this standard. It must have three significant figures. Now if I was to write it like such, I wouldn't know if that last zero was significant or not. So I will transfer this into scientific notation in order to identify the fact that now I for sure have three significant figures on the final part. See, we're given both addition and subtraction. 0.378 mil leaders is added to 42.3 mil leaders, which is then subtracted by 1.5833 mil leaders. When this is punching to a calculator, you will get a return of 41.9 47 mil leaders like I said before, in addition and subtraction, you cannot have any more decimal places than the fewest scene in the problem. This part had three decimal places. This one had one and this one had four. The fewest is one. Therefore, my answer must be rounded toe one decimal place. So look at the second decimal place and realized that's going to round up zero and therefore I will round this properly to 41.1 milliliters.

Okay, so starting with the first part A. We have 0.5 millimeter in this question, there is only one significant digit Option B. We have 0.50 ml. We have two significant digits options. See we have 5.0 millimeter. In this case we have three significant significant digits, and next is 500 millimeter, we have one significant digits.

Mhm. Okay again we have to calculate the number of significant digits. So starting with one 10 days to power negative one millimeter. We only have one significant digit Option B. We have 1.0 10 days to put negative to. In this case we have two significant digit options. See we have 1.0 10 days to power one Milito. We have three significant digits 10 and zero. And lastly we have four significant digits 1000.


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