6
Zeros between significant digits are significant
The last zero specifies the number more, so it is significant
Two - the trailing zeros are just placeholders.
There are 3 significant figures in this number.
There are 6 significant figures in this number.
To determine the number of significant figures in the product of 0.1400, 6.02, and (10^{23}), we need to identify the significant figures in each number. The number 0.1400 has four significant figures, 6.02 has three significant figures, and (10^{23}) has one significant figure (as it is a power of ten). The product will have the same number of significant figures as the term with the least significant figures, which is 6.02 with three significant figures. Therefore, the final product will have three significant figures.
12.5912
10 significant figures.
80.07749999999999
3 of them.
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
3
4
The least number of significant figures in any number of the problem determines the number of significant figures in the answer which in this case is 656.64
The least number of significant figures in any number of the problem determines the number of significant figures in the answer which in this case is 273.8
The least number of significant figures in any number of the problem determines the number of significant figures in the answer which in this case is 270.8
To determine the number of significant figures in the answer to the multiplication problem 532.1 x 101, you need to look at the number of significant figures in each of the numbers involved. The number 532.1 has four significant figures, while 101 has three significant figures. The final answer will have the same number of significant figures as the measurement with the least amount, which is three significant figures in this case.
It might have been possible to answer the question if the "following" multiplication had followed. But since you did not bother to make sure that it did, I cannot provide a more useful answer.
2.6