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Now, based on all these rules the number which is given that is 0.02 has only one significant figure because the preceding zeros are not considered. Thus, the correct answer is that there is one significant figure in 0.02.So 0.002 is to one significant figure as the zeros are not significant they are there to place the 2 in the correct position relative to the decimal point. and 0.102 is to three significant figures.All zeros at the left of the number are NOT significant. Example: 0.00032 has 2 sig. figs, 0.03 has 1 sig.
Contents
Does 0.002 have significant figures?
So 0.002 is to one significant figure as the zeros are not significant they are there to place the 2 in the correct position relative to the decimal point. and 0.102 is to three significant figures.
Is 0.03 A significant figure?
All zeros at the left of the number are NOT significant. Example: 0.00032 has 2 sig. figs, 0.03 has 1 sig.
Is 0.20 A significant figure?
(4) Trailing zeroes that are also to the right of a decimal point in a number are significant: 0.0230 mL has 3 significant figures, 0.20 g has 2 significant figures.
How many significant figures does .025 have?
For example 0.025 has two significant figures. Rule 3: Trailing zeros are significant if a decimal point is shown in the number, but may or may not be significant if no decimal point is shown. By convention, it is assumed that trailing zeros without a decimal point are not significant.
How many significant figures does 0.04 have?
All zeros on the right of a decimal point and also to the left of a non-zero digit are never significant. For instance, contained three significant digits.
How many significant digits does 1.0 have?
The number 1.0 also has two significant digits. So does the number 130, but 10 and 100 only have one “sig fig” as written. Zeros that only hold places are not considered to be significant.
Is 0.5 A significant figure?
The zero to the left of a decimal value less than 1 is not significant. All trailing zeros that are placeholders are not significant. Zeros between non-zero numbers are significant. All non-zero numbers are significant.
What are the significant figures in 0.03 40?
0.0340 = 3 significant figures because the point before there is a zero we do not consider it and after the point there is zero that is also we do not consider ( because of significant figure rules).
How many significant figures are present in the number 10450?
10450 has 4 significant figures.
How many significant figures does 0.05 have?
Examples: 0.05 – one significant figure (5); the leading zeroes are ignored.
How many significant figures does 120.0 have?
All non-zero digits are significant. (2.3, 22, and 120 all have two significant figures)
Why is .0 a significant figure?
Since the zeros to the right of the decimal place are not necessary as place-holders, their inclusion indicates they are significant. Zeros to the right of the decimal place that are NOT merely place holders are significant; these significant zeros will be to the right of non-zero significant digits.
Are preceeding zeros significant?
Leading zeros before a number are not significant.
Therefore, only the 2 & 5 are counted meaning it has two significant digits. The leading zeros are known as placeholder zeros as they do not add to the precision of the measurement, they simply occupy the Page 2 ones, tenths, and hundredths places.
How many significant figures does 0.020 have?
Explanation: 0.020 has two significant figures. The 2 is significant because all non-zero numbers are signficant. The second 0 is significant because all zeros at the end of a decimal are significant.
How many significant figures does 91630 have?
Number | 91630 |
---|---|
Sig Figs | 91630 91630 has 4 significant figures. Steps 91630 has no insignificant leading zeroes. Ignore trailing zeros in 91630 to get 9163. |
Decimals | 91630 has 0 decimals. |
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How many significant figures are in the number 002 class 11 chemistry CBSE
Hint:The answer is in the fact that significant figures take into consideration the non zero digit and the zeros of the one which is present after the non zero digit. Answer the question based on this fact.
Complete step by step solution:
We have studied the basic topic which is named as ‘some basic concepts of chemistry’ and have come across several basic calculations such as the significant figures, rounding off the digits etc.
Let us see what does significant figure mean and how can we calculate it on the above given question.
– Some of the rules to be followed while writing the significant figures is as follows:
– All non – zero digits are significant.
– Zeros preceding the first non – zero digit are not significant. Such zero indicates the position of the decimal point.
– Zeros between two non – zero digits are significant.
– Zeros at the end or right side of a number are significant provided they are on the right side of the decimal point.
Now, based on all these rules the number which is given that is 0.02 has only one significant figure because the preceding zeros are not considered.
Thus, the correct answer is that there is one significant figure in 0.02.
Note:
Note that counting numbers of objects have infinite significant figures like 2 has infinite figures as it can be written as 2.0000000 and the zeros after decimal continues infinitely which means that these types of counting objects can be written in the form of infinite significant figures.
Why aren’t the 0’s significant figures in 0.002?
$\begingroup$
tl;dr- Leading zeros aren’t significant because they trivially drop out. Still, if you prefer to keep ’em, that’s fine, too; you’ll just end up having a bunch of leading zeros.
And, no, this isn’t something anyone should really want to do. But, the logic behind it and the consequences probably help show why.
Background: Regarding the construction of numbers
First, to define numbers:
Natural numbers are defined through enumeration on a number line from $0 .$ Integers as defined as natural numbers extended with decrementation (inverse enumeration) on a number line, allowing negative values. Real numbers are defined as integers with interpolation, allowing decimal values.
Conceptually, it’d be simplest if we gave each integer its own, unique symbol. But since no one wants to have to memorize arbitrarily many symbols, we tend to construct numeric identifiers through a transform$$ n ~~ \Rightarrow ~~ \sum_{i} c_i \cdot {b}^{i} \, ,$$where
$c_i$ is a numeric symbol selected from a limited subset of “digits” $\in \left[0,~b\right) ;$
$b$ is the base (and usually $10$);
then emit this construction as a string,$$ \hspace{25px} \boxed{\begin{alignat}{7} &\texttt{for}~\left(\texttt{var}~i~=~i_{\text{max}};~~~i~\ge~i_{\text{min}};~~~i\text{–}\right) \\ &\{ \\ & \hspace{2em} \texttt{Print} \left( c_i \right) ; \\ \\ & \hspace{2em} \texttt{if} \left( i ~\text{==}~ 0 \right) \\ & \hspace{2em} \{ \\ & \hspace{4em} \texttt{Print} \left( “ . ” \right) ; \\ & \hspace{2em} \} \\ & \} \end{alignat}} _{~ \large{.}}$$
Background: Regarding multiplication
Since multiplication is distributive, then the product of two numbers written in the same base, $b ,$ is$$ \begin{alignat}{7} n^{\text{A}} \times n^{\text{B}} ~~ & \Rightarrow ~~ && \left( \sum_{i} c_i^{\text{A}} \cdot {b}^{i} \right) \times \left( \sum_{j} c_j^{\text{B}} \cdot {b}^{j} \right) \\[5px] & = && \sum_{i} {\sum_{j} {c_i^{\text{A}} \cdot c_j^{\text{B}} \cdot {b}^{i} \cdot {b}^{j}}} \\[5px] & = && \sum_{i} {\sum_{j} {c_i^{\text{A}} \cdot c_j^{\text{B}} \cdot {b}^{i+j}}} \, .\end{alignat} $$
Background: Regarding truncation
The above definitions are written for numbers that contain infinite information. In practice, computers (including humans) are finite (unless you find a hypercomputer), so we terminate the procedure at two ends:
We declare some minimal basis, $\cdot b^{i_{\text{min}}} ,$ past which we ignore all further bases, typically under the argument that they’re noisy (if from measurement/estimation) or to save on computation work (as computers do). We declare some maximal basis, $\cdot b^{i_{\text{max}}} ,$ past which we ignore all further bases. Usually, we choose to selected $i_{\text{max}}$ such that we truncate only terms in which $c_i = 0 ,$ since the zero-terms don’t affect anything, anyway.
First, we note that any basis $\cdot b^k$ is affected by a noisy term if $k < i_{\text{min}} + j_{\text{max}}$ or/and $k < i_{\text{max}} + j_{\text{min}}$ – ignoring cases in which $c_i c_j \geq b ,$ which I'll mention later. Second, we note that any basis $\cdot b^k$ is a zero-term if $k > i_{\text{max}} + j_{\text{max}} .$
Given these two constraints, we’re therefore only interested in bases$$ \cdot b^k ~~ \text{such that} ~~ k \in \left[ \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} , ~ i_{\text{max}} + j_{\text{max}} \right] \, .$$Since the number of elements in an inclusive range like this, i.e. $\left[ n_{\text{min}}, ~ n_{\text{max}}\right] ,$ is $1 + n_{\text{max}} – n_{\text{min}} ,$we’re therefore interested in$$ \begin{alignat}{7} 1 + i_{\text{max}} + j_{\text{max}} – \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} ~~ & = ~~ && 1 + \min{\left( i_{\text{max}} – i_{\text{min}} , ~ j_{\text{max}} – j_{\text{min}} \right)} \\[5px] & = && \min{\left(1+ i_{\text{max}} – i_{\text{min}} , ~1+ j_{\text{max}} – j_{\text{min}} \right)} \end{alignat} $$bases.
So, ya know how they say that, when you multiply two numbers with significant figures, the product has the lesser of the multiplicands’ significant figures? That’s because$$ \underbrace{1 + k_{\text{max}} – k_{\text{min}}}_{\begin{array}{c} \text{significant figures} \\ \text{in the product} \end{array}} ~~ = ~~ \min{(\underbrace{1+ i_{\text{max}} – i_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{first multiplicand} \end{array}} , ~ \underbrace{1+ j_{\text{max}} – j_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{second multiplicand} \end{array}} )} \, .$$This is, the product has the lesser of the number of significant figures from either multiplicand.
Except, there’s one problem here: the above logic assumed that bases don’t overflow. Which would be true if we were working in Base-2 (binary), but in Base-10 (decimal), we have cases in which $c_i c_j \geq b ,$ e.g. $5 \times 5 \geq 10 .$ Won’t discuss that here since it’s ignored by the standard rules, but I think the problem’s obvious enough. That said, significant figures are meant to be an easy trick rather than used for rigorous calculations, so that they’re a bit broken is kind of a given. (See also: my answer here.)
Considering leading-zeros significant
In the above derivation of significant-figure logic, we selected the rule that leading zeros are to be ignored. So, what happens if we do consider them to be significant?
Specifically, you’re asking about the case in which $i_{\text{max}}$ or $j_{\text{max}}$ is less-than-zero – e.g., as in $0.01 ,$ in which $i_{\text{max}} = -2$ – and then asking why we can’t consider the zeros that we still wrote, so presumably $i_{\text{max}} = 0 .$
So, let’s call your alterations $i_{\text{max}}^{*}$ and $j_{\text{max}}^{*} ,$ where$$ i_{\text{max}}^{*} ~ \equiv ~\max{\left(i_\text{max}, ~ 0\right)} ~~~~ \text{and} ~~~~ j_{\text{max}}^{*} ~ \equiv ~\max{\left(j_\text{max}, ~ 0\right)} \, .$$Then, we say that we’re interested in “significant” figures that include leading zeros, though we must still reference the prior notions of $i_{\text{max}}$ and $j_{\text{max}}$ because they’re important to the issue of tracking the propagation of noise in the calculation.
So then, we’re interested in the bases$$ \cdot b^k ~~ \text{such that} ~~ k \in \left[ \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} , ~ i_{\text{max}}^{*} + j_{\text{max}}^{*} \right] \, ,$$which, to redo the element-count calculation, contains$$ 1 + i_{\text{max}}^{*} + j_{\text{max}}^{*} – \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} $$members.
So, to derive the new rule for significant figures, we just need to rewrite $i_{\text{min}}$ and $j_{\text{max}}$ in terms of $i_{\text{min}}^{*}$ and $j_{\text{max}}^{*} ,$ and we’re done.
So, um. If $i_{\text{max}}^{*} ~ \equiv ~\max{\left(i_\text{max}, ~ 0\right)} ,$ and we know $i_{\text{max}}^{*} ,$ then how do we calculate $i_\text{max} ?$
I mean, obviously, $i_{\text{max}}$ is either going to be equal to $i_{\text{max}}^{*}$ if $i_{\text{max}}^{*} > 0 ,$ but if $i_{\text{max}}^{*} = 0 ,$ then all we know is that $i_{\text{min}} \leq i_{\text{max}} \leq 0 .$ Our problem is that this information is lost, such that merely knowing the number of “significant figures” is insufficient to establish how many we need.
But, screw it, significant figures are a hack anyway. And, so long as we respect the bound on the least-significant basis, we can keep extra zeros if we like. Since, ya know, they don’t affect anything.
So to avoid accidentally truncating leading non-zero digits, it’s left to us to write the rules such that there’re at least as many significant figures in the product as necessary to keep it consistent.
So, we need a number of significant figures equal to$$ \max{\left( 1 + i_{\text{max}}^{*} + j_{\text{max}}^{*} – \max{\left( i_{\text{min}} + j_{\text{max}} , ~ i_{\text{max}} + j_{\text{min}} \right)} ~~~~\forall \begin{array}{l} i_{\text{max}} \in \left[i_{\text{min}}, ~ \max{\left(0, ~ i_{\text{max}}^* \right)}\right] \\ j_{\text{max}} \in \left[j_{\text{min}}, ~ \max{\left(0, ~ j_{\text{max}}^* \right)}\right] \end{array} \right)} \, ,$$which reduces to$$ 1 + i_{\text{max}}^{*} – i_{\text{min}} + j_{\text{max}}^{*} – j_{\text{min}} \, ,$$or$$ \underbrace{1 + i_{\text{max}}^{*} – i_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{first multiplicand} \end{array}} + \underbrace{1 + j_{\text{max}}^{*} – j_{\text{min}}}_{\begin{array}{c} \text{significant figures in the} \\ \text{second multiplicand} \end{array}} – 1 \, .$$In other words, the new rule is that we need to retain a number of significant figures equal to the sum of the significant figures of the multiplicands, minus one, where the new “significant” digits, if any, are leading zeros.
Which is a rule you can have, but it then requires that you recount the significant figures of the product afterward before doing further operations, as this logic’s non-conservative.
Problem: Number of significant figures grows
To avoid improper truncation, we had to keep at least as many leading zeros as to ensure that nothing was dropped. If these calculations are repeated, the bloat of leading zeros may continue to grow. (Which I haven’t actually worked out; typing this all out took longer than I originally estimated, and, honestly, I’m bored. =P)
But, since the premise of this derivation is that we don’t mind doing trivial work, being why we rejected dropping terms $c_i c_j b^{i+j} = 0$ from the calculation, that’s presumably not an issue for someone who’d want to use this logic.
Conclusion
Long story short, you can consider leading zeros “significant” if you want, then maintain a bunch of leading zeros in front of numbers to maintain that logic.
It doesn’t really mean anything, as it’s basically just keeping extra zeros, but it’s a mathematically consistent calculation approach one could take if they were so inclined.
Significant Figures
Scientists imply the level of precision in measurements by how they report the number. Unlike in mathematics where 25 and 25.0 are identical, a measurement of 25 cm in science means something drastically different than a measurement of 25.0 cm.
Scientific measurements are reported to one digit more than what is known with certainty. A reported value of 25 cm implies that the actual value is somewhere between 24 cm and 26 cm, approximately. In contrast, a reported value of 25.0 cm implies that the actual value is somewhere between 24.9 cm and 25.1 cm, approximately.
Counting Significant Figures
To know which digits in a number are “significant”, first let’s go over some terminology. Non-zero digits are 1, 2, 3, 4, 5, 6, 7, 8, and 9 (any digit other than zero). Leading zeros are those that come before all the non-zero digits (but not necessarily before the decimal point). Trailing zeros are those that come after all the non-zero digits (but not necessarily after the decimal point).
Rule 1: Non-zero digits are always significant. For example, 25 has two significant figures.
Rule 2: Leading zeros are never significant. For example 0.025 has two significant figures.
Rule 3: Trailing zeros are significant if a decimal point is shown in the number, but may or may not be significant if no decimal point is shown. By convention, it is assumed that trailing zeros without a decimal point are not significant. For example, 250.0 has four significant figures, but 2500 only has two definitive significant figures. In these cases, it is best to write the number in scientific notation to avoid ambiguity.
Rule 4: Zeroes that are between non-zero digits are always significant.
Rounding
You will often need to round calculated numbers to the proper number of significant figures. For example, 25.18 rounded to one significant figure is 30, rounded to two significant figures is 25, and rounded to three significant figures is 25.2.
How would you round a number like 99.99 to three significant figures? You could either write 100. or 1.00×102.
Multiplication and Division of Measurements
The number of significant figures in the product or quotient of two or more measurements cannot be greater than that of the measurement with the fewest significant figures. For example: 25.0 × 1.0 = 25
Addition and Subtraction of Measurements
The number of digits after the decimal point in the sum or difference of two or more measurements cannot be greater than that of the measurement with the fewest digits after its decimal point. For example: 25.0 + 1 = 26
Logarithms and Antilogs
The number of digits after the decimal point of log(x) should be equal to the number of significant figures of x. For example: log (3.5×105) = 5.54
For an inverse log of x (10x), the number of significant figures in the answer should be equal to the number of digits after the decimal point in x. For example: 10–3.421 = 3.79×10–4
Multi-step Calculations
When doing a series of mathematical operations, you should not round at each individual step. You should keep track of the number of significant figures produced at each step, then round at the end.
For example: (30.62 ÷ 2.0) + 0.4 = 16
To arrive at this answer, you would first analyze each step…
1) 30.62 ÷ 2.0 = 15.31. If rounded, the result would have only have two significant figures, and thus no digits after the decimal point.
2) Take the unrounded value, 15.31, and add 0.4. 15.31 + 0.4 = 15.71. From step 1, we know the result should have no digits after the decimal point, and so we round to the ones place, 15.71 ≈ 16.
You may be wondering, “Why is 16 a better answer than 15.71? Isn’t the number more accurate with more digits?”. But remember, the original values (30.62, 2.0, and 0.4) each have some degree of error in them (since nearly everything in science is a measurement). If we report the answer as 15.71, we imply that we know it’s between 15.70 and 15.72, and that is simply not true. The accumulated error puts the ones place in doubt, so by reporting the answer as 16 we imply that it’s roughly between 15 and 17.
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