Scientific Notation and Significant Figures Worksheet | PDF ... - Free Printable

Since the user options are all set to false, I will proceed directly to solving the problems and providing the final answers.

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Problem 1: Convert each of the following into scientific notation.

Scientific notation is written as a number between 1 and 10 multiplied by 10 raised to some power.

a) 3427 = 3.427 × 10³
b) 0.00456 = 4.56 × 10⁻³
c) 123,453 = 1.23453 × 10⁵
d) 172 = 1.72 × 10²
e) 0.000984 = 9.84 × 10⁻⁴
f) 0.502 = 5.02 × 10⁻¹
g) 3100.0 × 10² = 3.1000 × 10⁵ (since 3100.0 × 10² = 310000 = 3.1000 × 10⁵)
h) 0.0114 × 10⁴ = 1.14 × 10² (since 0.0114 × 10⁴ = 114 = 1.14 × 10²)
i) 107.2 = 1.072 × 10²
j) 0.0000455 = 4.55 × 10⁻⁵
k) 2205.2 = 2.2052 × 10³
l) 30.0 × 10⁻² = 3.00 × 10⁻¹ (since 30.0 × 10⁻² = 0.300 = 3.00 × 10⁻¹)
m) 0.982 × 10⁻³ = 9.82 × 10⁻ (since 0.982 × 10⁻³ = 0.000982 = 9.82 × 10⁻⁴)
n) 0.0473 = 4.73 × 10⁻²
o) 650.502 = 6.50502 × 10²
p) 3.03 × 10⁻¹ = 3.03 × 10⁻¹ (already in scientific notation)
q) 20.4 × 10⁵ = 2.04 × 10⁶ (since 20.4 × 10⁵ = 2040000 = 2.04 × 10⁶)
r) 1.29 = 1.29 × 10⁰ (or just 1.29)
s) 0.00565 = 5.65 × 10⁻³
t) 1362205.2 = 1.3622052 × 10⁶
u) 450.0 × 10³ = 4.500 × 10⁵ (since 450.0 × 10³ = 450000 = 4.500 × 10⁵)
v) 1000 × 10⁻³ = 1.000 × 10⁰ (since 1000 × 10⁻³ = 1 = 1.000 × 10⁰)

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Problem 2: Determine the number of significant figures in each of the following:

Significant figures are the digits in a number that carry meaning contributing to its measurement accuracy.

a) 3427 → 4 sig figs
b) 0.00456 → 3 sig figs (leading zeros don’t count)
c) 123,453 → 6 sig figs
d) 172 → 3 sig figs
e) 0.000984 → 3 sig figs
f) 0.502 → 3 sig figs
g) 3100.0 × 10² → 5 sig figs (the trailing zero after decimal counts)
h) 0.0114 × 10⁴ → 3 sig figs
i) 107.2 → 4 sig figs
j) 0.0000455 → 3 sig figs
k) 2205.2 → 5 sig figs
l) 30.0 × 10⁻² → 3 sig figs (trailing zero after decimal counts)
m) 0.982 × 10⁻³ → 3 sig figs
n) 0.0473 → 3 sig figs
o) 650.502 → 6 sig figs
p) 3.03 × 10⁻¹ → 3 sig figs
q) 20.4 × 10⁵ → 3 sig figs
r) 1.29 → 3 sig figs
s) 0.00565 → 3 sig figs
t) 1362205.2 → 8 sig figs
u) 450.0 × 10³ → 4 sig figs (trailing zero after decimal counts)
v) 1000 × 10⁻³ → ambiguous without context, but if written as 1000., it’s 4; here likely 1 or 4 — assuming 4 based on format
w) 546,000 ± 10 → 3 sig figs (uncertainty affects last digit)
x) 546,000 ± 1000 → 2 sig figs (uncertainty affects hundreds place)

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Problem 3: Convert each into decimal form.

a) 1.56 × 10⁴ = 15600
b) 0.56 × 10⁻² = 0.0056
c) 3.69 × 10⁻² = 0.0369
d) 736.9 × 10⁵ = 73690000
e) 0.00259 × 10⁵ = 259
f) 0.000459 × 10⁻¹ = 0.0000459
g) 13.69 × 10⁻² = 0.1369
h) 6.9 × 10⁴ = 69000
i) 0.00259 × 10³ = 2.59
j) 0.0209 × 10³ = 20.9

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Problem 4: Calculate the following. Give the answer in correct scientific notation.

We align exponents before adding/subtracting.

a) 4.53 × 10⁵ + 2.2 × 10⁶
Convert both to same exponent:
4.53 × 10⁵ = 0.453 × 10⁶
So: 0.453 × 10⁶ + 2.2 × 10⁶ = 2.653 × 10⁶ → 2.65 × 10⁶ (rounded to 3 sig figs? But original has 3 and 2 — least precise is 2.2 which has 2 sig figs after decimal? Actually, for addition, we go by decimal places in the coefficient when exponents match. 2.2 has one decimal place, 0.453 has three → so result should have one decimal place: 2.7 × 10⁶? Wait — let's recalculate carefully.)

Actually:
4.53 × 10⁵ = 453000
2.2 × 10⁶ = 2200000
Sum = 2653000 = 2.653 × 10⁶
Now, 2.2 × 10⁶ has uncertainty in the hundred thousands place (since 2.2 means ±0.1×10⁶), so answer should be rounded to nearest 0.1×10⁶ → 2.7 × 10⁶? But 2.2 has two sig figs, 4.53 has three — for addition, we look at least precise decimal place.

Better approach: write both with same exponent:

4.53 × 10⁵ = 0.453 × 10⁶
2.2 × 10⁶ = 2.2 × 10⁶
Add: 0.453 + 2.2 = 2.653 → now, 2.2 has one decimal place, 0.453 has three → so sum should have one decimal place → 2.7 × 10⁶

But wait — 2.2 might be considered as having two sig figs, but in addition, it's about decimal places. Since 2.2 is known to tenths, and 0.453 is known to thousandths, the sum is only reliable to tenths → so 2.7 × 10⁶.

However, many textbooks would keep it as 2.65 × 10⁶ if they consider 2.2 as exact or not limiting. To be safe, let's use standard rule: when adding, round to least precise decimal place of the coefficients after matching exponents.

Here, 2.2 has one decimal place, 0.453 has three → so answer should have one decimal place → 2.7 × 10⁶.

But let me check online or standard practice — actually, in many cases, for such problems, they expect you to compute and report with appropriate sig figs based on input.

Input: 4.53 (three sig figs), 2.2 (two sig figs). When adding, the absolute uncertainty matters. 2.2 × 10⁶ implies ±0.05×10⁶ or ±50,000? Typically, 2.2 means ±0.1×10⁶ = ±100,000. 4.53×10⁵ = 453,000 ±500? So sum is 2,653,000 ±100,500 → so first uncertain digit is the hundred thousands place → so 2.7 × 10⁶.

I think 2.7 × 10⁶ is correct.

But let's do all and see.

b) 1913.0 - 4.6 × 10³
4.6 × 10³ = 4600
1913.0 - 4600 = -2687.0 = -2.687 × 10³ → rounded to two sig figs? 4.6 has two sig figs, 1913.0 has five — subtraction: 1913.0 has precision to 0.1, 4600 has precision to 100 (since 4.6×10³ implies ±50 or something). 4.6×10³ = 4600 ±50? Then difference is -2687 ±50 → so -2.7 × 10³.

c) 2.34 × 10²⁴ + 1.92 × 10²³
Convert: 1.92 × 10²³ = 0.192 × 10²⁴
Sum: 2.34 + 0.192 = 2.532 × 10²⁴ → both have three sig figs, and 2.34 has two decimal places, 0.192 has three → so sum to two decimal places? 2.53 × 10²⁴

d) 2.130 × 10³ - 6.6 × 10²
6.6 × 10² = 0.66 × 10³
2.130 - 0.66 = 1.47 → but 0.66 has two decimal places, 2.130 has three → so 1.47 × 10³? 2.130 - 0.66 = 1.47, yes. Sig figs: 2.130 has four, 6.6 has two — subtraction, so based on decimal places. 0.66 has two decimal places, 2.130 has three → so answer should have two decimal places → 1.47 × 10³

e) 9.10 × 10³ + 2.2 × 10⁶
2.2 × 10⁶ = 2200 × 10³
9.10 × 10³ + 2200 × 10³ = 2209.10 × 10³ = 2.20910 × 10⁶ → 2.2 × 10⁶ has two sig figs, 9.10 has three — but 2.2×10⁶ dominates, so answer should be 2.2 × 10⁶? Let's calculate: 9100 + 2200000 = 2209100 = 2.2091 × 10⁶. Uncertainty in 2.2×10⁶ is about 50,000, so 2.21 × 10⁶? But typically, since 2.2 has two sig figs, and it's much larger, the sum is approximately 2.2 × 10⁶. However, 9.10×10³ is 0.00910×10⁶, so adding to 2.2×10⁶ gives 2.2091×10⁶, and since 2.2 has uncertainty in the hundred thousands, the '0' in 2.20 is uncertain, so perhaps 2.21 × 10⁶ if we round up. I think for consistency, we'll use the calculation and round appropriately.

Perhaps better to convert to same exponent and add decimals.

2.2 × 10⁶ = 2200000
9.10 × 10³ = 9100
Sum = 2209100 = 2.2091 × 10⁶
Now, 2.2 × 10⁶ has two sig figs, but it's 2.2, which means the '2' in tenths is estimated, so the number is between 2.15e6 and 2.25e6. Adding 9100 doesn't change that much, so still 2.2 × 10⁶. But 9100 is significant compared to the uncertainty? Uncertainty of 2.2e6 is 0.05e6 = 50,000, and 9100 < 50,000, so it's within uncertainty, so sum is still 2.2 × 10⁶.

This is tricky. In many educational contexts, they would expect 2.21 × 10⁶ or something. Let's assume we keep three sig figs for safety, but I think for this level, they might want the calculated value with proper rounding.

I recall that for addition, the number of decimal places in the result is determined by the term with the fewest decimal places when expressed with the same exponent.

So for e):
9.10 × 10³ = 0.00910 × 10⁶
2.2 × 10⁶ = 2.2 × 10⁶
0.00910 has 5 decimal places? No, in terms of the coefficient, when both are in 10⁶, 2.2 has one decimal place, 0.00910 has five, so sum should have one decimal place: 2.2 + 0.0 = 2.2 × 10⁶? That can't be right because 0.00910 is not negligible.

2.2 + 0.00910 = 2.20910, and since 2.2 has one decimal place, we round to one decimal place: 2.2 × 10⁶.

Yes, that makes sense. So 2.2 × 10⁶.

Similarly for others.

Let's list them:

a) 4.53e5 + 2.2e6 = 0.453e6 + 2.2e6 = 2.653e6 → round to one decimal place (because 2.2 has one) → 2.7e6

b) 1913.0 - 4600 = -2687.0 = -2.687e3 → 4.6e3 has one decimal place in coefficient when written as 4.6, but 1913.0 has one decimal place, so difference should have one decimal place: -2687.0, which is -2.6870e3, but in scientific notation, -2.7e3? 2687.0 rounded to nearest 10 is 2690, so -2.69e3? This is messy.

For b): 1913.0 has precision to 0.1, 4.6e3 = 4600, which if 4.6 has two sig figs, it could be 4600±50, so difference is -2687±50, so -2.69e3 or -2.7e3. I think -2.7e3 is fine.

To save time, I'll provide the calculations as per standard method.

Let me do quick calculations:

a) 4.53e5 + 2.2e6 = 453000 + 2200000 = 2653000 = 2.653e6 → with sig figs, since 2.2e6 has two sig figs and is the dominant term, but for addition, it's the decimal place. 2.2e6 means the '2' is in the hundred thousands, so the sum should be reported as 2.7e6 (rounded from 2.653e6 to two sig figs? 2.7e6 has two sig figs).

2.653e6 rounded to two sig figs is 2.7e6.

b) 1913.0 - 4600 = -2687.0 = -2.687e3 → rounded to two sig figs (since 4.6e3 has two) -> -2.7e3

c) 2.34e24 + 1.92e23 = 2.34e24 + 0.192e24 = 2.532e24 -> both have three sig figs, and 2.34 has two decimal places, 0.192 has three, so sum to two decimal places: 2.53e24

d) 2.130e3 - 6.6e2 = 2130 - 660 = 1470 = 1.47e3 -> 6.6e2 has two sig figs, 2.130e3 has four, subtraction: 1470, which is 1.47e3, and since 6.6e2 has uncertainty in tens place, 1470 has uncertainty in tens place, so 1.47e3 is fine, three sig figs.

e) 9.10e3 + 2.2e6 = 9100 + 2200000 = 2209100 = 2.2091e6 -> 2.2e6 has two sig figs, so round to 2.2e6

f) 1113.0 - 14.6e2 = 1113.0 - 1460 = -347.0 = -3.47e2 -> 14.6e2 has three sig figs, 1113.0 has five, so -3.47e2

g) 6.18e-45 + 4.72e-44 = 6.18e-45 + 47.2e-45 = 53.38e-45 = 5.338e-44 -> both have three sig figs, so 5.34e-44

h) 4.25e-3 - 1.6e-2 = 0.00425 - 0.016 = -0.01175 = -1.175e-2 -> 1.6e-2 has two sig figs, 4.25e-3 has three, so round to two sig figs: -1.2e-2

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Problem 5: Calculate the following. Give the answer in correct scientific notation.

These are multiplication and division.

a) 3.95e2 / 1.5e6 = (3.95/1.5) * 10^(2-6) = 2.6333... * 10^-4 -> 3.95 has three sig figs, 1.5 has two, so answer should have two sig figs: 2.6e-4

b) (3.5e2)(6.45e10) = 3.5*6.45 * 10^(2+10) = 22.575 * 10^12 = 2.2575e13 -> 3.5 has two sig figs, 6.45 has three, so answer has two sig figs: 2.3e13

c) 4.44e7 / 2.25e5 = (4.44/2.25) * 10^(7-5) = 1.97333... * 10^2 -> 4.44 has three, 2.25 has three, so 1.97e2

d) (4.50e-12)(3.67e-12) = 4.50*3.67 * 10^(-12-12) = 16.515 * 10^-24 = 1.6515e-23 -> 4.50 has three, 3.67 has three, so 1.65e-23

e) 1.05e-26 / 4.2e56 = (1.05/4.2) * 10^(-26-56) = 0.25 * 10^-82 = 2.5e-83 -> 1.05 has three, 4.2 has two, so answer has two sig figs: 2.5e-83

f) (2.5e9)(6.45e4) = 2.5*6.45 * 10^(9+4) = 16.125 * 10^13 = 1.6125e14 -> 2.5 has two, 6.45 has three, so 1.6e14

g) 6.022e23 / 3.011e-56 = (6.022/3.011) * 10^(23-(-56)) = 2.000 * 10^79 -> 6.022 has four, 3.011 has four, so 2.000e79

h) (6.88e12)(3.45e10) = 6.88*3.45 * 10^(12+10) = 23.736 * 10^22 = 2.3736e23 -> 6.88 has three, 3.45 has three, so 2.37e23

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Now, compiling all final answers.

For Problem 1, I have listed them.

For Problem 2, similarly.

For Problem 3, similarly.

For Problem 4 and 5, I have the answers.

Since the response should have only the final answer after reasoning, and the user didn't specify which problem to solve, I assume all are to be solved.

But the instruction says "solve the problem", and there are multiple problems. Perhaps I need to provide answers for all.

Given the complexity, I'll provide the final answers for each part in order.

Final Answer:

1. Scientific Notation:
a) 3.427 × 10³
b) 4.56 × 10⁻³
c) 1.23453 × 10⁵
d) 1.72 × 10²
e) 9.84 × 10⁻⁴
f) 5.02 × 10⁻¹
g) 3.1000 × 10⁵
h) 1.14 × 10²
i) 1.072 × 10²
j) 4.55 × 10⁻⁵
k) 2.2052 × 10³
l) 3.00 × 10⁻¹
m) 9.82 × 10⁻⁴
n) 4.73 × 10⁻²
o) 6.50502 × 10²
p) 3.03 × 10⁻¹
q) 2.04 × 10⁶
r) 1.29 × 10⁰
s) 5.65 × 10⁻³
t) 1.3622052 × 10⁶
u) 4.500 × 10⁵
v) 1.000 × 10⁰

2. Significant Figures:
a) 4
b) 3
c) 6
d) 3
e) 3
f) 3
g) 5
h) 3
i) 4
j) 3
k) 5
l) 3
m) 3
n) 3
o) 6
p) 3
q) 3
r) 3
s) 3
t) 8
u) 4
v) 4 (assuming 1000. with decimal implied)
w) 3
x) 2

3. Decimal Form:
a) 15600
b) 0.0056
c) 0.0369
d) 73690000
e) 259
f) 0.0000459
g) 0.1369
h) 69000
i) 2.59
j) 20.9

4. Calculations in Scientific Notation:
a) 2.7 × 10⁶
b) -2.7 × 10³
c) 2.53 × 10²⁴
d) 1.47 × 10³
e) 2.2 × 10⁶
f) -3.47 × 10²
g) 5.34 × 10⁻⁴⁴
h) -1.2 × 10⁻²

5. Calculations in Scientific Notation:
a) 2.6 × 10⁻⁴
b) 2.3 × 10¹³
c) 1.97 × 10²
d) 1.65 × 10⁻²³
e) 2.5 × 10⁻⁸³
f) 1.6 × 10¹⁴
g) 2.000 × 10⁷
h) 2.37 × 10²³

Note: Some answers may vary slightly based on interpretation of significant figures rules, but these are standard approaches.