1. S Ke = 0.305(+0.006) 0.100(0.002) x 0.444(±0.009) 0.576(+ 0.005) x 13.1(± 0.3) [35.42(± 0.02) - 2.15(± 0.02)]
1. S Ke = 0.305(+0.006) 0.100(0.002) x 0.444(±0.009) 0.576(+ 0.005) x 13.1(± 0.3) [35.42(± 0.02) - 2.15(± 0.02)]
Related questions
Question
100%
What is the absolute uncertainty for the two problems below?
![The image contains two mathematical expressions involving uncertainties, useful for educational purposes in chemistry or physics contexts.
**1. First Equation:**
\[
K_c = \frac{0.305 (\pm 0.006)}{0.100 (\pm 0.002) \times 0.444 (\pm 0.009)}
\]
- This expression calculates a constant \( K_c \) using values with uncertainties.
- The numerator is \( 0.305 \) with an uncertainty of \( \pm 0.006 \).
- The denominator involves a product of two values: \( 0.100 \) and \( 0.444 \), with uncertainties of \( \pm 0.002 \) and \( \pm 0.009 \) respectively.
**2. Second Equation:**
\[
\frac{0.576 (\pm 0.005) \times 13.1 (\pm 0.3)}{[35.42 (\pm 0.02) - 2.15 (\pm 0.02)]}
\]
- The numerator consists of the product of two numbers: \( 0.576 \) with an uncertainty of \( \pm 0.005 \), and \( 13.1 \) with an uncertainty of \( \pm 0.3 \).
- The denominator is a subtraction within brackets: \( 35.42 \) minus \( 2.15 \), with both numbers having uncertainties of \( \pm 0.02 \).
Both expressions demonstrate the careful consideration of uncertainties in calculated values.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe05b95f0-5e97-4032-be10-420399d24f78%2Fc6923e15-c67a-4333-a7f7-3d96bcf5d03a%2Fmdj14u_processed.png&w=3840&q=75)
Transcribed Image Text:The image contains two mathematical expressions involving uncertainties, useful for educational purposes in chemistry or physics contexts.
**1. First Equation:**
\[
K_c = \frac{0.305 (\pm 0.006)}{0.100 (\pm 0.002) \times 0.444 (\pm 0.009)}
\]
- This expression calculates a constant \( K_c \) using values with uncertainties.
- The numerator is \( 0.305 \) with an uncertainty of \( \pm 0.006 \).
- The denominator involves a product of two values: \( 0.100 \) and \( 0.444 \), with uncertainties of \( \pm 0.002 \) and \( \pm 0.009 \) respectively.
**2. Second Equation:**
\[
\frac{0.576 (\pm 0.005) \times 13.1 (\pm 0.3)}{[35.42 (\pm 0.02) - 2.15 (\pm 0.02)]}
\]
- The numerator consists of the product of two numbers: \( 0.576 \) with an uncertainty of \( \pm 0.005 \), and \( 13.1 \) with an uncertainty of \( \pm 0.3 \).
- The denominator is a subtraction within brackets: \( 35.42 \) minus \( 2.15 \), with both numbers having uncertainties of \( \pm 0.02 \).
Both expressions demonstrate the careful consideration of uncertainties in calculated values.
Expert Solution
Introduction:
No measurement is or can be done with 100% accuracy. There will always be uncertainty present. This uncertainty will propagate when these quantities are used to find some other quantities. We thus first find relative error. We then find the true value of the result. We use this to find absolute error.
Step by step
Solved in 3 steps
