Use the References to access important values if needed for this question. It is often necessary to do calculations using scientific notation when working chemistry problems. For practice, perform each of the following calculations. (3.10 × 10-5) (1.00 × 10-8). = 3.55 x 104 2.75 x 10-4 - 7.55 × 105 (3.10 × 10-5) (2.75 × 10-4) 3.55 x 104

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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### Solving the Equation \( a = \frac{b}{c} \) for \( c \)

#### Equation to Solve:
\[ a = \frac{b}{c} \]

#### Steps to Isolate \( c \):
In order to solve the equation above for \( c \), you must multiply both sides of the equation by the same expression:

\[ a \times \boxed{c} = \frac{b}{c} \times \boxed{c} \]

By multiplying both sides by \( c \), the equation simplifies to:

\[ c = \frac{b}{a} \]

### Explanation:
1. **Original Equation**: Start with \( a = \frac{b}{c} \).
2. **Multiply Both Sides by \( c \)**: To eliminate the fraction, multiply both sides by \( c \), giving \( a \cdot c = b \).
3. **Isolate \( c \)**: To solve for \( c \), divide both sides by \( a \), resulting in \( c = \frac{b}{a} \).

This method ensures that \( c \) is isolated on one side of the equation, providing the solution \( c = \frac{b}{a} \).
Transcribed Image Text:### Solving the Equation \( a = \frac{b}{c} \) for \( c \) #### Equation to Solve: \[ a = \frac{b}{c} \] #### Steps to Isolate \( c \): In order to solve the equation above for \( c \), you must multiply both sides of the equation by the same expression: \[ a \times \boxed{c} = \frac{b}{c} \times \boxed{c} \] By multiplying both sides by \( c \), the equation simplifies to: \[ c = \frac{b}{a} \] ### Explanation: 1. **Original Equation**: Start with \( a = \frac{b}{c} \). 2. **Multiply Both Sides by \( c \)**: To eliminate the fraction, multiply both sides by \( c \), giving \( a \cdot c = b \). 3. **Isolate \( c \)**: To solve for \( c \), divide both sides by \( a \), resulting in \( c = \frac{b}{a} \). This method ensures that \( c \) is isolated on one side of the equation, providing the solution \( c = \frac{b}{a} \).
**Using Scientific Notation in Chemistry Calculations**

When working on chemistry problems, it's often necessary to perform calculations using scientific notation. For practice, complete each of the following calculations:

1. \[(3.10 \times 10^{-5}) \times (1.00 \times 10^{-8}) = \quad \text{[Provide your answer in the box]}\]
   
2. \[\frac{3.55 \times 10^4}{2.75 \times 10^{-4} - 7.55 \times 10^5} = \quad \text{[Provide your answer in the box]}\]

3. \[\frac{(3.10 \times 10^{-5}) \times (2.75 \times 10^{-4})}{3.55 \times 10^4} = \quad \text{[Provide your answer in the box]}\] 

These exercises will help you practice the skill of handling numbers in scientific notation, which is critical for various chemistry calculations.
Transcribed Image Text:**Using Scientific Notation in Chemistry Calculations** When working on chemistry problems, it's often necessary to perform calculations using scientific notation. For practice, complete each of the following calculations: 1. \[(3.10 \times 10^{-5}) \times (1.00 \times 10^{-8}) = \quad \text{[Provide your answer in the box]}\] 2. \[\frac{3.55 \times 10^4}{2.75 \times 10^{-4} - 7.55 \times 10^5} = \quad \text{[Provide your answer in the box]}\] 3. \[\frac{(3.10 \times 10^{-5}) \times (2.75 \times 10^{-4})}{3.55 \times 10^4} = \quad \text{[Provide your answer in the box]}\] These exercises will help you practice the skill of handling numbers in scientific notation, which is critical for various chemistry calculations.
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