$\begin{array}{l}\text { Step } 1 \frac{K}{0.12}=2.4 \\ \text { step } 2(0.12) \times\left(\frac{K}{0.12}\right)=2.4 \times 0.12 \\ \text { step } 3 \quad K=2.4 \times 0.12=0.288\end{array}$

Verified Answer :

  • Analyze the equation \(\frac{K}{0.12} = 2.4\) to identify the goal of isolating \(K\).
  • Apply the multiplication property of equality by multiplying both sides by \(0.12\).
  • Perform the multiplication \(2.4 \times 0.12\) to confirm the value of \(K\).
  • The calculated result is \(\boxed{0.288}\).

Explanation

  1. Analyzing the Equation

The problem asks us to verify the steps and calculation for the equation \(\frac{K}{0.12} = 2.4\). We are dealing with a simple linear equation where \(K\) is the variable we want to isolate. To solve for \(K\), we must isolate it on one side of the equation.

  1. Isolating the Variable K

The given steps utilize the Multiplication Property of Equality. In the original equation \(\frac{K}{0.12} = 2.4\), \(K\) is divided by \(0.12\). To isolate \(K\), we perform the inverse operation, which is multiplication. By multiplying both sides of the equation by \(0.12\), we get \((0.12) \times \left(\frac{K}{0.12}\right) = 2.4 \times 0.12\). This simplifies the left side to just \(K\).

  1. Final Calculation and Verification

Finally, we calculate the right side of the equation: \(2.4 \times 0.12\). Converting these decimals to fractions or performing standard multiplication, we have \(2.4 \times 0.12 = \frac{24}{10} \times \frac{12}{100} = \frac{288}{1000} = 0.288\). Therefore, the value of \(K\) is \(0.288\), and the provided steps are correct.

Examples

In chemistry, you might use an equation like this to calculate the amount of a reagent needed based on a concentration. If you know the concentration \(C\) in moles per liter and the volume \(V\) in liters, and the formula is \(K/V = C\), where \(K\) is the amount of substance in moles, you can find \(K\) by multiplying the concentration by the volume. This is exactly how proportions and concentrations are handled in laboratories and industrial processes to ensure precision.