An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
The problem requires finding the missing unit in a stoichiometry equation.
Isolate the missing unit by algebraic manipulation.
Simplify the equation and perform the calculation.
The missing unit is grams (g): g
Explanation
Understanding the Problem We are given an equation used in solution stoichiometry and need to find the missing unit. The equation is: ( 44.12 m o l g ) ( 7.8 L m o l ) ( 72.8 □ ) ( 1 0 − 3 L 1 m L ) = 2.1 × 1 0 2 m L Let's denote the missing unit as 'x'. Our goal is to find 'x'.
Isolating the Unknown We can rewrite the equation as: ( 44.12 m o l g ) ( 7.8 L m o l ) ( 72.8 x ) ( 1 0 − 3 L 1 m L ) = 2.1 × 1 0 2 m L To solve for 'x', we need to isolate it. First, multiply both sides of the equation by ( 44.12 m o l g ) ( 7.8 L m o l ) :
( 72.8 x ) ( 1 0 − 3 L 1 m L ) = ( 2.1 × 1 0 2 m L ) ( 44.12 m o l g ) ( 7.8 L m o l )
Simplifying the Equation Next, divide both sides by ( 1 0 − 3 L 1 m L ) :
72.8 x = ( 1 0 − 3 L 1 m L ) ( 2.1 × 1 0 2 m L ) ( 44.12 m o l g ) ( 7.8 L m o l ) Simplifying the right side, we get: 72.8 x = ( 2.1 × 1 0 2 ) ( 44.12 ) ( 7.8 ) × 1 0 − 3 g
Calculating the Missing Unit Now, divide both sides by 72.8: x = 72.8 ( 2.1 × 1 0 2 ) ( 44.12 ) ( 7.8 ) × 1 0 − 3 g Calculating the value of x: x = 72.8 ( 210 ) ( 44.12 ) ( 7.8 ) × 0.001 g x = 72.8 722.1816 × 0.01 g x ≈ 0.992 g Since the result is approximately 1, the missing unit is grams (g).
Final Answer Therefore, the missing part of the student's equation is 'g'.
Examples
In chemistry, stoichiometry is used to calculate the amounts of reactants and products in a chemical reaction. For example, if you are synthesizing a drug, you need to know exactly how much of each ingredient to use to get the desired amount of the final product. This equation helps ensure that the units are consistent and the calculation is accurate, which is crucial in pharmaceutical and chemical industries to avoid errors and ensure safety.
A charge of 450 coulombs flows through the device when a current of 15.0 A is sustained for 30 seconds. This charge corresponds to approximately 2.81 × 1 0 21 electrons. Therefore, around 2.81 × 1 0 21 electrons flow through the electric device in 30 seconds.
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