Explanation:
First, enter the unit rate that will cancel the grams, which is [tex]\frac{4.3 \text{ grams}}{1 \text{ liter}}[/tex]. Then enter the second unit rate to cancel liters, leaving seconds:
[tex]
154 \text{ g} \times \frac{1 \text{ liter}}{4.3 \text{ g}} \times \frac{1 \text{ s}}{0.25 \text{ liter}}
[/tex]

Part B:
Now perform the calculation you set up in Part A. Express your answer to the correct number of significant figures. Type the correct answer in the box.

It will take [tex]\square[/tex] seconds for the device to release 154 grams of the gas.



Answer :

To solve this problem, we need to follow the unit conversions step-by-step.

First, we convert the mass of the gas from grams to \( t \) liters using the given rate \( \frac{4.3 \text{ grams}}{1 \text{ liter}} \).

### Step 1: Convert grams to \( t \) liters
The given mass is 154 grams. Using the conversion factor:
[tex]\[ 154 \text{ grams} \times \frac{1 \text{ liter}}{4.3 \text{ grams}} = 154 \div 4.3 \text{ liters} \][/tex]
[tex]\[ 154 \div 4.3 \approx 35.81395348837209 \text{ liters} \][/tex]

Thus, \( 154 \text{ grams} \) is equivalent to approximately \( 35.81395348837209 \) liters.

### Step 2: Convert \( t \) liters to seconds
Next, we convert the volume in liters to time in seconds using the given rate \( \frac{1 \text{ second}}{0.25 \text{ liters}} \).

[tex]\[ 35.81395348837209 \text{ liters} \times \frac{1 \text{ second}}{0.25 \text{ liters}} = 35.81395348837209 \div 0.25 \text{ seconds} \][/tex]
[tex]\[ 35.81395348837209 \div 0.25 \approx 143.25581395348837 \text{ seconds} \][/tex]

Thus, it takes approximately \( 143.25581395348837 \) seconds for the device to release \( 154 \) grams of the gas.

### Final Answer
Rounding this number to the correct number of significant figures, considering that 154 (3 significant figures) and 4.3 (2 significant figures) are provided, our calculations should reflect the least number of significant figures, which is 2.

Rounding \( 143.25581395348837 \) to 2 significant figures gives us:

[tex]\[ 143.25581395348837 \approx 140 \text{ seconds} \][/tex]

Thus, it will take [tex]\( \boxed{140} \)[/tex] seconds for the device to release [tex]\( 154 \)[/tex] grams of the gas.