Isopropyl alcohol evaporated from a beaker at the rate of 1.057 mL/min. how many seconds would it take for 3.70 mL of water to be lost?
thanks for helping!
1.057 mL/min x ? min = 3.70 mL
Solve for ?min and convert to seconds.
To find the time it would take for 3.70 mL of water to be lost, we can use the given evaporation rate of isopropyl alcohol.
First, we need to calculate the time it takes for 1 mL of isopropyl alcohol to evaporate:
1 mL / 1.057 mL/min = 0.946 min
Now, we can calculate the time it would take for 3.70 mL of water to evaporate:
3.70 mL x (0.946 min/1 mL) = 3.4982 min
To convert the time to seconds, we multiply by 60:
3.4982 min x 60 s/min = 209.892 s
Therefore, it would take approximately 209.892 seconds for 3.70 mL of water to be lost.
To find out how many seconds it would take for 3.70 mL of water to evaporate, we can use the given rate of evaporation for isopropyl alcohol and relate it to the rate of evaporation for water.
First, let's convert the rate of evaporation for isopropyl alcohol from mL/min to mL/s:
1 min = 60 s
So, the rate of evaporation for isopropyl alcohol is 1.057 mL/min = 1.057 mL/60 s ≈ 0.0176 mL/s.
Now, we know that the rate of evaporation is directly proportional to the amount of liquid evaporated. So, we can set up a proportion to relate the amount of water evaporated to the time it takes:
(rate of evaporation for isopropyl alcohol) : (rate of evaporation for water) = (amount of isopropyl alcohol evaporated) : (amount of water evaporated)
0.0176 mL/s : x = 1.057 mL/min : 3.70 mL
Cross-multiplying and solving for x, we get:
x = (0.0176 mL/s * 3.70 mL) / 1.057 mL/min
To simplify the units, we need to convert 1.057 mL/min to mL/s:
1 min = 60 s
So, 1.057 mL/min = 1.057 mL/60 s ≈ 0.0176 mL/s.
Substituting this value into the equation, we have:
x = (0.0176 mL/s * 3.70 mL) / 0.0176 mL/s
x = 3.70 mL
Hence, it would take 3.70 mL of water the same amount of time (in seconds) as the isopropyl alcohol to evaporate.