150 lb of nacl are initially dissolved in 500 gal of a nacl solution in a container. water is continuously added to the container at a rate of 5 gal/min and the container content, which is kept uniform by mixing, is also withdrawn out of the container at 5 gal/min. determine how much nacl will be in the tank at the end of 1.5 hours.

After 90 minutes there will be 60.5 lb NaCl in the tank

Explanation:

Initially, you have 150 lb/500 gal = 0.300 lb/gal NaCl

After 1 minute, the volume will be the same but you will have less NaCl in the solution.

After the first minute, the mass of NaCl you will lose will be:

0.3 lb/gal · 5 gal = 1.5 lb

Then, the concentration of the solution will be:

(150 lb - 1.5 lb)/ 500 gal = 0.297 lb/gal

After the second minute:

mass of NaCl lost = 0.297 lb/gal · 5 gal = 1.485 lb

concentration of the solution = (148.5 lb - 1.485 lb) / 500 gal = 0.29403 lb/gal

After the third minute:

mass of NaCl lost = 0.294 lb/gal · 5 gal = 1.47015 lb

concentration of the solution = 147.015 lb - 1.47015 lb / 500 gal = 0.2910897 lb/gal

Notice that the solution is being diluted by 1% every minute. Then, the concentration of the solution after every minute will be 99% of the previous solution. For example, after the first minute, the concentration of the solution  will be 99% of the original solution (99% of 0.3 lb/gal = 0.297 lb/gal). Mathematically, we can express it as follows:

0.3  lb/gal · 0.99 = 0.297 lb/gal

After the second minute:

0.297 lb/gal · 0.99 = 0.29403 lb/gal

Since 0.297 lb/gal = 0.3  lb/gal · 0.99

we can write for the second minute:

0.3  lb/gal · 0.99 · 0.99 = 0.29403 lb/gal

And for the third minute:

0.3  lb/gal · 0.99 · 0.99 · 0.99 = 0.2910897 lb/gal

And for the nth minute:

0.3 lb/gal · 0.99ⁿ = concentration of the solution after n minutes

Now, we can find the concentration of the solution after 90 min:

0.300 lb/gal · 0.99^90 = 0.121 lb/gal

The mass of NaCl after 90 minutes will be:

0.121 lb/gal · 500 gal = 60.5 lb

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

mark brainliest and have a great day!

2b2t

explanation:

2b2t