If a 250.0 ml sealed plastic bag starts out at a temperature of 19.0 °c and after sitting in the sun reaches a temperature of 60.0 °c, what will the new volume of the bag be? assume the pressure inside the bag remains constant. a. 219 ml b. 267 ml c. 272 ml d. 285 ml

D

Explanation:

If the pressure remains constant then the temperature and Volume are all that you have to consider.

Givens

T1 = 19oC = 19 + 273 = 292o K

T2 = 60oC = 60 + 273 = 333oK

V1 = 250 mL

V2 = x

Formula

V1/T1 = V2/T2

250/292 = x/333                

Solution.

The solves rather neatly.  Multiplly both sides by 333

250*333 / 292 = 333 *x / 333

Do the multiplication

250 * 333 / 292 = x

83250 / 292 = x

Divide by 292

x = 285.1 mL

The answer is D

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.

answer; 4 electrons;