An analytical chemist wants to measure the concentration of lead in drinking water in their building. They dilute 1mL of the sample to 100mL of ultrapure water, then acidified (acidification does not affect concentration. It is just a step to make sure the metals are dissolved in the water) and quantified the metal concentration by Inductively coupled plasma - optical emission spectrometry. The result of the diluted sample measured 0.5ppm Pb. Which of the following represents the concentration of the undiluted sample? Is this water safe to drink? (select one or more) a. 50 ppb
b. 50 ppm
c. 2 x 10-4 M
d. 2 x 10-4 % mass
e. 5 ppm
f. 5 ppb
g. 5 x 10-4 M
h. 5 x 10-4 % mass
i. Yes
j. No

Lem 2 the data below are for the system ethyl propyl ether (1)-chloroform (2) at 0.5 bar. use the data to answer the following questions (all questions refer to p d 0: 5 bar). a) what are the boiling points of the pure components at 0.5 bar? b) a mixture with the overall composition z1 d 0: 1 is brought to 47.6ä±c, 0.5 bar. what is the phase? c) 100 mole of a mixture with z1 d 0: 1 (state a) is mixed with 22 mole of pure ethyl propyl ether vapor (state b). the mixing takes place at 47.6 ä±c, 0.5. bar. what is the phase of the resulting mixture (state c)? if the state is a v/l mixture report the number of moles and mole fractions in each phase. d) plot the txy graph and show states a, b and c. the graph must be done by computer and should be properly annotated. ethyl propyl ether (1) - chloroform (2) at 0.5 bar t ( ä±c) x1 y1 t ( ä±c) x1 y1 42.9 0.000 0.000 49.0 0.470 0.455 43.0 0.020 0.010 49.1 0.520 0.520 43.9 0.065 0.029 48.9 0.567 0.592 45.4 0.156 0.089 48.3 0.652 0.720 46.4 0.215 0.142 47.6 0.745 0.815 47.6 0.296 0.223 46.7 0.822 0.872 48.3 0.362 0.302 45.7 0.907 0.937 48.7 0.410 0.375 44.6 1.000

Two nitro no2 groups are chemically bonded to a patch of surface. they can't move to another location on the surface, but they can rotate (see sketch at right). it turns out that the amount of rotational kinetic energy each no2 group can have is required to be a multiple of ε, where =ε×1.010−24 j. in other words, each no2 group could have ε of rotational kinetic energy, or 2ε, or 3ε, and so forth — but it cannot have just any old amount of rotational kinetic energy. suppose the total rotational kinetic energy in this system is initially known to be 32ε. then, some heat is removed from the system, and the total rotational kinetic energy falls to 18ε. calculate the change in entropy. round your answer to 3 significant digits, and be sure it has the correct unit symbol.

Which statement best explains the relationship between an area is geography and the temperature of its surface water