Where are the electrons most probably located in a molecular bonding orbital?
a. in stationar...
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Chemistry, 21.06.2019 18:30
In a sample of oxygen gas at room temperature, the average kinetic energy of all the balls stays constant. which postulate of kinetic molecular theory best explains how this is possible?
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Chemistry, 22.06.2019 00:30
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
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Chemistry, 22.06.2019 12:00
Ican determine the molar mass of an element by looking on the under the atomic mass for the element. for example the molar mass of phosphorus is 30.974 grams/mole. avogadro’s number tells me the amount of representative particles in 1 mole of any substance. this means 12.011 gram sample of carbon and a 32.0 gram sample of sulfur have the same number of atoms.
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Chemistry, 22.06.2019 15:30
Count the number of each type of atom in the equation below, and then balance the equation. write in the numbers of atoms and coefficients. add a 1 if there should be no coefficient. cs2(l) + o2(g) → co2(g) + so2(g) c [ ] s [ ] o > c [ ] s [ ] o [ ] cs2(l) + [ ] o2(g) > [ ] co2(g) + [ ] so2(g)
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