Name: ___________________________________ Date: ______________ Practice Test #3 ____ 1. When a precipitation reaction occurs, the ions that do not form the precipitate A) evaporate B) are cations only C) form a second insoluble compound in the solution D) are left dissolved in the solution E) none of these 2. An aqueous solution of potassium chloride is mixed with an aqueous solution of sodium nitrate. The complete ionic equation contains which of the following species (when balanced in standard form)? A) B) C) D) E) ____ 3.
In the absence of other water molecules, you interact fine with butanol, but the attraction isn't strong. When more water is around, then your interaction with that is strong, and so the butanol is excluded. Do demo while describing Thus, it is not so much that non-polar compounds can't interact with water as much as that the interaction of water with other water molecules (hydrogen bonds) is much stronger. 15. Now explain why NaCl won't dissolve in hexane (a non-polar solvent) using a similar thought process.
2. In Experiment 1, explain why the membrane potential between the axon hillock and axon either changed or did not change with subthreshold stimulus. Differences of 1.0 mV or less are not significant Subthreshold is a change as it hasn’t reached the action potential level. 3. In Experiment 2, explain why the membrane potential between the axon hillock and axon either changed or did not change with threshold stimulus.
f. Placed 2 drops of sodium phosphate solution into vertical well plate. g. Placed 2 drops of sodium iodide solutions into vertical well plate h. Placed 2 drops of sodium sulfate solution into vertical well plate. i. Placed 2 drops of sodium chloride solution into vertical well plate. j.
Also, there is a limit of how much solute you can put in the solvent and you should not exceed that certain percentage. If you put too much solute and end up having more solute than the water, the freezing point will start going back up again because the solute won’t be able to dissolve within the solvent at such a big quantity. Another limitation is that the concentration of the solvent in the mixture should be at least 10% and if it’s lower, the freezing point depression might not be affected because the amount of solute used will be too low. Lastly, the solution must be cooled to a lower temperature than the pure solvent in order to freeze. If water was used as the solvent, it has a freezing point of 0 °C and the solution will freeze at even a lower temperature so you have to put it somewhere cool like a freezer.
MYTH! Drinking water will only absorb into your body which means water will only hydrate you. - Drinking lots of water leads to weightless. MYTH! It doesn't, it only leads to losing water weight.
On-line Science Simulations - Marble Chip Student Worksheet When hydrochloric acid is added to calcium carbonate (marble chips) carbon dioxide is evolved. In the experiment shown, 21.6 g of calcium carbonate is added to 200 cm3 of hydrochloric acid. The number of chips can be varied but the total mass is always 21.6 g and the volume of acid is always 200 cm3 but the concentration can be varied. The apparatus is placed on a balance that has been zeroed so that it always shows the same initial mass at the start of the experiment. During the reaction, carbon dioxide is evolved and the mass decreases.
Empirical formula: CH5N Steps for molecular formula: 1- Calculate the molar mass of the empirical formula. 2- Divide the known (given) molar mass by the calculated empirical formula molar mass to get a whole number 3- Multiply that whole number through subscripts of the empirical formula to obtain the molecular formula. Example CH5N 12.01 g C x 1 C= 12.01 g/mol 1.008 g H x 5 H = 5.040
This solution was placed in a burette and 18.4 cm3 was required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Deduce the molecular formula of the acid and hence the value of n. 5. Sodium carbonate exists in hydrated form, Na2CO3.xH2O, in the solid state. 3.5 g of a sodium carbonate sample was dissolved in water and the volume made up to 250 cm3. 25.0 cm3 of this solution was titrated against 0.1 moldm-3 HCl and 24.5 cm3 of the acid were required.
NaOH solution would be in excess and thus prepare 1 M of HNO3 solution in burette, which will be used in back-titration. 4. Determine the end point of the back-titration when NaOH solution changes its color into pink. Record the results of at least three titrations. (Make a rough titration first).