Biology 12

Today we took some short notes on how the kidneys regulate the pH of the blood.   We did the blue worksheet on Urinary system reflection questions.   These are due tomorrow.

Practice test was given out for your Respiratory/Excretory system test (on Thursday, June 1).

practice test answers:    1)A     2)D      3)A    4)D   5)B     6)D    7)A     8)A      9)C      10)D    11)omit
12)B    13)D     14)B    15)B    16)A    17)D    18)D     19)C    20)A    21)D    22)A    23)B    24)A
25)B     26)C     27)D    28)D    29)B    30)C     31)B    32)B    33)D     34)A   35)A    36)D    37)C
38)A     39)A     40)D    41)B    42)B    43)B     44)A    45)D    46)A     47)C   48)B    49)B    50)C

Chemistry 11

Today we reviewed the Solutions Unit by looking at solubilities of polar and non-polar solutes and solvents.  We did the lab:  16A Polar and Non-polar solutes and solvents.   Do all questions #1-5 at the end of the lab.  No conclusion or follow-up questions.

Some answers to selected Unit 6 Solutions Review Assignment:
2) Solvent is water and solute is salt.
3) Polar solutes and ionic solutes will dissolve in polar solvents, non-polar solutes will dissolve in non-polar solvents.

4a) polar solvent= water             c) a non-polar solute = iodine crystals          e) ionic solute = salt

5b) (NH₄)₃PO₄ -------3NH₄⁺ + PO₄^-3

6b) H₂SO₄ + H₂O -----------H₃O⁺ + HSO₄^- 

^{assume that only one H leaves the acid each time} 

7a) [Na⁺]= 2 x 0.25 M = 0.50 M  [CO₃^-2] = 0.25 M

8) Two kinds of double replacement reactions are precipitation reactions and acid/base neutralizations

9 and 10 a) 

Molecular:  Na₂SO_4(aq)  + Pb(NO₃)_2(aq) -----------PbSO_4(s) +  2NaNO_3(aq)

 

Total ionic:   2Na⁺_(aq) + SO₄^-2_(aq) + Pb⁺²_(aq) + 2NO₃^-_(aq) ---------PbSO_4(s)  +  2Na⁺ + 2NO_{3 (aq)}

  

Net ionic:  Pb⁺²_(aq) + SO₄^-2_(aq) ---------PbSO_4(s)

12) b is a acid       c and e are bases      d and f are neither  

14) ionic solutes will conduct electricity in solution so if you know that the solute is ionic then you know that the solution formed with this solute 

15) [NaCl] = 0.00085 M                    17) Final concentration = 0.13 M

19) 75.0 mL of 2.00 M HCl         21) 0.93 M
 

23) 22.8 g of NaOH