5: Reactions in Aq. Solutions

Electrolytes
compounds whose aqueous solutions conduct electricity
Strong electrolytes
Substances whose solutions are good electrical conductors  (such as sodium chloride)
weak electrolytes
poor electrical conductors (such as acetic acid)
solution
homogeneous mixture of two or more substances
solvent
the component of a solution present in greater quantity
solute

the component of a solution present in lesser quantity.  the solute is said to be dissolved in the solvent

electrolyte
a substance whose aqueous solutions contain ions. often these are ionic compounds
nonelectrolyte
a substance whose aqaueous solution does not contain ions. often these are molecular compounds
strong electrolyte
solutes that exist in solution completely or nearly completely as ions
weak electrolyte
solutes that exist in solution mostly in the form of molecules with only a small fraction in the forms of ions
chemical equilibrium
equal rates of the forward reaction of reactans converting into products and the reverse reaction of products reconverting back to reactants
How many mL of 1.50 M KOH solution are needed to provide 0.125 mol of KOH?
0.125mol KOH x   1 L solution   x  1000 mL      = 83.3 mL solution
                          1.50 mol KOH    1 L solution
You have a stock solution of 14.8 M NH3how many mL of this colution should you dilute to make 100.0 mL of 0.25 M NH3?
100mL x    1 L     x 0.25 M NH3   x   1 L stock solution x 1000 mL  = 1.69 mL
              1000 mL        1 L                  14.8 mol NH3        1 L ss.

^# moles needed for final sol.^    ^mL of ss. w/ desired # moles^

glacial acetic acid is a liquid with a density of 1.049 g/mL at 25 C. calculate the molarity of a solution of acetic acid made by desolving 20.0 mL of glacial acetic acid at 25 C in enough water to make 250.0 mL of solution.
20.0 mL glacial x 1000 mL x 1.049 g acetic acid   mol CH3COOH    =1.4 M
  250.0 mL               1 L            1 mL glacial         60.0520 g CH3COOH
calculate the KE in J of a 45 g of golf ball moving at 61 ms
E= 1 mv2
    
2

E = 1 (45g) x kg x(61m/s)2 =84 J
      2

convert 84 J to calories

84 J x   1 cal   = 20 cal
          4.184 J
A system releases 113 kJ of heat to the surroundings and does 39 kJ of work on the surroundings, endothermic or exothermic?
<>E = q + w = 1.62 kJ – 0.874 kJ = 0.75 kJ, endothermic (q>0)

         2 Mg(s) + O2(g) —-> 2 MgO(s)       <> h = -1204 kJ

(a) is this reaction exothermic or endo thermic?
(b) calulate the amount of heat transferred wh

(a) exothermic (<>H<0)
(b) 2.4g Mg(s) x     1 mol Mh    x -12204 kJ = -59kJ
                          24.3050 g Mg    2 mol Mg
(c) -96.0 kJ x 2 mol MgO x 40.3044 g MgO = 6.43 g MgO
                     -1204 kJ          1 mol MgO
(d) 7.50 MgO x      1 mol MgO   +1204 kJ  = 112 kJ
                         40.3044 g MgO    2 mol MgO

the specific heat of copper metal is 0.385 J/g K.  How many J of heat are necessary to raise the temperature of a 1.42 kg block of coppter from
25.0 C to 88.5 C?
<>T = 88.5 C -25.0 C = 63.5 C => 63.5 k

1.42 kg Cu x 1000 g Cu x 0.385 J x 63.5 k = 34700 J
                         1 g              g k

a 2.20 g sample of guinine (C6H4O2) is burned in a bomb calorimeter where total hear capacity is 2.854 kJ/ C.  the temperature f the calorimeter increases from 23.44 C to 30.57 C.  what is the heat capacity per gram of guinine?
<>T = 30.57 C -23.44 C = 7.13 C

q = 7.854 kJ/ C x 7.13 C x    1     = 25.45 kJ/g guinine
                                         2.20 g

25.45 kJ   x 108.0948 g guinine = 2750 kJ/mol guinine
g guinine          mol guinine

calculate the wavelength of the radiation released when an electron moves from n= 6 to n = 2
E = h(freq.) =      hc          => wavelength = hE
                              wavelength                      E(<>E=Ef-Ei=E(n=2)-E(n=6))
(6.06×10-34Js)(2.9979×108 ms-1) = -4.10×10-7m
(-5.45×10-19J) – (-6.06×10-20J)

wavelength = -410 nm, (neg. sign means radiation emitted)
wavelength(emitted) = 410nm

Heisenberg’s uncertaint principle:
a 1.50 mg mosquito moving with a speed of 1.40 m/s and the speed is known to + 0.01 m/s
(;;x)(;;mv);;;; h;;;;;;;;; -or-; (;;x) ;;;;;;;;;; h;;;;;;;;
;;;;;;;;;;;;;;;;;;;;;; 4(pie);;;;;;;;;;;;;;;;;;;;;;;;;; 4(pie)(;;mv)

;;x ;;;;;;;;;;; (6.626×10-34 Js);;;;;;;;;x 106 mg
;;;;;;;;;; 4(pie) (1.50 mg) (0.01 m/s);;;;;;; ;kg

;;x; 3.52×10-27 m

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