Q. If you hook up a pickle to a 110V power supply and then turn the electricity on, the pickle starts to glow with a yellow color. Why might this be happening? HINT: Pickles are made by soaking cucumbers in very concentrated salt solution.
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A. If you dip a clean wire into a clean salt solution and then place it into a flame [with precautions to protect your hand], it glows yellow because of the sodium in the salt. Sodium outdoor lights produce yellow light also by heating sodium vapor to a very high temperature.
The strange thing about the glowing pickle is that it is not very hot. Attempts to explain the yellow glow contain assumptions that have not been proven. Most of them assume that sodium atoms in the salt somehow become activated enough to give off yellow light. That sounds very reasonable and it is probably correct.
The problem is explaining the activation of sodium without raising the average temperature of the pickle very much. The water in the pickle is not even boiling. Yet, sodium atoms become activated enough to produce the yellow-orange glow.
Some Internet sites post explanations that promote the idea of mini electric arcs within the pickle: Multiple electric sparks jumping across little spaces formed by steam inside the pickle. However, no serious, systematic experimental study has been undertaken to solve the glowing pickle mystery. Take a look at the web pages below and the links in those pages:
http://www.exo.net/~pauld/activities/aaptelectric/glowingpickle.html
http://www.winzy.com/word/Pickle%20Light%20Experimen
http://home.nycap.rr.com/useless/pickle/index.htm
4/21/08
3/2/08
Unit Cell Volume
The smallest repeating unit of s crystal of table salt(the unit cell) is a cube with an edge of .0563nm. If the density of NaCl is 2.17 g/cm^3, what is the mass in grams of one unit cell?
==========================
The edge of a unit cell in meters is:
1 nm = 1.00x10^-9m
Since 1 m = 100 cm = 10^2 cm,
(1.00x10^-9m)(10^2cm/m) = 1.000x10^-7 cm
The volume of a unit cell is:
(1.00x10^-7cm)^3 = 1.00x10^-21 cm^3
mass(unit cell) = (2.17 g/cm^3)(1.00x10^-21 cm^3) = 2.17x10^-21 g
==========================
The edge of a unit cell in meters is:
1 nm = 1.00x10^-9m
Since 1 m = 100 cm = 10^2 cm,
(1.00x10^-9m)(10^2cm/m) = 1.000x10^-7 cm
The volume of a unit cell is:
(1.00x10^-7cm)^3 = 1.00x10^-21 cm^3
mass(unit cell) = (2.17 g/cm^3)(1.00x10^-21 cm^3) = 2.17x10^-21 g
Saline Solution
On 2/4/08 2:30 PM, "emaildave" wrote:
Physiological saline is a .9% solution of NaCl. (a) Express this in terms of molar or millimolar concentrations. (b) Make 125 mL of physiological saline.
=======================
(a) 0.9% means 0.90g NaCl / 100 mLs
The above concentration is equivalent to 9.0g / 1000mLs or 9.0g / liter
The molar mass of NaCl is 23.0 + 35.45 = 58.45 g/mole of NaCl
(9.0g/L) / (58.45 g/mol) = 0.154 moles NaCl / liter or 154 millimoles/liter
NOTE:
1 Mole of NaCl contains 2 moles of ions (Na+ and Cl-). The above molar concentration is equivalent to:
(0.154 mol/L)(2 moles of ions/mol) = 0.308 moles of ions / liter.
This adjusted molarity is called “osmolarity”. “Colligative” properties such as osmotic pressure, freezing point depression, boiling point elevation, vapor pressure, etc. depend on osmolarity rather than molarity.
(b) (0.125L)(0.154mol/L)((58.45 g/mol) = 1.125g/L NaCl
Physiological saline is a .9% solution of NaCl. (a) Express this in terms of molar or millimolar concentrations. (b) Make 125 mL of physiological saline.
=======================
(a) 0.9% means 0.90g NaCl / 100 mLs
The above concentration is equivalent to 9.0g / 1000mLs or 9.0g / liter
The molar mass of NaCl is 23.0 + 35.45 = 58.45 g/mole of NaCl
(9.0g/L) / (58.45 g/mol) = 0.154 moles NaCl / liter or 154 millimoles/liter
NOTE:
1 Mole of NaCl contains 2 moles of ions (Na+ and Cl-). The above molar concentration is equivalent to:
(0.154 mol/L)(2 moles of ions/mol) = 0.308 moles of ions / liter.
This adjusted molarity is called “osmolarity”. “Colligative” properties such as osmotic pressure, freezing point depression, boiling point elevation, vapor pressure, etc. depend on osmolarity rather than molarity.
(b) (0.125L)(0.154mol/L)((58.45 g/mol) = 1.125g/L NaCl
pH During Titration
Consider the titration of 100 mL of .500 M NH3 (kb=1.8x10^-5) with .500 M HCl. After 50.0 ml of HCL have been added, the [H+] of the solution is a) 1.8x10^-5 b) 5.6x10^-10 c) 1.2x10^-5 d) 1.0x10^-7 e) none of these
===================
NH3(aq) + H2O <==> NH4+(aq) + OH-(aq)
Kb = [ NH4+][ OH-]/[ NH3] = 1.8x10^-5
(0.100L)(0.50mol/L) = 0.050 moles NH3
(0.050L)(0.50mol/L) = 0.025 moles HCl
0.025 moles of NH3 will be converted to NH4+ when NH3 reacts with HCl
We will end up 0.025 moles of each, NH3 and H+, in 150 mLs total volume
The concentrations are:
[NH3] = [H+] = 0.025 moles / 0.150L = 0.167 M
[ 0.167][ OH-]/[0.167] = 1.8x10^-5, SOLVE for [OH-]
[OH-] = 1.8x10^-5
[H+][1.8x10^-5]=1x10^-14
[H+]=5.55x10^-10
===================
NH3(aq) + H2O <==> NH4+(aq) + OH-(aq)
Kb = [ NH4+][ OH-]/[ NH3] = 1.8x10^-5
(0.100L)(0.50mol/L) = 0.050 moles NH3
(0.050L)(0.50mol/L) = 0.025 moles HCl
0.025 moles of NH3 will be converted to NH4+ when NH3 reacts with HCl
We will end up 0.025 moles of each, NH3 and H+, in 150 mLs total volume
The concentrations are:
[NH3] = [H+] = 0.025 moles / 0.150L = 0.167 M
[ 0.167][ OH-]/[0.167] = 1.8x10^-5, SOLVE for [OH-]
[OH-] = 1.8x10^-5
[H+][1.8x10^-5]=1x10^-14
[H+]=5.55x10^-10
Dilution Calculations
You only have 2mL of a 1M NaCl stock solution. You need 100mL of a .005 M NaCl solution. Can it be done with the volume of stock solution you have? Show your calculations then do the dilution.
========================
1. A helpful formula for this type of calculation is:
V1M1 = V2M2
Substituting,
(2 mL)(1M) = (V2)(0.005)
V2 = 400 mLs (that we can prepare, more than enough for 100mls).
If we just want to prepare 100 mLs.,
(V1)((1M) = (100mLs)(0.005M)
V1 = (100mLs)(0.005M) / (1M) = 0.5 mL needed
========================
1. A helpful formula for this type of calculation is:
V1M1 = V2M2
Substituting,
(2 mL)(1M) = (V2)(0.005)
V2 = 400 mLs (that we can prepare, more than enough for 100mls).
If we just want to prepare 100 mLs.,
(V1)((1M) = (100mLs)(0.005M)
V1 = (100mLs)(0.005M) / (1M) = 0.5 mL needed
Average Isotopic Mass
Silver has two table isotopes. The actual isotopic mass of 107Ag is 106.9051 amu and the isotopic mass of 109Ag, 108.9047amu. What is the percent abundance of each isotope?
=============================
The average atomic mass of Ag is 107.8682 (you can look this up).
Let the fraction Ag-109 = x (this is really the % expressed as a decimal fraction)
Then, the fraction of Ag-107 would be 1-x (1 means 100%)
Now set up:
108.9047x + 106.9051(1-x) = 107.8682
(This is an expression for the “weighted average” which is given so we calculate the percent of each isotope)
We simplify the above equation and solve for x,
then we substitute below to get the percents
100x = % of Ag-109
(100)(1-x) = % of Ag-107
=============================
The average atomic mass of Ag is 107.8682 (you can look this up).
Let the fraction Ag-109 = x (this is really the % expressed as a decimal fraction)
Then, the fraction of Ag-107 would be 1-x (1 means 100%)
Now set up:
108.9047x + 106.9051(1-x) = 107.8682
(This is an expression for the “weighted average” which is given so we calculate the percent of each isotope)
We simplify the above equation and solve for x,
then we substitute below to get the percents
100x = % of Ag-109
(100)(1-x) = % of Ag-107
1/1/08
Q.
Show with diagrams why hydrochloric acid is used in the test for non-reducing sugars (e.g. sucrose)
A.
Sucrose is a combination of two simple sugars, fructose and glucose. The two simple sugars are combined in a way that gives a negative test for glucose or other reducing sugars. Hydrochloric acid will cause sucrose to break down into the two simple sugars from which it is composed. Study the diagrams below:
Glucose (or other reducing sugar) -----> Positive Benedict test
Sucrose (or other non-reducing sugar) ----> Negative Benedict test
Sucrose + HCl -------> glucose + fructose ----> Positive Benedict test
You can improve on the above diagrams if you spend enough time to understand them.
Show with diagrams why hydrochloric acid is used in the test for non-reducing sugars (e.g. sucrose)
A.
Sucrose is a combination of two simple sugars, fructose and glucose. The two simple sugars are combined in a way that gives a negative test for glucose or other reducing sugars. Hydrochloric acid will cause sucrose to break down into the two simple sugars from which it is composed. Study the diagrams below:
Glucose (or other reducing sugar) -----> Positive Benedict test
Sucrose (or other non-reducing sugar) ----> Negative Benedict test
Sucrose + HCl -------> glucose + fructose ----> Positive Benedict test
You can improve on the above diagrams if you spend enough time to understand them.
Related Reaction Rates
Q.
The average rate of disappearence of ozone in the reaction 2O3(g) --> 3O2(g) is found to be 3.0 x 10^-3 atm/sec over a certain interval of time. What is the rate of appearance of O2 during this interval?
A.
The mole ratio of O2(oxygen) to O3(ozone) is 3/2, the ratio of the two coefficients in the balanced chemical equation. Here, the partial pressure ratio, P(O2)/P(O3), of the two gases is also 3/2. The Ideal Gas Law written in subscript form is:
P1V1/P2V2 = n1T1/n2T2
If the temperatures and volumes are the same, this simplifies to:
P1/P2 = n1/n2 (pressure ratio = mole ratio)
or
P(O2)/P(O3) = n(O2)/n(O3) = 3/2
or
P(O2) = P(O3)*[ n(O2)/n(O3)]
or
P(O2) = P(O3)*[3/2]
(3.0 x 10^-3 atm.O3/sec)[(3 atm.O2/s)/(2 atm.O3/s)]=4.5x10^-3 atm.O2/s
=======
The average rate of disappearence of ozone in the reaction 2O3(g) --> 3O2(g) is found to be 3.0 x 10^-3 atm/sec over a certain interval of time. What is the rate of appearance of O2 during this interval?
A.
The mole ratio of O2(oxygen) to O3(ozone) is 3/2, the ratio of the two coefficients in the balanced chemical equation. Here, the partial pressure ratio, P(O2)/P(O3), of the two gases is also 3/2. The Ideal Gas Law written in subscript form is:
P1V1/P2V2 = n1T1/n2T2
If the temperatures and volumes are the same, this simplifies to:
P1/P2 = n1/n2 (pressure ratio = mole ratio)
or
P(O2)/P(O3) = n(O2)/n(O3) = 3/2
or
P(O2) = P(O3)*[ n(O2)/n(O3)]
or
P(O2) = P(O3)*[3/2]
(3.0 x 10^-3 atm.O3/sec)[(3 atm.O2/s)/(2 atm.O3/s)]=4.5x10^-3 atm.O2/s
=======
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