**Miscellaneous questions****Measurement****Atoms and Trends****Wave Mechanical Model****Molecules****Bonding and Geometry****Chemical Equations (including net ionic)****Thermochemistry****States of Matter**

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1. Can I use a TI-83 calculator?

No, you can not use a TI-83 since it has alphabetic memory. You need a TI 30 or 36 or its Casio equilivent. The calculator cannot have alphabetic memory. It should cost $9-12. Be sure to buy one with the keystrokes you like. For example: TI-83 require that you put in log then the number. If you want this same keystroke for your lower calculator, then at the store, hit the log button. If the calculator gives you an error, this calculator will require that you put in the number, then hit log. If the calculator says log and is ready for the number, this calculator has the same keystrokes as your TI-83.

1) In (16.122 + 2.3)/ 1.461, how do we decide significant figures?

The rule for adding and subtracting is that you use the smaller number of decimal places. Since 16.122 has 3 decimals and 2.3 has 1 decimal, the number on top will be counted as having 1 decimal. This is confusing since you really leave it all in your calculator. You consider the 1 decimal at the end when rounding the final answer. (16.122 + 2.3) = 17.422 , but really only 17.4 is significant. Again leave all (17.422) in the calculator and divide by 1.461. The rule for division and multiplication is that the answer will have the smaller number of sig figs (pay no attention to where the decimal is). So 17.4 is 3 and 1.461 is 4 sig. figs so the final answer will have 3 sig figs.

2) What is the formula for standard deviation?

3) Metric Prefixes on the OWL homework.

If you need the table for prefixes for those prefixes that you don't have to memorize, follow these steps. (Now you have been told to memorize kilo-, centi-, milli- and nano-) Hit the ebook link on the left side of the screen. This will popup in a new tab. Then go to page 26 in your textbook. There is the table of prefixes.

4) In the equation to convert Celsius to Kelvin, should we use "C+273=K" or "C+273.15=K" in our calculations?

Often it will not matter, due to sig figs, but C + 273.15 = K is more accurate and will give the most sig figs.

1) Can you go over calculating atomic mass from isotopes again?

Element X is 98.7% X-13 and 1.3% X-12

What is the atomic mass?

Step 1: Change % to decimal (98.7% --> .987 AND 1.3% --> .013)

Step 2: Multiply each decimal by the mass of the isotope

(.987 x 13) and (.013 X 12)

Step 3: Add the mass from each isotope to = the atomic mass

(12.831) + (0.156) = 12.987

2) How do you solve for % given the atomic mass? I know how to go from % to atomic mass, but this is reverse.

You know that the atomic mass from the chart is 6.941 and that this is made up of Li-6 and Li-7. The % of Li-6 + the % of Li-7 = 100. So you can let one of them = x % and the other 100-x%.

If you let Li-6 = x% and Li-7 = 100-x%, then

6.941 = (1.00 -x) 7.01600 + (x) 6.01512

Solve for x

3) How do you determine which element is largest for 2nd I.E.?

Write out the electron structure. Take away one electron. Now which would require the most energy to remove the 2nd electron. For example: Using the Bohr model Na is 2)8)1) so the Na ion would be 2)8). Using the Wave Mechanical model: Na is 1s^{2}, 2s^{2}, 2p^{6}, 3s^{1}The first electron is easy to remove, because it is in the 3s, the second electron is from a full p level and very hard to remove. The first ionization energy for Na is low (easy to remove), the 2nd ionization energy is high (hard to remove).

4) How do we explain the trends?

Energy levels increase as you go down a column, and effective nuclear charge increases as you go across a row. The trick is to apply these things to the trends that you memorized. Let me give you an example with atomic size and electronegativity.

You memorized that atomic size increases as you go down a column, which can be explained by the increasing # of electron shells. You memorized that electronegativity decreases down a column (increases as you go up), which can be explained by the idea that with increasing # of electron shells, the tendancy to gain an electron would decrease as the valence shell gets further from the nucleus.

You memorized that atomic size decreases across a row (increases to the left), which can be explained by the increasing effective nuclear charge across. With strong effect. nuclear charge, the pull on the valence shell will make the atom smaller. You memorized that electronegativity increases as you go across a row, which can be explained by the increasing effective nuclear charge as you go across a row. The more the nucleus pulls the valence shell, the more likely the atom will have a tendancy to gain an electron.

The only variations in this is the difference a fully-filled s orbital or 1/2 filled p or d orbital makes.

1) What is core notation?

When writing out the electron conf. of K (or anything thereafter) You have a choice:

The complete electron config for K is:

1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{1}

A condensed way to show K is to show the nearest noble gas in brackets. [Ar] 4s1

This works because [Ar] = the electron conf of Ar = 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6},

It can be done will all noble gases, so a condensed or core notation way to show Sr is: [Kr] 5s^{2}

instead of the full electron conf. 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{2}, 3d^{10}, 4p^{6}, 5s^{2}

2) Why does Ni end in a 4s^{2} instead of a full
3d^{10}?

It depends on whether you write it in the order the levels are from the nucleus OR in the order of increasing energy (they way electrons fill).

Ni = [Ar] 4s^{2}, 3d^{8}is the same as [Ar] 3d^{8}, 4s^{2}The first is in energy order, the second is in order from the nucleus.

3)Can you explain the electron configuration and orbital configuration for Sc & Cd ?

Sc = [Ar] 4s^{2}, 3d^{7}is the same as [Ar] 3d^{7}, 4s^{2}so you would show all orbitals filled with an up arrow and a down arrow, except 3d. For 3d you have 5 orbitals, but 7 electrons. Since electrons fill with 1 in each orbital, you will have 3 of the d orbitals with both arrows, and 3 of the d orbitals with a single arrow.

Cd =[Kr] 5s^{2}, 4d^{10}which means all filled with 2 arrows for the [Kr] part, then 2 arrows in the single 5s orbital and 2 arrows in the five d orbitals.

4) Why do the electron configurations for Tl and Bi contain 4f^{14}? I got the [Xe] and the last sublevels,
but not the f sublevel in my answer.

Ti is element # 81, so the nearest noble gas is [Xe], then 6s^{2}, then notice than that element number 57 is followed by element 58, which is in the lower f series. So Ti = [Xe] 6S^{2}, 4f^{14}, 5d^{10}, and 6p^{1}. Also note that Xe is element # 54 so

54 electrons (Xe) +2 (6s) + 14 (4f) + 10 (5d) + 1 (6p) = 81

If you leave out any of these you won't have 81 electrons.

5) Can you explain how to determine the orbital spin for a specific sublevel?

Each orbital can contain 2 electrons, one will spin one direction, one the opposite direction. If there is 1 electron in the orbital, we usually put it in an orbital diagram as an up arrow, but I would not expect you to say if it's spin quantum number is +1/2 or -1/2

6) How many orbitals in any atom can have the given quantum number or designtion D) n=5 ? I came up with 23 orbitals since energy level 5 has s, p, d, and f 1+3+5+7=16..how did solutions manual get 25

The 5th level has s, p, d, f so 1, 3, 5, and 7 = 16. The 5th level is predicted to have s, p, d, f, and g (although no known elements have the g). So 1, 3, 5, 7, and 9 =25.

7) What are the answers to the practice sheet that has 3
questions about electron configurations? Click here for the
quantum numbers worksheet!

** The answers are:**

1) The Cl- ion puts its last electron in the 3p level so:

n = 3;

l=1;

m_{l}= 1, 0, or -1;

m_{s}= +1/2 or -1/2

n = main shell;

l = shape or sublevel;

m_{l}= orientation;

m_{s}= spin

2) Cu = [Ar] 4s^{2, }3d^{9 }from the chart, but is an exception, so really Cu = [Ar] 4s^{1, }3d^{10}

The Cu^{1+}= [Ar] 3d^{10}and is diamagnetic;

3) size = Rb; trend is larger down & to left

size = Br^{-}; Rb^{+}, Sr^{2+}; and Br^{-}have same # of electrons and Br has fewer protons.

1st IE = O; trend is larger up and to right

2nd IE = K; both K and Rb have a noble gas config. after 1 electron has been removed, so MUCH energy to remove 2nd electron, but K is smaller and nucleus can pull harder

EA = O; trend is larger up and to left

1) Is there more practice on converting between molecules, moles, and grams? Here is a practice sheet that has 10 questions:

THE QUESTIONS ARE:

1. ? moles H

_{2}O = 36 g

2. ? molecules H_{2}O = 36 g

3. ? g H_{2}O = 0.5 mole

4. ? g H_{2}O = 1 x 10^{23}molecules

5. If 88 g of X = 2 moles, what is the molar mass of X?

6. If 56 g of Y = 1.2 x 10^{24}molecules, what is the molar mass of Y?

7. If 3.01 x 10 24 atoms is 95 g of an element, what is the atomic mass?

8. In 66 g of CO_{2}there are:

how many moles?

how many molecules?

how many oxygen atoms?

9. If 1215 AMU of an element is 8.306 x 10^{- 23}moles:

how many atoms are in the 1215 AMU?

what is the Atomic Mass?

how many grams are in the 1215 AMU?

10. What is the mass percent composition of Ca(HSO_{4})_{2}?

The answers are: (1) 2.0 mol H_{2}O; (2) 1.2 x 10^{24}molecules; (3) 9 g H_{2}O; 4) 3 g H_{2}O; (5) 44 g/mol; (6) 28 g/mol; (7) 19 amu/atom; (8) 1.5 mol CO_{2}; 9.0x 10^{23}CO_{2}molecules;18 x 10^{23}O atoms; (9) 50.02 atoms; 24.3 amu/atom; 202 x 10^{-23}g; (10) 17.1% Ca; 0.8% H; 27.4% S; 54.7% O

#5. solution.

If 88 g of X = 2 moles, what is the molar mass of X?

Molar mass is the # of grams in 1 mole. You know the grams (88) in 2 moles, so

88g/2mol = 44 g/mol = molar mass

#7. solution

If 3.01 x 10 24 atoms is 95 g of an element, what is the atomic mass?

Again, molar mass is the # of grams in 1 mole and is the same number as the atomic mass (only the units differ).

Here you know the grams (95) and the number of atoms. Since 6.02 e23 atoms makes up 1 mole, you can find the number of moles from the atoms given. 3.01 e 24 (1 mole / 6.02 e23 ) = 5.00 moles

So use the definition of molar mass: 95 g / 5.00 moles = 19 g/mol = molar mass

Then use the idea that molar mass and atomic mass are the same # only different units:

so atomic mass = 19 amu/atom

#9. solution.

8.306 e-23 moles (6.022e23 atoms/mole) = 50.02 atoms

atomic mass = amu/atom ...so 1215 AMU/50.02 atoms = 24.3 AMU/atom

24.3 AMU/atom would be 24.3 g/mol of atoms .....So 24.3g/mol ( 8.306 e-23 mol) = 201.8 e-23 g = 2.02 e-23gFor a pdf of the solutions to all 10 quesitons click here.

2) Can you give us more practice with naming?

See these 2 websites for more practice (both have a place you can click for the answers):https://quizlet.com/100989495/naming-chemical-compounds-worksheet-flash-cards/

https://1.cdn.edl.io/RYkJJWfYWAZA3D1GZdXucxbwuxg53pqSqym5aWQEquzKttWR.pdf This link may not have all of the solutions.

https://quizlet.com/37132294/naming-compounds-mixed-flash-cards/

http://lhsblogs.typepad.com/files/naming-acids-ws.pdf FOR ACIDS ONLY

http://www.kentschools.net/ccarman/cp-chemistry/practice-quizzes/compound-naming/ NO ACIDS

3) You mentioned in class that we should use the atomic mass rounded to the nearest tenths place. When we are adding the atomic mass of two or more elements, should we start our calculations by using the rounded number, or should we round after we add the numbers together?

for PCl

_{5}

1P=30.9738 or 31.0

5Cl=177.2635 or 177.3

mass="208.2373, rounding off to 208.2" or "208.3" if you used the masses with 1 decimal.

I would round with the atomic mass:

so P = 31.0

for Cl use 35.45 x 5 = 177.25 so the formula mass is 208.25 or really 208.3

For exams, we will usually use a table that has 1 decimal, sometimes 2, except for the nuclear chapter..

4) How do you find molecular formulas?

The empirical formula times a whole number = the molecular formula.

If you know the empirical and the molecular, then:

molecular formula mass/ empirical formula mass = whole number

This is the whole number that you multiply the empirical formula by: so if the empirical formula is HX and when you are given the molecular mass, you find the whole number is 3, the actual molecular formula is H_{3}X_{3}.

**5) Can you repeat the naming rules and give us the list of 10 ions to memorize?**

Click here for the naming rule and the list of 10 ions to memorize.

6) Why is H_{3}PO_{4} called
phosphoric acid rather than perphosphoric acid and why is H_{2}SO_{4}
called sulfuric acid rather than persulfuric acid?

See the handout about acid's. when the -ate ion is mixed with H+ to form an acid, it is called --ic acid. so SO_{4}(2-) is sulfate so H_{2}SO_{4}is sulfuric acid. Likewise the PO_{4}(-3) ion is called phosphate, so the acid form is phosphoric acid. It depends on what ion is mixed with H^{+}, not on the amount of O's or H's. H_{2}SO_{5}, if it exists, would be persulfuric acid because it is H^{+}added to the persulfate ion, which has one more oxygen than the sulfate ion that you memorized.

7) If you know that you have a hydrated compound of CoCl**_{2}** and that the mass of the hydrated compound was 31.04, but after heating the mass was 16.94, how do you find the actual formula of the hydrated compound

You know that the actual formula is CoCl* X H_{2}O, but you have to calculate the mole ratio of water to anydrous compound. You are to assume that all of the water was removed with heating. So you calculate the moles of the anydrous compound, then the moles of water_{2}

moles of CoCl= 16.94 g ( 1mol/129.84g) = 0.1305 mol CoCl_{2}_{2 }moles of water = 31.04-16.94 = 14.10g ( 1mol water / 18 g) = 0.7825 mol water

next divide each by the smallest, like you do empirical formulas

moles of CoCl= 0.1305 mol CoCl_{2}/ 0.1305 = 1_{2}

moles of water = 0.7825 mol water/ 0.1305 = 6

So the formula isCoCl* 6H_{2}O_{2}

1) I'm emailing you because I have a question
about the polarity of molecules. Is it set that, for example, all linear shaped molecules are nonpolar, all bent/angular shaped molecules are polar, all triangular plane
shaped molecules are nonpolar, etc?

This is basically true, if the atoms that attach to the central one are all the same. For example: CCl_{4}and CHCl_{3}are both tetrahedral, but CCl_{4}is nonpolar & CHCl_{3}is polar.

IF the atoms attached to the central on are all the same:

tetrahedral molecules are nonpolar

triangular bipyramidal are non polar

octahedral are nonpolar

Otherwise ( & for other shapes), consider each bond. Find the negative end of each bond. Use the "horse" analogy, assume that the 'horses' are all running toward the negative end. If the center will move, the molecule is polar. If the center will not move, the molecule is nonpolar.

2) What are the regions of electron density, the
shape, the angles and the polarity for PCl_{3}?

there are 4 regions of electron density,.

the structure is pyramidal (triangular pyramidal),

the bond angles are less than 109.5 degrees (the lone pair pushes the bonds closer),

the molecule is polar

3) See a good video on molecular bonding and geometry at: https://m.youtube.com/watch?v=f8FAJXPBdO

4) What can help with learning geometry?

Get some toothpicks and gumdrops (or gummy bears). Construct the different shapes so that you will be able to visualize these in your mind for the exam. You will have to be able to 'see' the base geometry in you mind, then remove the lone pairs. See this pic of what students constructed! Click here.

1) Can you give us more practice balancing equations? Click here for a pdf. Once you print the sheet, fold it in half. On one side you will have a set of equations to balance. The answers will be on the other side.

2) What are the answers to the practice sheet on stoichiometry with the questions below:

1. How many grams of O

_{2}are required to completely convert 12.0 moles of C into CO_{2}?

2. From: N_{2}(g) + 3H_{2}(g) --> 2NH_{3}(g) + 22 kcal How many mol of N_{2}will be needed and mol of NH_{3}will be produced from 7.5 moles of H_{2}?

3. By the reaction 2SO_{2}+ O_{2}--> 2SO_{3}how much SO_{3}could be made from 12.8 g SO_{2}and excess O_{2}?

4. By the reaction 2SO_{2}+ O_{2}--> 2SO_{3}how much SO_{3}could be made from 6.4 g O_{2}and excess SO_{2}?

5. From the reaction of Hg and S to produce Hg_{2}S, how much Hg would be needed to produce 217.0 g Hg_{2}S?

6. If 3.00 mole of C_{6}H_{6}and 10.0 mole of O_{2}are combined, how many moles of H_{2}O is produced? The balanced equation is :

2C_{6}H_{6}+ 15O_{2}--> 12 CO_{2}+ 6 H_{2}O

7. If 159 g Na_{2}CO_{3}are reacted with 146 g HCl via the unbalanced reaction:

Na_{2}CO3 + HCl --> NaCl + H_{2}O + CO_{2}, how many moles of CO_{2}are produced?

8. If 30.0 g NH_{3}are obtained by reaction of 1.0 mole of N_{2}with 3.0 moles of H_{2}via the reaction :

N_{2}(g) + 3H_{2}(g) --> 2NH_{3}(g) What is the % yield?ANSWERS:

The answers are: 1) 384g Oxygen ; 2)5.0mol NH_{3, }2.5mol of N_{2}; 3) 16.0g of SO_{3}; 4) 32g of SO_{3}; 5) 201.0g of Hg; 6) 4.00mol Water; 7) 1.50mol of CO_{2}; 8) 88.0%

CLICK HERE FOR THE SOLUTIONS TO THE 8 QUESTIONS in a pdf

3) Can you give us some practice with limiting reagents?

PRACTICE

Balance the following equation and answer questions 1-4 based on the balanced equation.

**
NaOH + H _{2}SO_{4} ---->
Na_{2}SO_{4}
+ H_{2}O**

1. 120 g NaOH + excess H_{2}SO_{4}yields how many gram of Na_{2}SO_{4}?

2. 120 g NaOH + 49 H_{2}SO_{4}yields how many grams of Na_{2}SO_{4}?

3. 120 g NaOH requires how many grams of H_{2}SO_{4}for neutralization?

4. If 80 g NaOH and 98 g H_{2}SO_{4}reacted, producing 30 g H_{2}O, what was the % yield of water?

The answers are: 1) 210 gNa_{2}SO_{4}_{; }2) 71 gNa_{2}SO_{4}; 3) 150 gH_{2}SO_{4}; 4) 83%.

CLICK HERE FOR THE SOLUTIONS TO THE 4 QUESTIONS.

4) I was just wondering which solubility guidelines we need to memorize in order to determine if a salt is soluble or not.

Good Question! I usually ask students to know the basics (almost all sodium, potassium, ammonium, chloride, and nitrate salts are soluble.) If you need more than that, the problem will say "soluble potassium hydroxide" or KOH (aq) or will give you a solubility table.

5) Describe what is meant by redox or oxidation/reduction?

When an atom or ion loses electrons, it is oxidized, it is the reducing agent, and it increases in oxidation number (the number becomes more positive). When an atom or ion gains electrons, it is reduced, it is the oxidizing agent, and it decreases in oxidation number (the number becomes more negative).

For example if Cu^{+}loses another electron to become Cu^{2+}, it is oxidized, it is the reducing agent, and its oxidation number goes from +1 to +2 or becomes more positive. IF Cu^{+}gains an electron to become Cu, it is reduced, it is the oxidizing agent, and its oxidation number goes from +1 to 0 or becomes more negative.

6) What about some __ practice
on net-ionic equations__ ?

Click Herefor a problem set with equations (a) through (h) AND the solutions.Remember that if a substance is aqueous, you split it into ions for the total or complete ionic equation, with the acception of aqueous weak acids and weak bases. To say it another way, DO NOT split solids, liquids, gases, aqueous weak acids, or aqueous weak bases into ions for the total ionic. By canceling substances on both sides of the arrow, you can get to the net ionic.

Weak acids are those NOT on the list of the 6-7 strong acids(HCl, HNO_{3}, HI, HBr, H_{2}SO_{4}, HClO_{4}---some books list HClO_{3}, but not Kotz or OWL).

Weak bases are thosethat are not hydroxides of first family metals or 2nd family metals beginning with Ca.Click Here for a solubility table to determine if salts are soluble.

7) Can you give us practice with using the reactivity chart and single replacement reactions?

Click hereto go to a website that will give you 10 practice questions and answers.

8) Can you give us practice limiting reagent problem?

Click here for a pdf of the

limiting reagent problem & solution.

1) What are the important things to remember about calorimetry?

the equation q = m s deltaT, sometimes written as q = m c deltaT

The delta means means change and this is final minus initial or after the reaction minus before the reaction.

q= heat - usually in Joules

m= mass

c or s = specific heat-usually in J/(g x degrees C) or in J/(g x K)q_{hotwater}= q_{coldwater}+ q_{calorimeter }The heat given off by the hot water is equal (but opposite in sign) to the heat gained by the cold water, plus the heat gained by the calorimeter itself.q_{hotwater}= q_{coldwater}+ q_{calorimeter }could be expanded to be:m s deltaT of the hot water =m s deltaT of the cold water + K(calorimeter constant) delta TNegative q means heat given off; postive means heat absorbed.a coffee cup calorimeter is at constant pressure so q = delta H.When you have a reaction, then q_{reaction}= q_{water}+ q_{calorimeter }which is expanded to:

delta H of the reaction =m s deltaT of thewater + K(calorimeter constant) delta Tsigns of deltaH_{rxn}and delta H_{soln}are reversed. The heat absorbed by the calorimeter plus heat gained by the solution is equal (but opposite in sign) to the heat lost by the reaction or process in an exothermic reaction.

2. How do you find the final temperature when a hot piece of metal is put into a liquid,?

You will use the idea that the heat the metal loses plus the heat the liquid gains = 0 (the change in the energy of the universe is zero.) It is true that you are ignoring any heat lost to the equipment, but we will ignore that.

q metal + q liquid = 0

ms(Tf-Ti) + ms(Tf-Ti) = 0

[mass of the metal x specific heat of the metal (Tf -Tinital of the metal) ] + [mass of liquid x specific heat of liquid (Tf -Tinital of the liquid) ] = 0

So multiply mass of the metal x specific heat of the metal to get a #.

If this # is 9.7, and the T initial is 80.0 degrees C, then you have 9.7 (Tf-80.0), when you distribute the 9.7 across, you will get 9.7Tf - 776

You will do the same for the liquid side.

If the multiplication of the mass of the liquid and the specific heat of the liquid gives you 4.5 and the initial temperature is 10.0, you get 4.5 (Tf-10.0).

When you distribute the 4.5 across, you will get 4.5Tf - 45.0

So you have [9.7Tf - 776]+[4.5Tf - 45] = 0

Add to get 14.2 Tf - 821 = 0

So 14.2 Tf = 821

Divide by 14.2 to get Tf = 57.816 or 57.8 for the final temp, which makes sense as the hot metal was a 80 degrees and the liquid was at 10 degrees, when combined the final temp is 57.8

**See the tanker car implode due to cooling vapor which causes lower pressure inside than the atmospheric pressure outside at: https://www.youtube.com/watch?v=E_hci9vrvfw

1) When determining the pressure of a certain gas over water, will the pressure of water always be given? Is it always 24 torr?

You will either have a chart or be given the vapor pressure of water at the temperature. Vapor pressure is related to temperature so the 24 torr is for a specific temperature.

2) Do the gas particles at one temperature have the same speed?

No, at the same temp all have the same KE; but not the same speed (or velocity).

KE = 1/2 mV^{2}

3)
In my notes from lecture I wrote that gas particles that take up
the most space (have a greater MW of # of electrons) will deviate
the most from ideal. Which will be nearest ideal
H** _{2}** , F

For now you can think of this as HF being a polar molecule, and having more attraction between molecules. Always use the type of IM force to separate first. H_{2}and F_{2}have only London forces, while HF has hydrogen bonding. ...so HF will deviate the most (more attraction between molecules) because of the stronger IM force. Then between the 2 with London forces, use the MW to determine that H_{2}will behave most like ideal because it is the lightest.

4) At 333K, which of the pairs of gases below would have the most nearly identical rates of effusion??

a) N_2_O and NO_2_

b) CO and CO_2_

c) CO and N_2_

d) N_2_ and O_2_

e) NO_2_ and N_2_O_4_

this question is from fall 2000, the correct answer is C..

rates of effusion depend on molar mass. (same KE, but KE = 1/2 mv2 ; heavy particles move more slowly, etc.)

a) 44 g/mol compared with 46

b) 30 g/mol to 44 g/mol

c) 28 g/mol compared to 28 -----has the same Molar Mass.

d) 28 g/mol to 32 g/mol

e) 46 g/mol to 92 g/mol

5)What is boiling?

Boiling is the point where the vapor pressure of a liquid equals the atmospheric pressure. A liquid with high vapor pressure at room temperature will need just a little added heat to increase the vapor pressure to the atmospheric, so boiling pt will be low. (higher IMF = higher bp).

6)** **What is viscosity, surface tension,
capilary action, and vapor pressure?

Viscosity is the resistance to flow of a liquid (higher IMF = more viscosity); surface tension is the 'beading' up of water, it is caused by the attraction of surface molecules to interior molecules (higher IMF = higher surface tension); capillary action is when a liquid climbs up a wick or rises in a glass tube, this depends on the IMF of the liquid molecules compared to those between the liquid and the glass. (higher IMF = lower capillary action); vapor pressure is the pressure created by gas molecules above the surface of the liquid from which they have escaped (higher IMF = lower vp or slower evaporation).

7) HEATING GRAPH SHOWING PHASE CHANGES: On the graph comparing the temperature of water at different states, why was the temperature leveled out for the solid and water phase?. Is that saying that ice water is a constant temperature of 0 degrees Celsius?

This is a good question. During the level parts of the graph, heat is being used to break up the IM forces to change the state rather than being used to increase the temperature.

8) How do the IM Forces compare? NOTE: these are intermolecular forces NOT intramolecular (those inside 1 molecule).

Ionic > hydrogen b.> dipole-dipole > London f.

Differentiate between IONIC by:

Charge on ions

Size

Requires ions

Differentiate between DIPOLE-DIPOLE by:

Charge on Dipole Moment

EN Difference

Requires polar molecules

Requires H bonded to F,O,or N for HYDROGEN BONDING

Differentiate between LONDON FORCES by:

# of electrons (Molar Mass)

Requires nonpolar molecules

9) What is the relationship between Intermolecular forces and physical properties?

**................Ionic > hydrogen b.>
dipole-dipole > London f.**

IF IMF is higher b.p. goes up m.p. goes upVapor pressure goes downViscosity goes upSurface Tension goes upCapillary action goes downEvaporation goes downdelta H vaporization goes updelta H fusion goes up