Where's our November 12 post?  On the lesson on colligative properties, the day before the test?  Colin?

Molarity & Stoichiometry Review

Scribe Post Author: Serene P.

11/8/13

To begin class today, Ms. Friedmann checked our previous night's homework assignments. Then, she informed us of our homework over the weekend.

1) Unit 4 Test has been moved to Wednesday!

-A study session will be held Wednesday morning at 7 am

- A review packet was given and a set of extra precipitation stoich problems (however, these are optional preparation items for the test)

2) Read sections 10.3 and 10.4 in your textbook and take notes in your journal-due Tuesday

-You need to know about Colligative Properties for the test, but you do not need to know how to calculate freezing point depression, boiling point elevation, or osmotic pressure.

3) WebAssign 10.3-due Monday night at 11:59 pm

During class, we also picked up a sheet titled, "Chemistry Scene Investigation~Trouble in the Chemistry Store Room." However, we were not able to get to this activity.

We went over the two homework sheets. Ms. Friedmann demonstrated how to do question 2a on the homework sheet titled 'Molarity.' Here is the work for the question:

2) What is the molarity of each ion present in aqueous solutions prepared by dissolving 15.00g of the following compounds in water to make 5.25 L of solution?

a) magnesium bromide?

15.00g MgBr2 x 1 mole MgBr2 = 0.08148 moles MgBr2

                        184.10g MgBr2   

0.08148 moles MgBr2   = 0.0155 M MgBr2 ←overall chemical molarity

5.25 L Solution        

Chemical concentration x number of particles

So…

0.0155 x 1M Mg2+ ions = 0.0155M

0.0155 x 2M Br- ions = 0.031 M

If you want the total ion concentration, do 0.0155 x 3 = 0.0465M

For question one of the ‘Molarity’ worksheet, the answer was 6.46M NH3. The work is shown below:

11.0g NH3 x 1 mole NH3  = 0.64554 moles NH3

                   17.04g NH3

100 mL x 1 L       = 0.1 L

            1000mL

0.64554 moles NH3  = 6.46 M NH3   

0.1 L NH3                     

       

On the ‘Concentration and Solution Stoichiometry’ sheet, the answer for question one should have been 0.519 M NaOH. To solve this question, you should have done the total number moles in solution divided by the total volume of the solution. The answer was a little closer to the higher molarity.

For question two on this worksheet, the answer was that both reactants were used up equally. There was not any limiting reactant because the amount that resulted from both Pb(NO3)2 and KI was 0.0875g PbI2.

The remaining answers for questions 2b, 2c, and 3 on the ‘Molarity’ worksheet will be found on the key that should be posted on moodle.

Ms. Friedmann also discussed the differences of molecular compounds and ionic compounds getting dissolved in water. She explained that the ionic compounds will be able to split into more particles when dissolved in water. This is because the water molecules will be able to attach to the ends of particles with opposing charges. An example is MgBr2.

Ionic compounds are able to conduct electricity as well.

However, molecular compounds in water will act differently. They will not conduct electricity and will not break apart. One particle of the compound will become one particle in solution. However, from being a solid, the compound will be referred to as aqueous because water gets in between molecules of the compound. It will separate molecules of the compound, not ions. Ionic compounds will separate into ions when in solution.

We also learned how colligative properties depend on the number of particles in solution. Water boils at a higher temperature when particles are added. Particles with charges that hold onto water molecules keep them from escaping into a gaseous state. Therefore, you must increase the heat energy to reach the boiling point of the water(boiling point elevation). If there is a lot of solute present, the water can boil at a temperature higher than a 100 degrees.

At the end of class, we had some fun with Georgia’s large hot pack.

The next scribe post author will be Colin S.

More Solution Stoichiometry

Fun with Homework 

To start off class today Mrs. Friedmann checked in the three worksheets we had for homework last night.  We then went over these three sheets.  If you missed class, the key is posted on the moodle page under Unit 4 keys.

Fun with Molarity

Next, Mrs. Friedmann passed back our Concentration and Molarity Simulation packets.  We went over these, and the key is also posted on the moodle page under Unit 4 Keys.  

Fun with Supersaturation

After that, Mrs. Friedmann showed us the demonstration of the supersaturated sodium acetate solution.  The sodium acetate packet was put in boiling water to heat it up so the the water could dissolve all of the sodium acetate.  Then the packet was slowly cooled down until it became supersaturated.  The liquid was very syrupy and slow moving.  When Mrs. Friedmann flipped the metal disk around that was inside the packet, the sodium acetate went out of solution and crystalized.  The packet became warm because all of the heat energy we put into it when putting it in the boiling water is now being released.  Below is a video of this happening:

  

More Fun with Homework

This brought us up to the end of class.  Our homework for tonight is the two worksheets that were handed out in class today which are the Molarity Thinking Problems worksheets.  These can be found in the Unit 4 handouts folder.  If you are struggling with these problems, post them on the blog!  Try to check the blog again by 10 PM and see if you can answer anyone's questions.

Fun with Blogging

The next blogger is...Valerie K

Solutions & Stoichiometry

It's finally here... Solutions + stoichometry!

Okay, so you might not be so excited about this. But hopefully this blog

can clear up any confusions you have about blending what we already

know about stoichometry with this unit's topic: solutions.

• Today, we started class by turning in our Molarity Stiumlations packets, which we're on schedule to go over tomorrow. If you need one of these, they're in the Unit 4 Handouts folder!

• Then, we started on the torture part of this unit: Solution stoich. 

Here's a brief overview of the notes we took and the problems we did!

Disclaimer: Ms. Friedmann's notes, which she wrote in class today, are in the Unit 4 Notes folder 

under  'Notes on Solution Stoichiometry'. Go check these out, as well.

First, we redefined stoichometry: looking at one "thing" in a reaction (either a reactant or a product) and figuring out how much of another "thing" in the same reaction will be used up or produced based on the measurement of "thing" 1. 

This definition might seem a little extensive, but to remind everyone...

A stoichiometry set-up looks just like this! 

Now, we all remember doing (a little bit too many of) these problems!

So, where do solutions come in?

Well, to solve stoichometry problems involving Molarity, we need new algorithms.

Our old algorithm looked like this:

We've used this algorithm to solve all of our stoichiometry problems to date. But now,
with solutions, we need a new algorithm involving molarity.
The new algorithm:

Obviously, the new parts of this algorithm are:

• Calculations using the volume to find other measurements. Use the molarity to convert from volume to moles, and vice versa.
• Calculations using the molarity to find other measurements. Use the volume to convert from molarity to moles, and vice versa.

Let's do a problem:

This is the first problem on the worksheet entitled 'solution stoich', located in the Unit 4 Handouts folder.

CaCl2 (aq) + 2 AgNO3 (aq) -----> Ca(NO3)2 (aq) + 2 AgCl (s)

What volume of 0.150 M AgNO3 is needed to react with 45.0 mL of 1.50 M CaCl2?

What is "thing 1"? CaCl2, because we are given its molarity and volume.

What is "thing 2"? AgNO3, because we are asked to calculate the volume of it needed to react.

There are two ways to start this problem, either with the molarity or the volume of CaCl2. We chose to start with the volume of CaCl2.

Since the volume is in mL, we must convert to L just by shifting the decimal 3 places to the left.

Knowing our first conversion factor must be 1.50 moles CaCl2/1L, because molarity is moles of solute over liters of solvent, we can begin to set this problem up.

0.045 L * 1.50 moles CaCl2/1L

Now, we need to convert from moles of CaCl2 to moles of AgNO3. To do that, we look at the balanced equation. There are 2 moles of AgNO3 for every 1 mole of CaCl2. So...

0.045 L * 1.50 moles CaCl2/1L * 2 moles AgNO3/1 mole CaCl2

Next, we need to get from moles to volume of AgNO3. So, knowing the molarity of AgNO3 and using the corresponding conversion factor...

0.045 L * 1.50 moles CaCl2/1L * 2 moles AgNO3/1 mole CaCl2 * 1L/0.150 moles AgNO3 =

0.90 L of AgNO3!

And that's it! The rest of this sheet, along with the two others recieved in class today, are for homework. Good luck, and feel free to ask questions in the comments section!

Here's an awesome Khan Academy video on solution stoichiometry if you're still confused:

Next scribe is Mary L!

So, What is Molarity Anyway?

So, What is Molarity Anyway?

Scribe: Grace K.

Date: November 5th, 2013

We all know it's a bummer when you can't make it to Mrs. Friedmann's fifth period class. It is the best class of the day after all! However, don't worry about the content you missed. This blog will get you all caught up on everything you need to know!

Handouts

As usual, Mrs. Friedmann had an assortment of handouts for us to pick up when we walked into class today. These included a packet titled Concentration and Molarity PhET-Chemistry Labs and a notes sheet on how to mix a solution given a value of molarity properly. Both can either be found using the links included here or in the Unit 4 Handouts and Unit 4 Notes, respectively.

Last Night's Homework

Mrs. Friedmann checked in the homework due for today in our journal. This included two molarity worksheets that can either be found by this link or in the Moodle handouts folder and the molarity packet that we began in class yesterday.

At this time, we spent a few minutes reviewing question 21 in the Molarity Packet. An in depth explanation can be found in Mrs. Friedmann's key. The main idea is this: molarity compares moles of solute to liters of SOLUTION. This solution consists of both the solute and the solvent-NOT JUST THE SOLVENT! So, in question 21, the mistake was that the student needed 50mL of solution, not 50 mL of water. We will explore this concept further in the simulation section.

The BIG Idea

The in-class notes today were on the main ideas of molarity. These notes can be found either by this link or in the Unit 4 Moodle notes folder. Here's a video to help you understand this very important topic. 

Mixing Things Up

After we understood, this whole "molarity" thing, we decided to "mix things up a bit". It was at this point we looked at the handout on how to mix solutions given the value of molarity. This can be found either by this link or in the Unit 4 Moodle notes folder.

The handout prompts, "Suppose you work in a chemistry lab. Your boss tells you to make 0.50 liters of a 0.75 M solution of sodium chloride. How would you do it??". Mrs. Friedmann gave us five minutes with a partner to discuss possible solutions and this is what the class came up with:

Class Brainstormed Solution:

Step 1: Find moles of sodium chloride.

Step 2: Convert moles to grams so we can actually measure it in the lab.

Step 3: Add water until you get 0.50 Liters.

As it turns out, we are a very smart class and our solution was very correct! A more in depth answer can be found on the answer key. Some students in the class noted that they used the proportionality approach to compute the calculations. However, Mrs. Friedmann stated that she encourages us to use dimensional analysis, as it will be more useful to us in the long run.

"I won't believe it until I see it!"

To make sure we understood the process explored in the notes above, Mrs. Friedmann performed a simulation for the class. She first put 22 grams of sodium chloride in a small container, by zeroing the container on the scale and measuring the substance accordingly. She filled a volumetric flask (see below) with less than 5oo mL of water and added the 22 grams of sodium chloride. Mrs. Friedmann shook the mixture to allow it to dissolve. She also added water to assist in the dissolving process, being sure she did not add too much water to the mixture as such a mistake is hard to reverse. It is best practice to eye the volumetric flask from the level of the meniscus, to make sure one is as accurate as possible. Right before our eyes, Mrs. Friedmann had created 0.50 liters of a 0.75 M solution of sodium chloride, using our calculation of grams of solute necessary. 

Let Us Not Whine About Wine...

Volumetric flasks happen to be one of Mrs. Friedmann's favorite pieces of equipment. (I smell a potential Christmas present! A bejeweled volumetric flask perhaps? ) Today, we learned that the device is apparently used by some people in their kitchen to decant wine. We also learned a bit on the process of wine oxidation. Volumetric flasks are meant to be created very exactly. Once a potential flask is made, it is filled with highly concentrated water. A special machine laser marks the meniscus of the water at an exact measurement, ensuring accuracy. You learn something new everyday!

Last Night's Molarity Homework

Towards the end of the period, we spent a bit of time discussing the Molarity calculations we performed in our journal. The answer key can be found by this link or in the Unit 4 Moodle keys folder. Here are some ideas that may help you to understand this worksheet.

1c. Remember that molecular compounds don't dissociate. The only way to show solid sugar vs. sugar water is using the aqueous symbol (aq) and the solid symbol (s).

3c. Remember that if given mL of solution, you must convert to L using a conversion factor.

6 and 7. These problems are known as dilution problems. This equation will be very useful in solving problems such as these:

M1 x V1 = M2 x V2

....where M stands for molarity and V stands for volume. 

SIG FIGS: When computing calculations of this liking (given a quantity of solute and solution), use the significant figures from the measurement with the smallest number of significant figures given.

Homework

1) Complete the Concentration and Molarity PhET Simulation Packet (in the Unit 4 Handouts folder).  You will need to click on the link posted in the Unit 4 box to access the simulations...the link will take you to a page with 8 or 9 simulations; you will only need to access the ones called "Concentration" and "Molarity".  Due tomorrow.

2) WebAssign 10.2 - Solubility.  Due tonight by 11:59 pm.

THE NEXT SCRIBE IS CAMERON B.

November 4, 2013

Kevin M

Mrs. Friedmann

DA Overview

Today, upon entering the classroom, we took into our possession a packet and two sheets to be completed for the Homework. Although, due to time constraints, we were forced to finish the packet for homework as well (our lives are terrible). We completed the first two pages but there is still much left.

Today during class we spent most of DA time reviewing the homework. DA two homework assignments due today were the Webassign (which, sadly, is back) and the Solubility Curve Practice Problems (two pages). The answers for the Solubility Curve problems can be found in this link ( http://gbs-moodle.glenbrook225.org/moodle/file.php/12015/1314_Unit_4/Unit_4_Keys/1_Key_to_Solubility_Curve_Practice_Probs.pdf ). After reviewing the Homework, we moved onto Molarity

DA Molarity

The equation to find the Molarity of a solution can be found in this equation:

Moles of Solute

Liters of Solution

This equation is used to find the ratio of Moles to Liters. We had to complete the worksheet for homework

Da Game

DA BEARS managed to upset the Packers 27-20. Many thought that the Bears were not a strong enough opponent for Aaron Rodgers and the Packers (What is a packer?). HAH. Those people were wrong. Bears back-up quarterback Josh McGown was able to throw for 272 yards and 2 TD's. Bears are now 5-3 tied with the Packers and the Lions in this tricky NFC North division. Any person willing to argue with me about who the best team in the Division is may find me throughout the day. Enjoy my favorite fight song:

http://www.youtube.com/watch?v=SG8OpQyY6TI

And my favorite picture:

Solubility of Common Compounds

Friday, November 1, 2013

Overview

Today in class, we spent time discussing solubility and rules of solubility, by going over the homework and a solubility graph (got in class today). In addition Mrs.Friedmann conducted a cool experiment that illustrated how supersaturated solutions occurred. 

 Homework Discussion

First, we were shown the answers to the questions and made correction to our homework. For every equation a reaction occurred so you had to write a balanced equation and a net ionic equation for each question. This is because it is very rare for two soluble elements to combine in water and form another soluble element, it is usually a precipitate that forms. 

  Here is a problem from the homework...

1) LiCl(aq)  +  Na3PO4(aq)  >>> Li3PO4(s) + Na3Cl(aq)

balanced: 3LiCl(aq)  +  Na3PO4(aq)  >>> Li3PO4(s) + 3Na3Cl(aq)

Net equation:  3Li(aq)  +  PO4(aq)  >>> Li3PO4(s) 

 Review:

-net ionic equation shows only what has reacted to form the solid.

- molecular equation is what we have been doing so far, it shows molecules that reacted and those that did not react expressed in molecular form.

- complete ionic equation shows each ion in the equation separately. 

Wikipedia Controversy...

Well, we all heard our teachers say "Wikipedia is BAD" well, today in class we discussed the reliability of wikipedia when we searched the solubility rate for Lithium. It was said that "nerdy" topics such as solubility rates or " the reasoning behind calculus" can be trusted. This is because it highly unlikely for someone to change such small and "nerdy" detail just to be sneaky. However if something was written about Justin Bieber being awesome on wikipedia then it should be assumed that some obsessed fan wrote it, and that it does not reflect true facts (like really, its not true!).

Cool Experiment...

It wasn't really cool, but hot. Mrs. Friedmann boiled a solution that was supersaturated in a plastic bag, and then at the end of the class we tried to cool it down while trying to prevent it from precipitating.

- This was based on our lesson today in which we discussed a few rule about solubility by studying a graph about the solubility of Sodium acetate...

- based off the chart we can see that as the temperature increased, the solubility of Sodium acetate also increased

- In order to create a super saturated solution which by definition a solution that has more of a solute than the can be dissolved, well then how does it get dissolved and turn into a supersaturated solution.. Simple, bu increasing the temperature. This works because when the water molecules heat up ther begin to move extremely fast and collide into one another, allowing them to dissolve more solvent. Then the solution is cooled down BUT the solute or Sodium acetate still stays dissolved if done carefully!

- A unsaturated solution is one that is not completely full with solvent or in this case Sodium acetate

- A saturated solution is one that is completely full with solvent( not too much or too little)

Homework:

- Complete the Solubility curve worksheet passed out in class a couple of days ago 

- Webassign is due by Sunday night 

- enjoy your halloween candy!

Fun facts for the weekend...

- Elephants are the only mammals that can't jump.

-Every time you lick a stamp, you're consuming 1/10 of a calorie.

-If NASA sent birds into space they would soon die; they need gravity to swallow.

-A 'jiffy' is an actual unit of time: 1/100th of a second.

-Ketchup was sold in the 1830s as medicine

-Coca Cola was originally green.

-Every person has a unique tongue print.

it's magical...

next blogger is- kevin mihelic