Ms. D. Beckingham - Archived 09/23

Principal of DRHS

Posted: May 12, 2020

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Posted: May 12, 2020

Hey everyone. Sorry for the lateness of this post.

This week we will continue on the topic of solutions. I am attaching word problems and a solubility data table for you to use. This table is used to determine whether of not a compound will dissolve in water.

Explanation on how to use the solubility data table:

Anions are negative ions. The top row of this chart indicates possible anions in solution (eg. Cl-, S2-, CO32-, etc...) 

Below the word anion, you will see "high solubility (aq)", then low solubility (s). High solubility means it will dissolve easily in water. (aq) means it's dissolved in water (aqueous). Low solubility means it doesn't dissolve well in water. (s) means it remains a solid in water.

Cations are positively charged ions. If you go across the chart in the same line with high solubility or low solubility, you will see some the cations that when mixed with the anions on top of them in the chart, will either dissolve well in water or not.

Example 1: Take Na2S. To determine if we write (aq) or (s) after it, we look at the chart. Na2S contains Na+ and S2-. You find the negative charge first, the S2-. Look underneath it and find the positive ion. Na is in group 1 on the periodic table. So Na with S is highly soluble. After the compound, you would write Na2S(aq).

Example 2: Take Ag2SO4. Again, find the anion first. Then look under it to see where the cation fits in. According to the chart, silver with sulfate has low solubility. It will not dissolve well in water, so we write Ag2SO4(s).

Example 3: When you look at Cl-, Br-, I-, these 3 anions combined with most cations have high solubility in water (aq), except the ones on the bottom (s) Ag+, Pb2+, Tl+, etc... So AgCl(s) but NaCl(aq).

For a flame test or solution colour test, just look at teh chart to see what colour the flame would be and what colout the substance would be in solution. This is a tool to help identify what is inside a solution.

A Dissociation Reaction identifies what happens to a compund when it is placed in water. it shows whether or not the compund "dissociates" into its anion and cation. See last week's power point notes for examples.

 

Posted: May 3, 2020

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Posted: May 3, 2020

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File chemical_solutions.pptx85.58 KB

Posted: May 3, 2020

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Posted: May 3, 2020

Hi everyone. I placed for the past 2 weeks information on properties of gases. For the next 2 weeks you will learn about solutions. I assigned one video from Khan academy for you to watch this week. I also uploaded some power point slides for you to look at on here. I need to be at the school to scan some data charts (solubility, flame and colour) for you to be able to do any word problems. For now, just take a look at the video and notes. I will post again mid-week with more info for you to work on. This is a section that you may need more explanation from me. I will most likely video a lab demonstration from school if I can. Skype sessions are available if you are interested.

Posted: April 29, 2020

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Posted: April 26, 2020

21 questions to complete. I will post the answers after you have had time to work on them. Submit your work by email if you are working towards a credit.

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Posted: April 26, 2020

This week you can work on a practice sheet for gas laws. You can find it in the document section of my teacher page. There are 21 questions. This will complete the gase section. Next week we work on solutions. I can do a Skype session on Wednesday at 8:00 p.m. for those who want help. Just send me your email address so I can invite you in. Submit your work for theses sheets to me by email if you are working for credit. dawn.beckingham@nbed.nb.ca

Videos were assgined last week to watch if you need a refresher. Use the formulas that were given to you. They are also found below and on the sheets themselves. You will be calculating pressure, temperature or volume of gases. Remember that:

P1V1 = P2V2      P÷ T1 = P2 ÷ T2    V1 ÷ T= V2 ÷ T2    (P1V1) ÷ T1 = (P2V2) ÷ T2      

 

PV = nRT      n=m/M       C= V/n (for gases)

 

R is a constant (8.31 k•PaL/mol•K)

At STP: Temperature=0°C and pressure =101.3 kPa; c = 22.4 L/mol

At STAP: Temperature= 25°C and pressure =100kPa; c = 24.8 L/mol

Temperature in the formulas for gases must be in Kelvin: Add 273 to the celcius value to get Kelvin. To change kelvin to celcius, subtract 273 for the kelvin value.

0°C = 273 K 

Posted: April 22, 2020

Yes you can take this course as a credit.

Pages

Documents

Solubility data table
Practice Sheet 1 on Solutions
concentration of solutions
chemical solutions