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## Week 6 Chapter 6 and Review

by: carolyn martinez

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# Week 6 Chapter 6 and Review CHEM 1311

carolyn martinez
UTB-TSC
GPA 3.58

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Notes and Examples in Class over chapter 4
COURSE
General Chemistry I
PROF.
Vanessa Garcia
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Chemistry

This 2 page Class Notes was uploaded by carolyn martinez on Sunday October 9, 2016. The Class Notes belongs to CHEM 1311 at University of Texas at Brownsville and Texas Southmost College taught by Vanessa Garcia in Fall 2016. Since its upload, it has received 3 views. For similar materials see General Chemistry I in Chemistry at University of Texas at Brownsville and Texas Southmost College.

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Date Created: 10/09/16
Slide 39 Aqueous Solutions and Chemical Analysis Quantitative Analysis when to find how much of a substance is present. Useful when finding concentration of a substance in a solution. Quantitative Analysis Gravimetric Analysis: Titration: Calculation of a Use the mass as a base solute’s concentration in a as knowledge solution. Finding gravimetric mass: - Example: A 0.8633 g sample of an ionic compound containing Cl ions and an unknown metal cation is dissolved in water and treated with an excess of AgNO . If 1.5615 g of AgCl solid forms, what is mass % of Cl in the original 3 compound? 1 (35.45g) Determine the Cl % in AgCl: x 100% = 24.73% (35.45g+107.9g) Mass of Cl in the precipitate: 0.2473 x 1.5615 g = 0.3862 g 0.3862g Cl% = x 100% = 44.73% 0.8633g ****Indicator is used in a solution to determine the equivalence point of the reaction. The known solutions is called standard solution, with that we are able to find the unknown solution. Example: If 45.7 mL of 0.500 M H SO is r2qui4ed to neutralize 20.0 mL of NaOH, what is the concentration of the NaOH solution? 1. H2SO 4aq) + 2 NaOH (aq) --> 2 H O (l)2+ Na SO (aq2 4 0.500mol H SO 2mol NaOH 2. 0.0457 L x 2 4 x = 4.57 x 10 moles NaOH L 1 mol H 2SO 4 4.57x10 molesNaOH 3. = 2.29 M NaOH 0.0200L Titration: 1. Equation written out 2. Start with the volume sample and balance to moles 3. With the moles, you balance it out with the volume to get the molar mass. Examples: + - NaCH C3O (S) Na (AQ) CH 3OO (AQ) HClO 4H +(aq)+ ClO 4 (aq) + -2 K 2O 32K (aq) CO 3 Sr(OH) =2Insoluble (NH 4 2O = 3oluble AgI = Insoluble BaCl 2 =soluable NaNO 3(AQ)+ NiSO 4(AQ) NA 2O 4(AQ) Ni(NO )3 2(AQ) KOH (AQ)+ Cu(NO ) 3 2 (AQ) KNO 3(aq) Cu(OH) 2(s)PRECIPITATE OCCUR 2HBr (aq) Ca(OH) 2 (S) 2H O2+(l)Br 2 (aq) + - +2 - 2H (aq) +Br (aq)+ Ca(OH) 2 (S)  2 +(L) (aq) Br 2 (aq) 2H +(AQ)+ CA(OH) 2 (S)2H O2+ (L) 2 (aq) Mn +(s)SO 2 4(aq) MnSO 4(aq) H 2(aq) o +2 - Oxidized: Mn  Mn + 2e Reduced: H SO 2 H 4 2 Fe (s) H S2 4(AQ) FeSO 4(aq)+ H 2(g) CIE: Fe (s)+ H 2(aq) SO 4(aq) Fe +2(aq)+ SO 4 (aq) H 2 (g) + +2 NIE: Fe (s)+ H 2  Fe (aq)+H 2 (g) 2.50moles 0.175 L x = 0.437 moles x 98.0 g 0.437 moles x 1 moles = 42.9 g

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