Save O OfExpt. 5 Quiz (3) for chemistry lab - Saved to this PC -P Search (Alt +Q)HomeInsertnerbsl@yahoo.com NDrawDesignLayoutReferencesMailingsReviewViewHelp8 ShareCalibri (Body)- 14 A A Aa- ApE-E-E, E E IAaBbCcDc AaBbCcDc AaBbC AaBbCcCPFind -BIU- ae x, x A- D. AReplace1 Normal1 No Spac. Heading 1HeadingDictateSensitivityEditorard sA SelectFontParagraphStylesEditingVoiceSensitivityEditor9) In the reversible reaction Fe 3+ (ag) + SCN (ag) E → FESCN2+ (aq), the reactants are yellowish and theproduct is red.Why are iron (III) nitrate and potassium thiocyanate used to create this reaction mixture?a) iron and thiocyanate are spectator ionsb) iron and thiocyanate are colorlessc) nitrate and potassium are spectator ionsd) nitrate and potassium are purple* Accessibility: InvestigateO Focusage 4 of 5810 wordsText Predictions: On7:43 AMP Type here to search12/1/2021G.KAltCidAlt

Spectator ions are ions that take part in the reaction but doesn't change their state during the…

9) In the reversible reaction Fe 3+ (ag) + SCN: (aq) E → FeSCN2* (aq), the reactants are yellowish and the product is red. Why are iron (1II) nitrate and potassium thiocyanate used to create this reaction mixture? a) iron and thiocyanate are spectator ions b) iron and thiocyanate are colorless c) nitrate and potassium are spectator ions d) nitrate and potassium are purple

9) In the reversible reaction Fe 3+ (ag) + SCN: (aq) E → FeSCN2* (aq), the reactants are yellowish and the product is red. Why are iron (1II) nitrate and potassium thiocyanate used to create this reaction mixture? a) iron and thiocyanate are spectator ions b) iron and thiocyanate are colorless c) nitrate and potassium are spectator ions d) nitrate and potassium are purple

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter2: Basic Statistical Analysis With Excel

Section: Chapter Questions

Problem 12P

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Chemistry (i) For T = 298K ΔG° = -8.314 x 298 x ln (2.83x10-3) = 14.524 KJ/mol (ii) For T = 308.15K ΔG° = -8.314 x 308.15 x ln (9.619x10-3) = 11.891 KJ/mol (iii) For T = 318.15K ΔG° = -8.314 x 318.15 x ln (3.14x10-2) = 9.147 KJ/mol (iv) For T = 328.15K ΔG° = -8.314 x 328.15 x ln (8.82x10-2) = 6.621 KJ/mol (v) For T = 338.15K ΔG° = -8.314 x 338.15 x ln (2.42x10-1) = 3.990 KJ/mol Based on the measured ΔG° values, is this equilibrium spontaneous at room temperature? Which factor, entropy, or enthalpy, has the greater impact on spontaneity in this case? Explain your answers.