post-lab 6 ch 204

post-lab 6 ch 204 - obennoskey (blo242) – Post-lab 6 –...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: obennoskey (blo242) – Post-lab 6 – Lyon – (52635) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Give the oxidation numbers for copper in the compound Cu 2 O and for Mo in the compound Mo 2 O 2- 7 , where oxygen maintains an oxida- tion number of- 2. 1. 0; 0 2. +1; +12 3. 0;- 2 4. +2; +6 5. +1; +1 6. +2; +7 7. +1; +6 correct Explanation: Oxidation numbers are per atom. The sum of oxidation numbers in a neutral molecule is zero and in a polyatomic ion is equal to the charge on the ion. We know the oxidation number of O is- 2. We use these values and set the sum of the oxidation numbers equal to the appropriate value. The oxidation number of Cu and Mo is symbolized by x : 2 x + (- 2) = 0 Cu 2 O : 2 x = +2 x = +1 2 x + 7(- 2) =- 2 Mo 2 O 2- 6 : 2 x- 14 =- 2 2 x = 12 x = +6 002 10.0 points If the oxidation state of H is +1 and O is- 2, what is the oxidation state of C in C 2 H 4 O? 1. +2 2. +1 3.- 1 correct 4.- 1 2 5.- 2 Explanation: To calculate oxidation numbers, remember that the the sum of the numbers must equal the total charge on the molecule or in the case of neutral species, zero. Let x be the oxidation number of C. 2 x + 4(+1) + (- 2) = 0 C 2 H 4 O : 2 x + 2 = 0 2 x =- 2 x =- 1 003 10.0 points Consider the process in which H 2 SO 3 is con- verted to H 2 SO 4 . The oxidation state of oxy- gen remains- 2, and the oxidation state of H remains +1. Select the correct statement. 1. S is reduced and gains electrons. 2. S is oxidized and loses electrons. correct 3. S is reduced and loses electrons. 4. S is oxidized and gains electrons. 5. S neither gains nor loses electrons in this reaction. Explanation: Oxidation numbers are per atom. The sum of oxidation numbers in a neutral molecule is zero and in a polyatomic ion is equal to the charge on the ion. We know the oxidation number of O is- 2. We use these values and set the sum of the oxidation numbers equal to the appropriate value. The oxidation number of S is symbolized by x : 2(+1) + x + 3(- 2) = 0 H 2 SO 3 : obennoskey (blo242) – Post-lab 6 – Lyon – (52635) 2 2 + x- 6 = 0 x = +4 2(+1) + x + 4(- 2) = 0 H 2 SO 4 : 2 + x- 8 = 0 x = +6 Sulfur goes from +4 to +6 for a loss of two electrons....
View Full Document

This note was uploaded on 04/14/2010 for the course CH CH302 taught by Professor Mccord during the Spring '10 term at University of Texas at Dallas, Richardson.

Page1 / 6

post-lab 6 ch 204 - obennoskey (blo242) – Post-lab 6 –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online