homework platinum 2 unit 1

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"Can i not be my self with this hair?"

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Exercício 2 ainda da página 163

1 (primeira fala): I am going to be a police officer

1 (segunda fala): Of course I am not

2 (primeira fala): I want to break up our relationship

2(segunda fala): No, I will not reconsider

3 (primeira fala): I will be a solider and fight in some war

3 (segunda fala): Just try it

Use of English:

Darkmoon86:

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The truth is that I’m afraid to change.

I’m not sure what to do.

I need to change but something prevents me. I have many doubts.

1) A: I’m going to be a police officer

A: I'm just thinking about it.

2) A: I want to break up our relationship

A: This my call to make

3) A: I’ve decided to be a soldier and fight in some war.

A: You can’t control me!

1)Ran up against

2) already made up her mind

3) The time has come

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HW Solutions #1

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Redox Basics

Redox equations, electrochemistry, spontaneity, half-cell potentials.

1.) Xenon forms several nonionic compounds with F and O. Give the oxidation number of the central Xe atom in each of the following compounds: XeO 4 , XeF 2 , XeO 3 , XeF 4 , and XeF 6 .

2.) Xenon can also form a number of ionic compounds, including CsXeF 7 , and Cs 2 XeF 8 . What are the ions in each compound? What is the formal charge and oxidation number of Xe in each ion containing Xe?

a.) CsXeF 7 → Cs + + XeF 7 - . Xe has +6 oxidation state and -1 formal charge.

b.) CsXeF 8 → 2Cs + + XeF 8 2- . Xe has +6 oxidation state and -2 formal charge.

4.) What is the oxidation number of nitrogen in each of the following ions or molecules: NH 3 , N 2 H 4 , NO, NO 2 , NO 2 - , NO 3 - ?

NH 3, x + 3(1) = 0, x = -3

N 2 H 4 , 2x + 4(1) = 0, 2x = -4, x = -2

NO, x + -2 = 0, x = 2

NO 2 , x + 2(-2) = 0, x = 4

NO 2 - , x + 2(-2) = -1 , x = 3

NO 3- , x + 3(-2) = -1, x = 5

5.) What is the oxidation number of platinum in the complex ion PtCl 4 2- ?

Pt + 4(-1) = -2

Pt – 4 = -2

6.) Assign oxidation numbers to the atoms in the following chemical species:

• iodine, I 2
• barium chloride, BaCl 2
• ethane, C 2 H 6
• stannous oxide, SnO
• stannic oxide, SnO 2
• nitrous oxide, N 2 O
• phosphorus pentoxide, P 2 O 5
• magnesium hydroxide, Mg(OH) 2
• sulfurous acid, H 2 SO 3
• telluric acid, H 6 TeO 6
• hypochlorous acid, HClO
• perchloric acid, HCIO 4
• dichromate, Cr 2 O 7 2-
• cyanide, CN -

7.) What is the oxidation number of the underlined element in each ion or molecule: V O 2 + , P 2 O 7 4- , P H 3 , K N O 2 , H 2 O 2 , L iH , Mg 3 N 2 , N F 3 , I Cl 5 , A g(NH 3 ) 2 + ?

V O 2+ : 4+ P 2 O 7 4- :5+ P H 3 : 3- K N O 2 : 3+ H 2 O 2 : 1- Li H : 1- Mg 3 N 2 :2+ N F 3 : 3+ I Cl 5 :5+ Ag (NH 3 ) 2+ : 2+

8.) When the equation

MnO 2 + I - + H + → Mn 2+ + I 2 + H 2 O

is balanced, what is the net charge on each side of the equation? What element is oxidized? What element is reduced?

Mn0 2 + I- + H+ → Mn 2+ I 2 + H 2 0

Mn0 2 → Mn 2+

oxidation state of Mn goes from 4 to 2

Its reduced

I - → I 2

Oxidation state I goes from - 1 to 0

Hence oxidized

H is a spectator ion , oxidation state is 1 on both the reactant and product side. for redox reaction to occur there has to be a gain or loss of electrons

8b: Multiple Choice: In the following reaction, which element that is oxidized and which is the oxidizing agent.

2 Al + 3 Br 2 → 2 AlBr 3

(a) aluminum…..aluminum (b) bromine…..bromine (c) aluminum…..bromine (d) bromine…..aluminum

2Al + 3Br 2 → 2AlBr 3

Oxidate of Al = 0, Oxiditaion state of Br 2 = 0 ( pure form as solid metal and Gas)

2Al + 3Br 2 → 2AlBr 3 [for Al 2[x + 3(-1) ] = 0, 2x = 6, x = 3 for Al

0 0 Al = 3 and Br 2 = -1

Al is Oxidized and Br is reduced

Al is the reducing agent and Br 2 is the oxidizing agent

9) For each of the following reactions, list the substance reduced, the substance oxidized, the reducing agent, and the oxidizing agent.

• 6H + + MnO 4 - + 5SO 3 2- → 5SO 4 2- + 2Mn 2+ + 3H 2 O
• 8H + + Cr 2 O 7 2- + 6HI → 2Cr 3+ + 31 2 + 7H 2 O
• 3Cl 2 + 6OH - → CIO 3 - + 5Cl - + 3H 2 O

a.) Mn 7+ is reduced to Mn 2+ . S 4+ is oxidized to S 6+ . MnO 4 - is the oxidizing agent. SO 3 2- is the reducing agent.

b.) Cr 6+ is reduced to Cr 3+ . I - is oxidized to I 0 . Cr 2 O 7 2- is the oxidizing agent. HI is the reducing agent.

c.) Cl 2 is both oxidized and reduced, so acts as the oxidizing and reducing agent.

10.) Phosphine, PH 3 , is a colorless, highly toxic gas that smells like rotten Fish and is produced in small amounts when animal and vegetable matter decay in moist situations such as damp graveyards, Traces of P 2 H 4 are produced simultaneously and cause the PH 3 to ignite in air to give pale, flickering lights commonly called “corpse candles" or "will-o’-the-wisps." In the laboratory, phosphine can be prepared by adding water to calcium phosphide. Write a balanced equation for the reaction. Assign oxidation numbers to each of the elements present.

Ca 3 P 2 + 6 H 2 O → 3 Ca(OH) 2 + 2 PH 3

Oxidation number Ca +2 P -3 H +1 O -2

11.) Balance the reaction

MnO 4 - + H + + H 2 S → Mn 2+ + H 2 O + S

What is the oxidation number of Mn in MnO 4 - ? What element is oxidized? What element is the oxidizing agent? Is this latter element oxidized or reduced?

Step 1:Write half reactions separately by identifying oxidation and reduction species

Eq.1 MnO 4 - → Mn 2+

Mn in Mn 2+ has an oxidation state = +2

Oxidation State Change: +7 → +2

Eq. 2 H 2 S → S

S is being oxidized (loss of 2 electrons)

Step 2: Balance elements other than H or O, if any

Both sides consist of 1 Mn, therefore, Mn is balanced

Both sides consist of 1 S, therefore, S is balanced

Step 3: Balance O by adding H 2 O molecules

Eq.1 (old) MnO 4 - → Mn 2+

Eq. 2 (old) H 2 S → S

Eq. 2 (new) H 2 S → S

No oxygen in Eq. 2

Step 4: Balance H by adding H + molecules

Eq.1 (old) MnO 4 - → Mn 2+ + 4H 2 O

Eq.1 (new) 8H + + MnO 4 - → Mn 2+ + 4H 2 O

Eq. 2 (new) H 2 S → S + 2H +

Step 5: Balance charge by adding electrons to more positive side

Eq.1 (old) 8H + + MnO 4 - → Mn 2+ + 4H 2 O

Total charge on reactants side 8H + + MnO 4 - : 8(+1) + 1(-1) = +7

Total charge on products side Mn 2+ + 4H 2 O: 1(+2) + 4(0) = +2

Reactants are more positive therefore, 5 electrons will be added left

Eq.1 (new) 5e - + 8H + + MnO 4 - → Mn 2+ + 4H 2 O

Eq. 2 (old) H 2 S → S + 2H +

Total charge on reactants side H 2 S: 1(0) = 0

Total charge on products side S + 2H + : 1(0) + 2(+1) = +2

Product side is more positive, therefore 2 electrons will be added to the right

Eq. 2 (new) H 2 S → S + 2H + + 2e -

Step 6: Cancel electrons by multiplying the whole equation by a whole number

Eq.1 (old) 5e - + 8H + + MnO 4 - → Mn 2+ + 4H 2 O

Eq.1 (new) [5e - + 8H + + MnO 4 - → Mn 2+ + 4H 2 O] x 2

Eq.1 (new) 10e - + 16H + + 2MnO 4 - → 2Mn 2+ + 8H 2 O

Eq. 2 (old) H 2 S → S + 2H + + 2e -

Eq. 2 (new) [H 2 S → S + 2H + + 2e - ] x 5

Eq. 2 (new) 5H 2 S → 5S + 10H + + 10e -

With 10 electrons on both sides, electrons can be canceled!

Step 7: Add half reactions and cancel common terms on both sides

Eq.1 (old) 10e - + 16H + + 2MnO 4 - → 2Mn 2+ + 8H 2 O

Eq. 2 (old) 5H 2 S → 5S + 10H + + 10e -

Adding both half-reactions together and canceling 10 electrons on both sides gives:

16H + + 2MnO 4 - + 5H 2 S → 2Mn 2+ + 8H 2 O + 5S + 10H +

Canceling H + that are common to both sides of the equation gives the final balanced equation:

6H + + 2MnO 4 - + 5H 2 S → 2Mn 2+ + 8H 2 O + 5S

What is the oxidation number of Mn in MnO 4 - ?

Mn + 4(-2) = -1

Mn – 8 = -1

What element is oxidized?

S is oxidized (refer to Step 1 above for explanation)

What element is the oxidizing agent?

Mn is being reduced and thus, it is the oxidizing agent

1.) A standard Cl 2 | Cl - half-cell has been What is ΔG (in joules) for the spontaneous coupled to a Cl 2 (1.00 atm) l Cl - (0.010M) half-cell. What is the cell voltage? Determine ΔG (at room temperature) for this reaction.

2. The following cell

Ag | Ag + (0.10M) || Ag + (1.0M) | Ag

is a concentration cell and is capable of electrical work.

(a) What is the cell potential?

(b) Which side of the cell is the cathode and which is the anode?

(c) What is the ΔG (in joules) for the spontaneous cell reaction?

3. Two copper electrodes are placed in two copper sulfate solutions of equal concentration and connected to form a concentration cell. What is the cell voltage?

Because the half reactions are the same, there will be no change. The cell voltage will be zero. This can be proved with the Nernst Equation.

E = E o - (0.0592V/2)log(Cu 2+ dilute /Cu 2+ concentrated ) E o = 0 because half reactions are the same. Cu 2+ dilute /Cu 2+ concentrated = 1 because concentrations are the same. log(1) = 0 so the equation becomes: E = 0 - (0.0592V/2)(0) E = 0

One of the solutions is diluted until the concentration of copper ions is one-fifth its original value. What is the cell voltage after dilution?

E can be calculated using the Nernst equation: E = E o - (0.0592V/2)log(Cu 2+ dilute /Cu 2+ concentrated ) E o = 0 because half reactions are the same. Cu 2+ dilute /Cu 2+ concentrated = (1/5)/(1) = 0.2/1 = 0.2 log(0.2) = -0.69897 so the equation becomes: E = 0 - (0.0592V/2)(-0.69897) E = +0.021

6. Assuming unit activities for all substances, determine which of the following reactions will be spontaneous:

These can be figured out using the Activity Series or a Table of Standard Electrode Potentials a. Zn + Mg 2+ → Zn 2+ + Mg Non-Spontaneous.Mg is a stronger reducer than Zn

b. Fe + Cl 2 → Fe 2+ + 2Cl - Spontaneous. Fe is a stronger reducer than Cl -

c. 4Ag + O 2 + 4H + → 4Ag + +2H 2 O Spontaneous. Ag is a stronger reducer than H 2 0

d. 2AgCl → 2Ag + Cl 2 Non-Spontaneous. Ag is a stronger reducer than Cl -

7.) What are the potentials for the following cells or half-cells?

a) E o for the cell: Zn(s) | Zn 2+ || Cu 2+ l Cu(s)

b) E for the half-cell: Zn(s) l Zn 2+ (0.001M)

c) E for the half-cell: Cu 2+ (1O -36 M) | Cu(s)

a.) E o = E(right)-E(left) = E(Cu 2+ /Cu)-E(Zn 2+ /Zn) = 0.340 - (-0.763) = 1.103 V

b.) E = E o (Zn 2+ /Zn) - 0.0592/2 log(1/0.001M) = -0.852 V

c.) E = E o (Cu 2+ /Cu) - 0.0592/2 log (1/10-36 M) = -0.726 V

8.) What are the standard potentials, E o , for the following half-cells?

• S 2- | CuS (s) | Cu(s)
• NH 3 (aq), Zn(NH 3 ) 4 2+ | Zn(s)

Neglect the ligands:

a.) Cu 2+ (aq) + 2e - → Cu (S)

b.) Zn 2+ (aq) + 2e - → Zn (s)

Eq.1 (new) MnO 4 - → Mn 2+ + 4H 2 O

S in H 2 S has an oxidation state = -2

S in S has an oxidation state = 0

Oxidation State Change: -2 → 0

Mn is being reduced (gains 5 electrons)

Mn in MnO 4 - has an oxidation state = +7

Cna inter 2 (progression 2) unit 1

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Homework CNA Inter 2

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CNA Platinum 1 Units 1,2 and 3.

16 questions

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Which expressions are used to express low frequency.

nearly always, often, more often than not.

Sometimes, every now and then

Hardly ever, rarely, every now and then.

Hardly ever, rarely, seldom.

Which expressions express high frequency?

Often, every once in a while, every now and then.

Seldom, rarely, hardly ever, every other day.

More often than not, often, nearly always

None of the other options.

Which expressions express medium frequency.

Every once in a while, every now and then, sometimes

Sometimes, often, nearly always.

Hardly ever, rarely, sometimes

As _________ as I'd like to help the environment, I just can't go to work by bycicle. It's too far.

as often as

I always avoid using the heater or the air conditioner as ______ as I can.

What words are used to summarize ideas?

It seems to me that, in the end, basically.

I tend to think, in short, in a nutshell.

Basically, Essentially, In a nutshell, in short

Which expressions express opinions.

it seems to me that, that way I see it, basically.

To me, I tend to think, essentially.

I seems to me that, the way I see it, from my perspective.

None of the above.

Cooking is a great activity. The word cooking is a...

None of the others.

Ing words can be...

nouns, adjectives, verbs.

nouns, adjectives, verbs and objects.

Nouns and verbs.

none of the options.

To take after somebody means to...

To care for someone

To be angry at someone.

to look or behave like an older family member.

none of the other options.

To take someone through means to...

to start doing someone regularly.

to accept an offer or invitation.

take control.

to explain something in detail.

To take someone up on something means to...

Accept an offer or invitation.

to start doing something regularly.

To take control.

to remove a piece of clothing,

To take over something means to...

Receive and understand information.

accept an offer or invitation.

Wish + past perfect is used to express...

Something about the past you wish you could change.

Something about the present you wish to be different.

Something about the past that you are happy with.

Wish + simple past expresses...

Something about the present that you like.

Something about the past.

Something you want to be different and hope for other people to change.

Wish + would is used to express...

Something about the past you want to be different.

Something about the present and don't like.

Something you want to be different and hope for other people to change/future.

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Equations and Inequalities Homework (Algebra 2 - Unit 1)

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Equations and Inequalities Homework Bundle:

This resource is a bundled set of homework practice sets and daily content quizzes for Unit 1: EXPRESSIONS, EQUATIONS & INEQUALITIES designed for Algebra 2 Honors students. The file includes 10 pages of homework assignments and two different forms of a daily content quiz that you can use as a homework check, group work, or exit tickets.

The unit includes the following topics:

1) Properties of Real Numbers

2) Algebraic Expressions

3) Solving Equations

4) Solving Inequalities

5) Absolute Value Equations and Inequalities

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