CHEMISTRY HIGH SCHOOL

The arrangement of particles is least ordered in a sample ofa. NaCl(aq)
b. NaCl()
c. NaCl(g)
d. NaCl(s)

Answers

Answer 1

Answer:

Answer:

The arrangement of particles is least ordered is a > c > b > d

Explanation:

Given that,

The arrangement of particles is least ordered in a sample of NaCl.

We know that,

The least ordered arrangement are.

The form of state of particle in least ordered.

So, The least ordered arrangement will be

aqua > gas > liquid > solid

The arrangement of particles is least ordered in a sample of NaCl

$NaCl(aq) > NaCl(g) > NaCl(l) > NaCl(s)$

Hence, The arrangement of particles is least ordered is a > c > b > d

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77. A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25°C. The initial concentrations of Ni2+ and Zn2+ are 1.50 M and 0.100 M, respectively. a. What is the initial cell potential? b. What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M? c. What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?

Answers

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^(2+)$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^(2+)$ and $Zn^(2+)$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_([Ni^(2+)/Ni])=-0.23V$

$E^0_([Zn^(2+)/Zn])=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^(2+)+2e^-$ $E^0_([Zn^(2+)/Zn])=-0.76V$

Reaction at cathode (reduction) : $Ni^(2+)+2e^-\rightarrow Ni$ $E^0_([Ni^(2+)/Ni])=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^(2+)(aq)\rightarrow Zn^(2+)(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_(cathode)-E^o_(anode)$

$E^o=E^o_([Ni^(2+)/Ni])-E^o_([Zn^(2+)/Zn])$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_(cell)$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_(cell)=0.53-(0.0592)/(2)\log ((0.100))/((1.50))$

$E_(cell)=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^(2+)$ has fallen to 0.500 M.

New concentration of $Ni^(2+)$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^(2+)$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])$

Now put all the given values in the above equation, we get:

$E_(cell)=0.53-(0.0592)/(2)\log ((1.1))/((0.500))$

$E_(cell)=0.52V$

(c) Now we have to calculate the concentrations of $Ni^(2+)$ and $Zn^(2+)$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)+x])/([Ni^(2+)-x])$

Now put all the given values in the above equation, we get:

$0.45=0.53-(0.0592)/(2)\log ((0.100+x))/((1.50-x))$

$x=1.49M$

The concentration of $Ni^(2+)$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^(2+)$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

How many electrons in an atom can have the n = 5, l = 2 designation?

Answers

Answer:

10 electrons

Explanation:

https://socratic.org/

Which condensed structural formula represents an unsaturated compound?(1) CH3CHCHCH3
(2) CH3CH2CH3
(3) CH3CH3
(4) CH4

Answers

Answer: The correct answer is option (1).

Explanation:

The general formula of alkanes (Saturated compounds) is given as: $C_nH_(2n+2)$

The general formula of alkene (Unsaturated compounds)is given as: $C_nH_(2n)$

The general formula of alkynes (Unsaturated compounds)is given as: $C_nH_(2n-2)$

1. $CH_3CHCHCH_3:C_4H_8$ unsaturated hydrocarbon

2. $CH_3CH_2CH_3:C_3H_8$ Saturated hydrocarbon

3. $CH_3CH_3:C_2H_6$ Saturate hydrocarbon

4. $CH_4$ Saturated hydrocarbon

Number(1).A carbon atom makes 4 bondings.As the second Carbon in the conpuond has only one hidrogen,it has to make other 3,two with the other carbon atoms,one of them Being a double,making the compound unsaturated.PS:Sorry for my english,um fron Brazil

What is the molarity of 1.5 liters of an aqueous solution that contains 52 grams of lithium fluoride, LiF, (gram-formula mass = 26 grams/mole)?

Answers

Molarity is simply the number of moles dissolved per liter of solution.
Converting the mass of LiF to moles:
52 g / 26 g/mol = 2 mol LiF

Next, simply divide the number of mole by the volume
2 mol / 1.5 L = 1.3333 mole/L

So, the molarity of the given solution is 1.3333 M

Which of these is an appropriate instrument to measure mass?balance
scale
ruler
graduated cylinder

Answers

Answer: Balance

Explanation:

Answer:

balance

Explanation:

because

Josh prepares a solution by mixing sodium nitrate with water. To calculate the molarity of the solution, josh should calculate the moles of which substance,and why?

Answers

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. Josh should calculate the moles of the sodium nitrate by dividing the mass with the molar mass of sodium nitrate. This is because it is the solute in this problem.

Answer:

the answer is sodium nitrate because it is solute

Explanation

thats what is correct I assure you.

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