What particle diagram represent a sample of one compound only

1. What is an isomer? How many possible isomers of hexane are there? What are the structural differences between these isomers?

UNO [17]

When two compounds have same molecular formula but different structural formula, they are said to be isomers of each other and this phenomenon is called as isomerism.

Hexane has six isomers.

The isomers of hexane have a difference in their structure due to parent chain length (chain isomerism). They are also differentiated on the basis of straight chain as in n-hexane and branched chain as in remaining five. Structures along with names are shown below.

8 0

4 years ago

A sample of an ideal gas has a volume of 0.500 L at 25 degrees celcius and 1.20 atm pressure. What is its volume at 75 degrees c

NemiM [27]

The ideal gas law is PV= nRT.
First you need to manipulate the equation to splice for volume,
Which will be V= nRT/P

Now you need to input the numbers for each variable. Make sure to remember what the value R equals and it’s units. R= 0.08206 L•atm/n•K

4 0

3 years ago

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 Zn

nlexa [21]

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 :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

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:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}$

$E_{cell}=0.49V$

(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 :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

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

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}$

$E_{cell}=0.52V$

(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 :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}$

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

$0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}$

$x=1.49M$

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

5 0

4 years ago

How many moles of dinitrogen tetroxide (N2O4) are in 6.49 x 109 particles?

yan [13]

Answer: There are $1.08 \times 10^{-14}$ moles $N_{2}O_{4}$ present in $6.49 \times 10^9$ particles.

Explanation:

According to the mole concept, 1 mole of a substance contains $6.022 \times 10^{23}$ particles.

Hence, number of moles present in $6.49 \times 10^9$ particles are calculated as follows.

No. of moles = $\frac{6.49 \times 10^9}{6.022 \times 10^{23}}$ mol

= $1.08 \times 10^{-14}$ mol

Thus, there are $1.08 \times 10^{-14}$ moles $N_{2}O_{4}$ present in $6.49 \times 10^9$ particles.

6 0

3 years ago

Read 2 more answers

Heyyyywooo Friends Can Yall Plz Help

VARVARA [1.3K]

Answer:

I think the answer is A

3 0

3 years ago