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2 years ago

It _______ moving right now.

Chemistry

2 answers:

Arte-miy333 [17]2 years ago

7 0

Answer:

Umm what do i do to help you??

ValentinkaMS [17]2 years ago

7 0

Answer:

It is moving right now

Explanation:

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What is the second quantum number of the 3p1 electron in aluminum 1s22s22p63s23p1?

Brut [27]

There are 4 quantum numbers that can be used to describe the space of highest probability an electron resides in.
First quantum number is the principal quantum number- n , states the energy level.
Second quantum number states the angular momentum quantum number - l,
states the subshell and the shape of the orbital
values of l for n energy shells are from 0 to n-1
third is magnetic quantum number - m, which tells the specific orbital.
fourth is spin quantum number - s - gives the spin of the electron in the orbital

here we are asked to find l for 3p1
n = 3
and values of l are 0,1 and 2
for p orbitals , l = 1
therefore second orbital for 3p¹ is 1.

5 0

3 years ago

Read 2 more answers

Andru [333]

Answer:

Where is the question?

7 0

2 years ago

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

$Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)$

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

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

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

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

where,

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

$E_{cell}$ = 0.507 V

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

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

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

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

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

where,

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

$E_{cell}$ = 0.507 V

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

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

6 0

3 years ago

A textbook measures 250 mm long, 225 mm wide and 50 mm thick. What is the volume of this book in mm3? What is the volume of this

Dvinal [7]

Answer:

2.81 × 10⁶ mm³

2.81 × 10⁻³ m³

Explanation:

Step 1: Given data

Length (l): 250 mm

Width (w): 225 mm

Thickness (t): 50 mm

Step 2: Calculate the volume of the textbook

The book is a cuboid so we can find its volume (V) using the following expression.

V = l × w × t = 250 mm × 225 mm × 50 mm = 2.81 × 10⁶ mm³

Step 3: Convert the volume to cubic meters

We will use the relationship 1 m³ = 10⁹ mm³.

2.81 × 10⁶ mm³ × 1 m³ / 10⁹ mm³ = 2.81 × 10⁻³ m³

6 0

3 years ago

Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min

Olenka [21]

(80+125+45) / 10 = 250/10 =25
25 meters per minute= 0.41 meters/second

the direction and stopping time is irrelevant to the problem.

6 0

3 years ago