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prisoha [69]
3 years ago
15

if the spin of one electron in an orbital is clockwise , what is the spin of the other electron in that orbital

Chemistry
1 answer:
nata0808 [166]3 years ago
6 0

Here we have to get the spin of the other electron present in a orbital which already have an electron which has clockwise spin.

The electron will have anti-clockwise notation.

We know from the Pauli exclusion principle, no two electrons in an atom can have all the four quantum numbers i.e. principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s) same. The importance of the principle also restrict the possible number of electrons may be present in a particular orbital.

Let assume for an 1s orbital the possible values of four quantum numbers are n = 1, l = 0, m = 0 and s = \frac{+}{-}\frac{1}{2}.

The exclusion principle at once tells us that there may be only two unique sets of these quantum numbers:

1, 0, 0, +\frac{1}{2} and 1, 0, 0, -\frac{1}{2}.

Thus if one electron in an orbital has clockwise spin the other electron will must be have anti-clockwise spin.    

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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
Nataly_w [17]
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
<span>            (M)     (M)          (M/s) </span>
<span>1         0.50    0.010      3.0×10−3 </span>
<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
8 0
3 years ago
Read 2 more answers
Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.
olga55 [171]

Answer:

At the start of the process, the volume not occupied by the water is 2 m3

Explanation:

At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).

Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).

The volume in time will be

V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3]

8 0
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The Mass of the mineral is 9.6 grams . The mineral is placed in a graduated cylinder containing 8.0 ml of water . The water rise
ryzh [129]

Answer:

1.2 g/ ml

Explanation:

The volume of the mineral = increase in volume of the water whuich is 16 - 8 = 8mls.

Therefore the  mineral's density = 9.6 / 8

=  1.2 g/ ml  answer

7 0
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max2010maxim [7]
B. Copper since the rest of the choices are compounds. Copper is an element.
5 0
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The electron pair in a C-F bond could be considered Question 3 options: closer to C because carbon has a larger radius and thus
Leokris [45]

Answer:

closer to F because fluorine has a higher electronegativity than carbon

Explanation:

Electronegativity refers to the ability of an atom in a bonding situation to draw the shared electrons of the bond closer to itself.

Electronegativity increases across the period and decreases down the group. A highly electronegative atom draws the shared electron pair of a bond towards itself.

When two atoms are bonded together, the electron pair is always drawn closer to the atom that has a higher electronegativity.

Hence, the electron pair in a C-F bond could be considered closer to F because fluorine has a higher electronegativity than carbon.

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