A direct variation function contains the points (–8, –6) and (12, 9). Which equation represents the function?

Give the set of four quantum numbers that represent the last electron added (using the aufbau principle to the cl atom.

qaws [65]

An element has four quantum numbers associated with it. We have

1. principal quantum number ( $n$ )
2. azimuthal quantum number ( $l$ )
3. magnetic quantum number ( $m_{l}$ )
4. spin quantum number ( $m_{s}$ )

To look for each quantum number of Chlorine (Cl), we mus first recall its electronic configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$ .

1. For $n$ , we need to look for its highest energy level or by simply checking the period that the element belongs to. So for this case, it's $n = 3$ .

2. For $l$ , we need to look for the sublevel where the last electron was assigned. The sublevels, p, d, f are assigned to a value of 0, 1, 2, and 3, respectively. Since the last electron was assigned in p, then $l = 1$ .

3. And for $m_{l}$ , the number of orbitals present in Cl is $2l+1 = 3$ . The possible magnetic quantum numbers for $p$ are -1, 0, +1. Since there 5 electrons we count starting at -1. So, we have $m_{l}=+1[tex].\\4. Lastly, for [tex]m_{s}$ , we need to determine the spin. Since we have 5 electrons, all the arrows upward are filled in. That means the fifth electron has a spin of +1/2.

Answers: $n =3, l = 1, m_{l} = 0, m_{s} =+1/2$

8 0

3 years ago

Plzzz help as soon as possible

Anika [276]

Answer:

Step-by-step explanation:

8 is not a function

9 is a function

10 is a function

4 0

3 years ago

-5 (1-5x)+5 (-8x-2)=-4x-8x ahhh plz send help

Yanka [14]

2x√8-x - x/2√8-x is the answer

7 0

4 years ago

ALGEBRA 2

vladimir1956 [14]

Answer:

$\left\{ \begin{array}{ll} a+s+r=350 \\ 4a+2.5s+2r=1095\\r=s-40 \end{array} \right.$

Step-by-step explanation:

Let's define our variables. Let a represent the number of adult tickets sold, s represent the number of senior tickets sold, and r represent the number of senior tickets sold.

We know that a total of 350 tickets were sold. So, the number of adult, student, and senior tickets sold must total 350. Therefore, we can write the following equation:

$a+s+r=350$

We know that each adult ticket costs $4, each student ticket costs $2.50, and each senior ticket costs $2. We are given that a total of $1095 was collected. Therefore, the number of tickets multiplied by their respective price must equal $1095. So, we can write the following equation:

$4a+2.5s+2r=1095$

Finally, we know that 40 fewer senior tickets were sold than student tickets. So, however many students tickets were sold, we can subtract 40 to get the number of senior tickets sold. Therefore:

$r=s-40$

So, our system of equations is:

$\left\{ \begin{array}{ll} a+s+r=350 \\ 4a+2.5s+2r=1095\\r=s-40 \end{array} \right.$

And we're done!

7 0

3 years ago

4. [5 pts] Describe how and why the formula for permutations differs from the formula for combinations.

ivann1987 [24]

Answer:

They are different because of the order in the permutation matters. In combination, the order doesn't matter. In other words in a permutation 123 and 132 are different but in a combination are the same group (they have the same digits 1,2, and 3).

Step-by-step explanation:

The formula of the permutation is $P(n,r)=\frac{n!}{(n-r)!}$ , when you are performing a permutation you pick r objects from a total of n, for the first pick you can choose from n, but for the second you have n-1, and this continues to your pick number r in which you will choose from n-r+1, and the total of permutation is the multiplicación of this number of choices for each pick, like this:

$n(n-1)(n-2)...(n-r+1)$

If $n!=n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)$ and $(n-r)!=(n-r)(n-r-1)(n-r-2)...(1)$

$\frac{n!}{(n-r)!}=\frac{n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)}{(n-r)(n-r-1)(n-r-2)...(1)}$

The factor equals above and under cancel each other.

$\frac{n!}{(n-r)!}=\frac{n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)}{(n-r)(n-r-1)(n-r-2)...(1)}\\\frac{n!}{(n-r)!}=n(n-1)(n-2)...(n-r+1)$

In combination, the order of the element isn't important, so from the total of permutation you have to eliminate the ones with the same objects with different order and counting just once each group, when choosing r objects the total of permutation for a single group of r objects is: $r(r-1)(r-2)...(1)=r!$ . If you divide the total of permutations of n taking r by r! you get the combinations (where the order is not important). The formula of the combination is $C(n,r)=\frac{n!}{r!(n-r)!}$ .

4 0

4 years ago