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3 years ago

The filling of which orbital is represented by the transition metals in period 4?

2 answers:

8 0

<span>The 4s orbital has a lower energy than the 3d, and so fills next. That fits with the chemistry of potassium and calcium and the entire period. So the answer is 3d</span>

Sindrei [870]3 years ago

3 0

Answer: 3d

Explantion:

1) Period 4 contains the elements with atomic numbers 19 through 36.

2) The elements with atomic numbers 19 (K) and 20 (Ca) fill the orbital 4s.

3) After that, as Aufbau's rule may help you to remember, the energy of the orbitals 3d is lower than the energy of the orbtitals 4p. So, the element 21 (Sc) start fillind the orbital 3d.

There are ten 3d orbitals, so the elements 21 through 30 fill the 3d orbitals.

Those elements, called transition metals are: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, and Zn.

When the 3d orbitals are full, the next elements in the same period 4, fill the six 4p orbitals.

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BRAINLIESTTT ASAP!!!! PLEASE HELP ME :)

Elan Coil [88]

Answer:

1s² 2s²2p³

Explanation:

If the atom has seven electrons, it is Element 7 (nitrogen).

In the Periodic Table, you count the electrons in all the subshells up to

No. 7.

In the first Period, you have filled the 1s level (2 electrons).

In the second Period, you have filled the 2s subshell (2 electrons) and put three electrons in the 2p subshell.

Thus, the electron configuration is

1s² 2s²2p³

Note how the superscripts tell you the number of electrons in each subshell: <em>2 + 2 + 3 = 7</em>.

6 0

3 years ago

Which statement is true for all minerals

nekit [7.7K]

What are the answer choices

8 0

3 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500=193000C$ is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

$14\times 1000g=14000g$ of copper is deposited by = $\frac{193000}{63.5}\times 14000=42551181 C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

$40.0=\frac{42551181 C}{t}\\\\t=1063779sec$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $1063779s\times \frac{1hr}{3600s}=295hr$

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0

3 years ago

What volume of a 0.550 M solution of potassium hydroxide (KOH) can be made with 19.9 g of potassium hydroxide?

Natalija [7]

Answer:

0.645 L

Explanation:

To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.

(Step 1)

Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol

Molar Mass (KOH): 56.104 g/mol

19.9 grams KOH 1 mole
-------------------------- x ----------------------- = 0.355 moles KOH
56.014 grams

(Step 2)

Molarity = moles / volume <----- Molarity ratio

0.550 M = 0.355 moles / volume <----- Insert values

(0.550 M) x volume = 0.355 moles <----- Multiply both sides by volume

volume = 0.645 L <----- Divide both sides by 0.550

6 0

1 year ago

Can someone help me with number 1 and 2 plz!

GalinKa [24]

Answer:

1) 0 N

2) 8 N

Explanation:

The net force is the sum of all of the forces acting on the object.

For question 1, we can see that there is a force of 5 N acting to the right and 5 N acting to the left. If we define the right to be positive and the left to be negative, then the net force equals:

Fnet = 5N - 5N = 0 N

Therefore, the net force in question 1 is 0 N.

For question 2, the process is very similar. We want to find the sum of the forces acting on the object. In this case, there are forces of 3 N and 5 N acting to the right.

Fnet = 3 N + 5 N = 8 N

Therefore, the net force in question 2 is 8 N.

Hope this helps!

3 0

3 years ago