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Snowcat [4.5K]
3 years ago
12

In the fourth period of the periodic table, how many elements have one or more 4p electrons

Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
3 0
The fourth period, or fourth row, of the periodic table contains the elements from Z = 19 (potassium) to Z = 36 (kripton).

The abbreviated electron configuratin of potassium is K = [Ar] 4s.

Where [Ar] is 1s2 2s2 2p6 3s2 3p6.

The next element, Calcium, Z = 20, has electron configuration Ca = [Ar] 4s2.

Starting the next element, Scandium, Z = 21, the atoms start to fill the 3d  orbitals. Scandium has electron configuration Sc = [Ar] 4s2 3d.

The 3d orbitals permit to accomodate 10 electrons. So, the filling of 3d orbitals end with the element Z = 30, wich is zinc. The electron configuration of zinc is Zn =[Ar] 4s2 3d10.

So, starting with the element Z = 31 (gallium) until Z = 36, this is 6 elements, will fill the 4p orbitals.Those are Ga, Ge, As, Se, Br and Kr.

The answer is that six elements in the fourth period of the periodic table have one or more 4p electrons. You can find those elements in the right side of the period and are from 13 group to 18 group.
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What is the volume of a sample of liquid mercury that has a mass of 53.8 g, given that the density of mercury is 13.6 g/mL? Answ
Liula [17]

The formula for density is:

density=\frac{mass}{volume}

We know the density for mercury is 13.6 g/mL, and we know the mass of the sample is 53.8 g. Thus, we can plug these numbers into our equation and solve for volume.

The volume is unknown, so we can simply denote it as "x"

13.6 g/mL=\frac{53.8 g}{x}

multiply both sides by x

(x)(13.6 g/mL)=\frac{53.8 g}{x}(x)

The x's cancel out on the right side and you are left with

(x)(13.6 g/mL)=53.8 g

From here, simply divide both sides of the equation by 13.6 g/mL and solve for x.

\frac{(x)13.6 g/mL}{13.6 g/mL}=\frac{53.8 g}{13.6 g/mL}

x=3.955882353 mL

Round to 3 significant figures, and your final answer is:

x=3.96 mL

The volume of the sample of mercury was 3.96 mL.


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A conjugate acid of a base results when the base accepts a proton.

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Ammonium, NH4+, acts as a conjugate acid to ammonia, NH3.

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