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3 years ago

List six elements with only 1 electron in an s orbital

Chemistry

1 answer:

Verdich [7]3 years ago

6 0

Answer:

$Lithium=Li2s^1\\Sodium=Na3s^1\\Potassium=K4s^1\\Rubidium=Rb5s^1\\Cesium=Cs6s^1\\Francium=Fr7s^1$

Explanation:

When it comes to electron configuration and orbitals, it's important to first identify what exactly we are trying to identify. Below is a given example:

$He1s^2$

$He=element$

$1=level$

$s=orbital$

$(exponent)^2=electrons$

Looking at the periodic table, identify the alkali metal family on the periodic table, or group one elements:

$Lithium=Li2s^1\\Sodium=Na3s^1\\Potassium=K4s^1\\Rubidium=Rb5s^1\\Cesium=Cs6s^1\\Francium=Fr7s^1$

Notice how each configuration has an exponent of $1$ , representative of a single electron in their s-orbital.

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put the contributions to the understanding of the atomic structure in order from most recent at the top to the earliest at the b

r-ruslan [8.4K]

Answer:

From Top to Bottom:

- Democritus coming up with the concept of an atom

- Dalton discovering that atoms are the smallest part of an element

- Rutherford discovering the nucleus of an atom

- Thomson discovering electrons

- Bohr modeling electrons orbiting the nucleus

- Schrodinger modeling electrons in the electron cloud

Explanation:

The best way to think about this is from the inside out. Democrats (who lived long before any of the other scientists mentioned) was the one who thought of the idea of the atom. - Therefore, this must be first because all other choices are elaborations on the idea that atoms exist. Next must be Dalton. Dalton saw atoms as "cannonballs" if you will; a solid mass. So then after that, Rutherford and his gold foil experiment (he discovered that some rays he shot through gold foil were deflected back; ie the existence of concentrated areas in an atom, ie the nucleus). Then we get into the information on electrons. We must start with discovery (Thomson). Heres where it gets complicated. Electrons don't actually orbit the nucleus, they exist in electron clouds. So it would be Bohr, who came up with the idea that electron exist outside the nucleus, then Schrodinger, who elaborated on Bohr's theory. Hope this helps!

Nat, Junior

Accel + AP Chem student

5 0

1 year ago

Which is more acidic among the following?

GaryK [48]

Answer:

Explanation:

7 0

2 years ago

6. Which example has BOTH ionic and covalent bonds? *

Fed [463]

Answer:

It’s either A or D buttttt probably d

Explanation:

7 0

2 years ago

Of the colors on the visible light spectrum, which one has the longest wavelength, lowest frequency, and lowest energy?

Elenna [48]

Answer:

Red

Explanation:

Violet - shortest wavelength, around 400-420 nanometers with highest frequency. They carry the most energy.

Indigo - 420 - 440 nm

Blue - 440 - 490 nm

Green - 490 - 570 nm

Yellow - 570 - 585 nm

Orange - 585 - 620 nm

Red - longest wavelength, at around 620 - 780 nanometers with lowest frequency and least amount of energy

Therefore, red is the answer you're looking for.

I hope this helps and that you have a great day! :)

5 0

2 years ago

One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO

zhenek [66]

Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)

The stoichiometric ratio between Na₃(CO₃)(HCO₃)·2H₂O(s) and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
= 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
= 1000 x 10³ g / 226 g mol⁻¹
= 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x $\frac{3}{2}$
= 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
= 703540.02 g
= 703.540 kg

Hence, the theoretical yield of Na₂CO₃ = 703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
= 92.34%

7 0

3 years ago