Answer:-
Carbon
[He] 2s2 2p2
1s2 2s2 2p2.
potassium
[Ar] 4s1.
1s2 2s2 2p6 3s2 3p6 4s1
Explanation:-
For writing the short form of the electronic configuration we look for the nearest noble gas with atomic number less than the element in question. We subtract the atomic number of that noble gas from the atomic number of the element in question.
The extra electrons we then assign normally starting with using the row after the noble gas ends. We write the name of that noble gas in [brackets] and then write the electronic configuration.
For carbon with Z = 6 the nearest noble gas is Helium. It has the atomic number 2. Subtracting 6 – 2 we get 4 electrons. Helium lies in 1st row. Starting with 2, we get 2s2 2p2.
So the short term electronic configuration is [He] 2s2 2p2
Similarly, for potassium with Z = 19 the nearest noble gas is Argon. It has the atomic number 18. Subtracting 19-18 we get 1 electron. Argon lies in 3rd row. Starting with 4, we get 4s1.
So the short electronic configuration is
[Ar] 4s1.
For long term electronic configuration we must write the electronic configuration of the noble gas as well.
So for Carbon it is 1s2 2s2 2p2.
For potassium it is 1s2 2s2 2p6 3s2 3p6 4s1
Answer:
Are sessile, have tentacles, mouths open downwards
Explanation:
Medusae may have many tentacles with which they catch their prey. Sometimes, people bathing in the ocean are touched by a medusa. Their tentacles are long and thin and covered in venomous stinging cells (known as nematocysts). They inject poison which immobilizes small fish. People who are stung should seek medical attention. There have been some deaths after severe medusa stings.
Answer:
9.430 * 10¹⁷ protons per second whill shine on the book from a 62 W bulb
Explanation:
To answer this question, first let's calculate the energy of a single photon with a wavelength (λ) of 504 nm:
E = hc/λ
Where h is Planck's constant (6.626*10⁻³⁴ J·s) and c is the speed of light (3*10⁸ m/s).
E = 6.626*10⁻³⁴ J·s * 3*10⁸ m/s ÷ (504*10⁻⁹m) = 3.944 * 10⁻¹⁹ J.
So now we can make the equivalency for this problem, that
<u>1 proton = 3.944 * 10⁻¹⁹ J</u>
Now we convert watts from J/s to proton/s:
1
= 1 W
Solving the problem, a 62 W bulb converts 5% of its output into light, so:
3.1 watts are equal to [ 2.535*10¹⁸ proton/s * 3.1 ] = 7.858 * 10¹⁸ proton/s
Of those protons per second, 12% will shine on the chemistry textbook, thus:
7.858 * 10¹⁸ proton/s * 12/100 = 9.430 * 10¹⁷ protons/s