Answer:
Explanation:
Given that,
The speed of light is 3×10⁸ m/s
We need to convert it into mi/hr.
Since, 1 mile = 1610 m
1 hour = 3600 s
So,
So, the required speed of light is 
.
<u>Rules to write the electronic configuration
:</u>
Electrons complete orbitals in a way to reduce the energy of the atom. Therefore, the electrons in an atom complete the principal energy levels in order of rising energy (the electrons are getting distant from the nucleus). The order of levels filled appearances like the following
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p
One way to recognise this pattern, apparently the simplest, is to refer to the periodic table and remember where each orbital block drops to rationally understand this pattern. Different way is to make a table like the one below and use vertical lines to determine which subshells resemble with each other.
- S block: The S obstruct in the periodic table of components known as gatherings 1 and 2. There is a limit of two electrons that can possess the s orbital.
- P Block: The P square contains group of 13, 14, 15, 16, 17, and 18, except for Helium.
- D Block: The D block elements are found in groups 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 of the periodic table.
2.94
×
10
24
⋅
molecules
6.022
×
10
23
⋅
molecules
⋅
mol
−
1
×
142.3
⋅
g
⋅
mol
−
1
≅
700
⋅
g
the Equator receives the most solar energy
The question is incomplete, the complete question is;
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.
Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answer:
3.81 g of lead
Explanation:
The equation of the reaction is;
Pb^2+(aq) + 2e ---->Pb(s)
Quantity of charge = 96.0 A * 37.0 seconds = 3552 C
Now we have that 1F = 96500 C so;
207 g of lead is deposited by 2 * 96500 C
x g of lead is deposited by 3552 C
x = 207 * 3552/2 * 96500
x = 735264/193000
x = 3.81 g of lead
