Explanation:
The problem here is to find the atomic number of each of the element given.
Sum the powers of the configuration.
a- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
Atomic number is = 2 + 2 + 6 + 2 + 6 + 1 = 19
b- 1s² 2s² 2p⁶ 3s² 3p⁴
Atomic number = 2 + 2 + 6 + 2 + 4 = 16
c- 1s¹
Atomic number = 1
1) 1 second = 1000000 microseconds
2) 186000 / 1.61 = 115527.9503 Km
3) We use the equation:
Speed = distance / time
4) rearrange for distance by multiplying both sides by time to get;
Distance = speed x time
5) plug in the values:
Distance = 115527.9503 x 0.000001
= 0.1155279503 Km
The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂<span>(g)
the limiting reactant in the equation is glucose as the whole amount of glucose is used up in the reaction.
the amount of </span>C₆H₁₂O₆ used up - 13.2 g
the number of moles reacted - 13.2 g/ 180 g/mol = 0.073 mol
stoichiometry of glucose to CO₂ - 1:6
then number of CO₂ moles are - 0.073 mol x 6 = 0.44 mol
As mentioned this reaction takes place at standard temperature and pressure conditions,
At STP 1 mol of any gas occupies 22.4 L
Therefore 0.44 mol of CO₂ occupies 22.4 L/mol x 0.44 mol = 9.8 rounded off - 10.0 L
Answer is B) 10.0 L CO₂
Answer:
There is 5.56 g of gold for every 1 g of chlorine
Explanation:
The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction 
You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:
The proportion is the equal relationship that exists between two reasons and is represented by: 
This reads a is a b as c is a d.
To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:
Solving for the mass of gold gives:
mass of gold= 5.56 grams
So, <u><em>there is 5.56 g of gold for every 1 g of chlorine</em></u>
The s orbitals are not symmetrical in shape is a FALSE statement.
An s orbital is so symmetric, more specifically spherically symmetric that it looks the same from all directions.
- The atomic orbitals in the atoms of elements differ in shape.
In essence, the electrons they describe have varying probability distributions around the nucleus. The spherical symmetry of s orbitals is evident in the fact that all orbitals of a given shell in the hydrogen atom have the same energy.
- All s orbitals are spherically symmetrical. Put simply, an electron that occupies an s orbital can be found with the same probability at any orientation (at a distance) from the nucleus.
The s orbitals are therefore represented by a spherical boundary surface which is a surface which captures a high proportion of the electron density.
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