To solve this we use the
equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the
volume of the stock solution, M2 is the concentration of the new solution and
V2 is its volume.
14 M x V1 = 4.20 M x 200 mL
V1 = 60 mL needed of the concentrated solution
Answer:
The empirical formula of this substance is:

Explanation:
To find the empirical formula of this substance we need the molecular weight of the elements Carbon, Hydrogen and Oxygen, we can find this information in the periodic table:
- C: 12.01 g/mol
- H: 1.00 g/mol
- O: 15.99 g/mol
With the information in this exercise we can suppose in 100 g of the substance we have:
C: 48.64 g
H: 8.16 g
O: 43.2 g (100 g - 48.64g - 8.16g= 43.2 g)
Now, we need to divide these grams by the molecular weight:

We need to divide these results by the minor result, in this case O=2.70 mol

We need to find integer numbers to find the empirical formula, for this reason we multiply by 2:

This numbers are very close to integer numbers, so we can find the empirical formula as subscripts in the chemical formula:

Here, the three different notation of the p-orbital in different sub-level have to generate
The value of azimuthal quantum number (l) for -p orbital is 1. We know that the magnetic quantum number
depends upon the value of l, which are -l to +l.
Thus for p-orbital the possible magnetic quantum numbers are- -1, 0, +1. So there will be three orbitals for p orbitals, which are designated as
,
and
in space.
The three p-orbital can be distinguish by the quantum numbers as-
For 2p orbitals (principal quantum number is 2)
1) n = 2, l = 1, m = -1
2) n = 2, l = 1, m = 0
3) n = 2, l = 1, m = +1
Thus the notation of different p-orbitals in the sub level are determined.
Light acts as a wave so when you burn a certain element it generates a specific wavelength which represents a specific color light. ^-^
None of the above.
1 mole filled with gas at STP occupies
=22.4 L
∴ 3mole of kr gas at STP occupies
= 3 × 22.4
= 67.2 L