You should get 5.05 x 10^23 moles.
<u>Answer:</u> The formula of the compound formed between rubidium and fluorine is RbF
<u>Explanation:</u>
Ionic bond is defined as the bond which is formed by complete transfer of electrons from one atom to another atom.
The atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom. This bond is usually formed between a metal and a non-metal.
Rubidium is the 37th element of the periodic table having electronic configuration of 
This will loose 1 electron to form
ion
Fluoride is the 9th element of the periodic table having electronic configuration of 
This will gain 1 electron to form
ion
To form
compound, 1 rubidium ion is needed to neutralize the charge on fluoride ion
The formation of the given compounds is shown in the image below.
The empirical formula is the same as the molecular formula : C₁₀H₅O₂
<h3>Further explanation</h3>
Given
Molecular formula : C₁₀H₅O₂
Required
The empirical formula
Solution
The empirical formula (EF) is the smallest comparison of atoms of compound forming elements.
The molecular formula (MF) is a formula that shows the number of atomic elements that make up a compound.
(empirical formula) n = molecular formula
<em>(EF)n=MF
</em>
(EF)n = C₁₀H₅O₂
If we divide by the number of moles of Oxygen (the smallest) which is 2 then the moles of Hydrogen will be a decimal number (not whole), which is 2.5, then the empirical formula is the same as the molecular formula
Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M