Answer:
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First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5
The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm
Answer:
The Relative Formula Mass of Fe(NO₃)₂ is 179.8524 grams
Explanation:
The Relative Formula Mass is the mass of one mole of a compound expressed in grams, obtained by adding together the Relative Atomic Masses, RAM, of the elements which makes the compound
The Relative Formula Mass of a compound is the same as its Relative Molecular Mass
The relative formula mass of Fe(NO₃)₂ is given as follows;
The relative atomic mass of Fe = 55.845 amu
The relative atomic mass of nitrogen, N = 14.0067 amu
The relative atomic mass of oxygen, O = 15.999 amu
Therefore, we have;
The formula mass of Fe(NO₃)₂ = (55.845 + 2×(14.0067 + 3×15.999)) amu = 179.8524 amu
The Relative Formula Mass of Fe(NO₃)₂ = 179.8524 grams.
answer : D= m/v
Explanation:
all about density and mass and size of the molecules in a liquid and how close they are packed together
Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:

1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

10 ml 17.50 ml
(x) M 0.200 M
Molarity = 
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
= 
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration = 
Molar Concentration = 
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M