Answer:
The fraction of the volume of the atom that is taken up by the nucleus is
.
The density of a proton is
.
Explanation:
Diameter of the atom ,d = 2.50 Å
Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å
Volume of the sphere= 
Volume of atom = V
..[1]
Diameter of the nucleus ,d' =
Radius of the nucleus ,r' = 0.5 d'=
Volume of nucleus = V'
..[2]
Dividing [2] by [1]



The fraction of the volume of the atom that is taken up by the nucleus is
.
Diameter of the proton ,d = 
1 m = 100 cm
Radius of the proton,r = 0.5 d=
Volume of the sphere= 
Volume of atom = V

Mass of proton, m = 1.0073 amu = 

Density of the proton : d

The density of a proton is
.
Answer:
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Explanation:
Answer:
52.2 g
Explanation:
Step 1: Write the balanced equation
3 KOH + H₃PO₄ ⟶ K₃PO₄ + 3 H₂O
Step 2: Calculate the moles corresponding to 89.7 g of KOH
The molar mass of KOH is 56.11 g/mol.
89.7 g × 1 mol/56.11 g = 1.60 mol
Step 3: Calculate the moles of H₃PO₄ needed to react with 1.60 moles of KOH
The molar ratio of KOH to H₃PO₄ is 3:1. The moles of H₃PO₄ needed are 1/3 × 1.60 mol = 0.533 mol.
Step 4: Calculate the mass corresponding to 0.533 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
0.533 mol × 97.99 g/mol = 52.2 g
Answer:
See explaination
Explanation:
please kindly see attachment for the step by step solution of the given problem
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