The basic difference between thin layer chromatography (TLC) and paper chromatography (PC) is that, while the stationary phase in PC is paper, the stationary phase in TLC is a thin layer of an inert substance supported on a flat, unreactive surface. ... Paper chromatography is performed using paper.
Answer:
The range of atoms = (30-300 pm) depending upon the element
Explanation:
The Atomic radii of the atom is the distance from the center of the circle to the outermost orbital.
The center of the circle is the nucleus and the radii is the outermost boundary.
The actual size of the atom is decided on the basis of the Zeff . Also known as <em>effective nuclear charge.</em>
<em>Zeff: It is the net positive charge felt by the outermost electron by the nucleus.</em>
<em>The value of Zeff depends upon the shielding constant. More the shielding less will be the Zeff . Hence the size of the atom increases.</em>
Due to shielding the outermost electrons feel less pull of nucleus.
<em>The greater the Zeff , the smaller the radius of the atom.</em>
The formula used to calculate the atomic mass is :
pm
Here "pm"= picometers

<u>The size of the smallest atom H-atom = 120 pm</u>
<u>The range of atoms = (30-300 pm)</u>
Answer:
25 mL
Explanation:
Step 1: Given data
- Concentration of the concentrated solution (C₁): 2 M
- Volume of the concentrated solution (V₁): ?
- Concentration of the diluted solution (C₂): 0.1 M
- Volume of the diluted solution (V₂): 0.500 L
Step 2: Calculate the volume of the concentrated NaCl solution
We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.1 M × 0.500 L / 2 M
V₁ = 0.025 L = 25 mL
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
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