Answer:
![\boxed{ \sf \: R_f \: value \: of \: sample \: 1 =0.3142}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Csf%20%5C%3A%20R_f%20%20%5C%3A%20value%20%5C%3A%20of%20%5C%3A%20sample%20%5C%3A%201%20%3D0.3142%7D)
<h3>
Explanation:</h3>
In Analytical Chemistry chromatography is widely used for the separation of samples.
- In thin layer chromatography, the mixture of components are separated on the basis of their polarity.
- The solvent solution(mobile phase) that we use are non polar & silica gel( TLC paper made of/stationary phase) are polar.
- Consider the mixture we have taken consist of two samples having large polar difference.
- Due to opposite nature of silica gel(polar) & solvent solution (non polar) the movement become easy & due to capillary action solvent solution rise to the top.
- The mixture of sample we have taken, the sample have less polarity have high peak or they travel more distance than that of more polar sample when they dipped into the solution.
In the given diagram, mixture of 8 samples are separated on the basis of their polarity, the distance travelled by solvent is 35 mm, distance travelled by sample 1 is 11 mm & similarly distance travelled by sample 2,3,4,5,6,7 are 15,31,4,22,25,33 in mm respectively.
Rf Value: Rf value is retention factor which tells about relative absorption of each sample & range of Rf value is 0-1.
Formula to calculate Rf value is
![\sf R_f \: value = \frac{distance \: moved \: by \: sample}{distance \: moved \: by \: solvent}](https://tex.z-dn.net/?f=%20%5Csf%20R_f%20%20%5C%3A%20value%20%3D%20%5Cfrac%7Bdistance%20%5C%3A%20moved%20%5C%3A%20by%20%5C%3A%20sample%7D%7Bdistance%20%5C%3A%20moved%20%5C%3A%20by%20%5C%3A%20solvent%7D%20)
Now, solving for Rf value of sample 1
<em>Given:</em>
Distance moved by sample 1 = 11 mm
Distance movedby solvent = 35 mm
<em>To find:</em>
Rf value of sample 1 = ?
<em>Solution:</em>
Substituting the given data in above formula,
![\small \sf R_f \: value = \frac{distance \: moved \: by \: sample \: 1}{distance \: moved \: by \: solvent} \\ \small \sf R_f \: value = \cancel\frac{11 \: mm}{35 \: mm} = 0.3142](https://tex.z-dn.net/?f=%20%5Csmall%20%5Csf%20R_f%20%20%5C%3A%20value%20%3D%20%5Cfrac%7Bdistance%20%5C%3A%20moved%20%5C%3A%20by%20%5C%3A%20sample%20%5C%3A%201%7D%7Bdistance%20%5C%3A%20moved%20%5C%3A%20by%20%5C%3A%20solvent%7D%20%20%20%5C%5C%20%20%5Csmall%20%5Csf%20R_f%20%20%5C%3A%20value%20%3D%20%20%5Ccancel%5Cfrac%7B11%20%20%5C%3A%20mm%7D%7B35%20%5C%3A%20%20mm%7D%20%20%3D%200.3142)
![\small \boxed{ \sf \: R_f \: value \: of \: sample \: 1 =0.3142}](https://tex.z-dn.net/?f=%20%5Csmall%20%5Cboxed%7B%20%5Csf%20%5C%3A%20R_f%20%20%5C%3A%20value%20%5C%3A%20of%20%5C%3A%20sample%20%5C%3A%201%20%3D0.3142%7D)
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Answer:
The pH of the solution is 3.0
Explanation:
Considering:
Or,
Given :
For
:
Molarity = 0.0030 M
Volume = 100 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 100×10⁻³ L
Thus, moles of
:
Moles of
= 0.0003 moles
Volume added = 200 mL
Total volume = 200 + 100 mL = 300 mL = 0.3 L
Thus,
SInce, it is strong acid. Thus, [H⁺] = 0.001 M
pH is defined as the negative logarithm of the concentration of hydrogen ions.
Thus,
pH = - log [H⁺] = - log 0.001 = 3.0
<u>The pH of the solution is 3.0</u>
Answer is: solubility of silver chromate is 2.2 × 10-2 g/l.
Chemical reaction (dissociation) of silver chromate
in water: <span>
Ag</span>₂CrO₄(s) → 2Ag⁺(aq) + CrO₄²⁻(aq).<span>
Ksp(Ag</span>₂CrO₄) = [Ag⁺]²·[CrO₄²⁻].<span>
[CrO</span>₄²⁻] = x.<span>
[Ag</span>⁺] =
2[CrO₄²⁻] = 2x<span>
1,1</span>·10⁻¹² = (2x)² · x = 4x³.
x = ∛1,1·10⁻¹² ÷ 4.
x = 6,5·10⁻⁵ M.
solubility of silver chromate: 6,5·10⁻⁵ mol/L · 331,73 g/mol = 0,022 g/L.
<span>
Ksp is the solubility product constant for
a solid substance dissolving in an aqueous solution.
[Ag</span>⁺]
is equilibrium concentration of silver cations.<span>
[CrO</span>₄²⁻] is equilibrium concentration of chromate
anions.
Explanation:
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period.
But when we move from top to bottom in a group then there occurs an increase in size of the atoms. Hence, ionization energy decreases along a group.
(a) As Sb, Sn and I are all period 5 elements. Hence, these elements are arranged in order of increasing
as follows.
Sn < Sb < I
(b) As Sr, Ca, and Ba are all elements of group 2a. Hence, these elements are arranged in order of increasing
as follows.
Ba < Sr < Ca