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MatroZZZ [7]
3 years ago
11

Match the terms in the first list with the characteristics in the second list.1. adsorption chromatography2. partition chromatog

raphy3. ion-exchange chromatography4. molecular exclusion chromatography5. affinity chromatography
A. Ions in mobile phase are attracted to counterions covalently attached to stationary phase
B. Solute in mobile phase is attracted to specific groups covalently attached to stationary phase
C. Solute equilibrates between mobile phase and surface of stationary phase
D. Solute equilibrates between mobile phase and film of liquid attached to stationary phase
E. Different-sized solutes penetrate pores in stationary phase to different extents. Large solutes are eluted first.
Chemistry
1 answer:
wariber [46]3 years ago
5 0

Answer: Please See answers below

Explanation:

A. Ions in mobile phase are attracted to counterions covalently attached to stationary phase  -----Ion Exchange Chromatography

B. Solute in mobile phase is attracted to specific groups covalently attached to stationary phase -----Affinity Chromatography

C. Solute equilibrates between mobile phase and surface of stationary phase  -----Adsorption Chromatography

D. Solute equilibrates between mobile phase and film of liquid attached to stationary phase  --- Partition Chromatography

E. Different-sized solutes penetrate pores in stationary phase to different extents. Large solutes are eluted first. ---Molecular Exclusion Chromatography

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The first energy shell around a nucleus can hold _______ electrons. The second energy shell can hold ________.
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Hi there,

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4 0
2 years ago
By using the following reactions, calculate the heat of formation of pentane (C5H12) gas from carbon and hydrogen gas. C (s) + O
Llana [10]

Answer: ΔH = 145 kJ/mol


Explanation:


1) Equations given:


C (s) + O₂ (g) → CO₂ (g), ΔH = 395 kJ/mol


H₂ (g) + 1/2O₂ (g) → H₂O (l), ΔH = 285 kJ/mol


C₅H₁₂ (g) + 8O₂ → 5CO₂ (g) + 6H₂O (l), ΔH = 3.54 × 10³ kJ/mol


2) Target equation:


5C(s) +6 H₂(g) → C₅H₁₂(g)


3) Hess law


As per Hess law you can handle the chemical equations (add, subtract, reverse, multiply by a coefficient) to find the target equation, and then the net sum of the heat changes of the individual reactions equals the heat change of the target reaction.


4) Now, you must figure out how to transform the individual reactions to end up with the final equation:


Here is how:


i) Multiply the first one by 5:


5C (s) + 5O₂ (g) → 5CO₂ (g), ΔH = 5×395 kJ/mol


ii) Multiply the second by 6:


6H₂ (g) + 3O₂ (g) → 6H₂O (l), ΔH = 6×285 kJ/mol


iii) Reverse the third:


5CO₂ (g) + 6H₂O (l) → C₅H₁₂ (g) + 8O₂ , ΔH = - 3.54 × 10³ kJ/mol


iv) Add them up:


5C (s) + 5O₂ (g) → 5CO₂ (g), ΔH = 5×395 kJ/mol


6H₂ (g) + 3O₂ (g) → 6H₂O (l), ΔH = 6×285 kJ/mol


5CO₂ (g) + 6H₂O (l) → C₅H₁₂ (g) + 8O₂ , ΔH = - 3.54 × 10³ kJ/mol

----------------------------------------------------------------------------------------------

5C (s) + 6H₂ → C₅H₁₂ (g),


ΔH = 5×395 kJ/mol + 6×285 kJ/mol - 3.54 × 10³ kJ/mol


⇒ ΔH = 145 kJ/mol

6 0
4 years ago
Read 2 more answers
solution was prepared by dissolving 43.0 g of KCl in 225 g of water. Part A Calculate the mole fraction of the ionic species KCl
notka56 [123]

<u>Answer:</u> The mole fraction of KCl in the solution is 0.044

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For KCl:</u>

Given mass of KCl = 43.0 g

Molar mass of KCl = 74.5 g/mol

Putting values in equation 1, we get:

\text{Moles of KCl}=\frac{43g}{74.5g/mol}=0.58mol

  • <u>For water:</u>

Given mass of water = 225 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{225g}{18g/mol}=12.5mol

To calculate the mole fraction of a substance, we use the equation:

\chi_A=\frac{n_A}{n_A+n_B}

Moles of KCl = 0.58 moles

Moles of water = 12.5 moles

Putting values in above equation:

\chi_{KCl}=\frac{n_{KCl}}{n_{KCl}+n_{\text{water}}}

\chi_{KCl}=\frac{0.58}{0.58+12.5}\\\\\chi_{KCl}=0.044

Hence, the mole fraction of KCl in the solution is 0.044

6 0
4 years ago
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