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Tom [10]
3 years ago
11

Limitations of paper chromatography​

Chemistry
2 answers:
Ksju [112]3 years ago
5 0

Answer:

Limitations of paper chromatography​ is described below in complete details.

Explanation:

Limitations of Paper Chromatography

  • A huge number of representations cannot be employed in paper chromatography.
  • In a quantitative investigation, paper chromatography is not sufficient.
  • The complex compound cannot be separated by paper chromatography.
  • Less Accurate correlated to HPLC or HPLC.

castortr0y [4]3 years ago
4 0

Answer:

Limitations of Paper Chromatography.

1.Large quantity of sample cannot be applied on paper chromatography.

2.In quantitative analysis paper chromatography is not effective.

3.Complex mixture cannot be separated by paper chromatography.

4.Less Accurate compared to HPLC or HPTLC.

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Why do some transition metals have multiple oxidation states? Select the correct answer from each drop down menu
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Answer:

Transition metals can have multiple oxidation states because of their electrons. The transition metals have several electrons with similar energies, so one or all of them can be removed, depending the circumstances. This results in different oxidation states. please mark me as brainellist

4 0
3 years ago
What are ionic compounds ​
musickatia [10]
Ionic compound is a chemical compound composed of ions held together by electrostatic forces termed ionic bonding
5 0
3 years ago
What is the mass of 3.0 x x times 10^23 atoms of neon?
Sladkaya [172]

Answer:

10.09 grams

Explanation:

First you need to know the number of moles you are dealing with.

If you know that each mole has 6.022x10²³ of something (in this case of atoms), you can divide 3x10²³ atoms of neons by 6.022x10²³ to obtain the number of moles.

You have 0.5 moles of Neon, so then by the periodic table, you see that the molar mass of neon is 20.18g/mol, so by each mole you have 20.18 grams of neon. Multiply 20.18 grams by 0.5 moles and you got 10.09 grams of Neon

7 0
3 years ago
1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

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