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2 years ago

Limitations of paper chromatography

Chemistry

2 answers:

Ksju [112]2 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]2 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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Pentane is a small liquid hydrocarbon, between propane and gasoline in size at C5H12. The density of pentane is 0.626 g/mL and i

Goshia [24]

Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and $3.073\times 10^4kJ/L$ respectively.

Explanation :

Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).

As we are given that:

$\Delta H^o_{comb}=-3535kJ/mol$

Fuel value = $\frac{\Delta H^o_{comb}}{\text{Molar mass of pentane}}$

Molar mass of pentane = 72 g/mol

Fuel value = $\frac{3535kJ/mol}{72g/mol}$

Fuel value = 49.09 kJ/g

Now we have to calculate the fuel density of pentane.

Fuel density = Fuel value × Density

Fuel density = (49.09 kJ/g) × (0.626g/mL)

Fuel density = 30.73 kJ/mL = $3.073\times 10^4kJ/L$

Thus, the fuel density of pentane is $3.073\times 10^4kJ/L$

4 0

3 years ago

The reaction for photosynthesis producing glucose sugar and oxygen gas is:

Anvisha [2.4K]

1.5 grams of glucose is produced from 2.20 g of CO₂.

To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

CO₂: 6 moles
H₂O: 6 moles
C₆H₁₂O₆: 1 mole
O₂: 6 moles

So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:

$moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}$

moles of CO₂= 0.05 moles

Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

$moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}$

moles of C₆H₁₂O₆= 8.33*10⁻³

Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:

$mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}$

mass of glucose= 1.5 grams

Then, 1.5 grams of glucose is produced from 2.20 g of CO₂.

5 0

3 years ago

Three beakers contain clear, colorless liquids. one beaker contains pure water, another contains salt water, and another contain

tiny-mole [99]

First, we'll identify the beaker containing pure water as follows:
We'll take equal masses from each of the three beakers and measure the mass of each.
We'll then identify the density of each by using the rule : density =mass/volume
Pure water will be the liquid having density equal to 1 gm/cm^3

Then, we'll differentiate between the salt and sugar solution by measuring the conductivity of each solution. Salt solution is a good conductor while solution of sugar is a bad conductor.

8 0

3 years ago

A monatomic ideal gas expands slowly to twice its original volume, doing 340 JJ of work in the process. Part APart complete Find

Dafna1 [17]

Answer:

A) if the system is isothermal then all the heat added to the system will be used to do work (since none is used to raise the temperature of the gas). The heat added will be equal to the work done = 340 J

B) change in internal energy of the system of the process is isothermal will be zero, since there is no rise in temperature.

C) an adiabatic process is one involving no heat loss or gain through the system, Therefore heat gain will be zero

D) if the process is adiabatic then there is no heat loss or gain through the system and hence there is no change in temperature. Change in internal energy will be zero

E) if the process is isobaric then, there is no work done and the total heat to the system is equal zero

F) if there is no work done, and no heat added, then the internal energy will be equal zero.

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3 years ago

Explain heat transfer in your own words

timofeeve [1]

Answer: thermal conductivity

Explanation:

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3 years ago

Limitations of paper chromatography​

Limitations of paper chromatography