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3 years ago

In a chromatography experiment, chlorophyll pigments are separated using paper. What is the stationary phase in this experiment?

2 answers:

4 0

Answer:
The stationary phase in chromatography experiment is paper.

Explanation:
In chromatography experiment, the stationary phase is defined as the fixed substance that is necessary to start chromatography. In our case, this fixed substance is paper, so that makes paper our stationary phase.

Hope this helps :)

xxTIMURxx [149]3 years ago

4 0

Answer:

Chromatography paper.

Explanation:

Chromatography may be defined as the technique used for the separation of molecules on the basis of their size, shape and color. The chromatography requires the mobile phase as well as the stationary phase.

The different pigments of chlorophyll can be separated by the chromatography technique. Here, the chromatography paper work as stationary phase and the drop of cholorophyll is placed on the paper. The water is used as mobile phase and as it moves over the paper separates the component of cholorophyll.

Thus, the answer is chromatography paper.

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Iron is biologically important in the transport of oxygen by red blood cells from the lungs to the various organs of the body. I

Aneli [31]

Answer : The number of iron atoms present in each red blood cell are, $1.077\times 10^9$

Explanation :

First we have to calculate the moles of iron.

$\text{Moles of iron}=\frac{\text{Mass of iron}}{\text{Molar mass of iron}}=\frac{2.90g}{55.85g/mole}=0.0519moles$

Now we have to calculate the number of iron atoms.

As, 1 mole of iron contains $6.022\times 10^{23}$ number of iron atoms

So, 0.0519 mole of iron contains $0.0519\times 6.022\times 10^{23}=3.125\times 10^{22}$ number of iron atoms

Now we have to calculate the number of iron atoms are present in each red blood cell.

Number of iron atoms are present in each red blood cell = $\frac{\text{Number of iron atoms}}{\text{Total number of red blood cells}}$

Number of iron atoms are present in each red blood cell = $\frac{3.125\times 10^{22}}{2.90\times 10^{13}}$

Number of iron atoms are present in each red blood cell = $1.077\times 10^9$

Therefore, the number of iron atoms present in each red blood cell are, $1.077\times 10^9$

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2 years ago

The number of protons in an atom's nucleus determines its

Nataly [62]

Number of proton present in the nucleus determines the atomic number of an element. It determines <span>chemical properties, which is why all atoms with proton count (atomic number) 6 are carbon</span>

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3 years ago

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PLEASE HELP ME ASAP What is the volume of 28.9 g of mercury if the density is 13.8 g/mL? *

strojnjashka [21]

Answer:

2.09 mL

Explanation:

density = mass/volume

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13.8v = 28.9

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2 years ago

A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10

lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

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2 years ago

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