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bekas [8.4K]
2 years ago
5

What are the difficulties encountered in precipitation titration

Chemistry
1 answer:
Mumz [18]2 years ago
8 0

One difficulty encountered in precipitation titration is that it is hard to determine the exact end point of its reaction.

Precipitation titration is a titration in which a  reaction  occurs from the  analyte and titrant  to form an insoluble precipitate.

With the use of silver for the titrations, (argentometric) we are able to develop many precipitation reactions.  

The  precipitation titrimetry  methods with the use of argentometry includes  

• Mohr’s Method

• Fajan’s Method

• Volhard’s Method

Difficulties encountered in precipitation titration includes

  • Getting the exact end point is hard.

  • it is a very slow titration method.

  • it includes periods of filtration and cooling thereby reducing the reactions available for this type of titration.

See more on Precipitation: brainly.com/question/20628792

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They are gases at room temperature

Hope this helps:)

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3 years ago
How many molecules are in 100 g of C6H120,?*​
Grace [21]

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

6 0
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After mitosis, you have 2 cells that are the same as the parent cell. Why is it that your first cell was not the same as your pa
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2 years ago
Identify the element from its location on the periodic table:<br><br> a. 4p^5
mafiozo [28]
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2 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
2 years ago
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