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JulsSmile [24]
3 years ago
8

A student did not read the directions to the experiment properly and mixed up where to place the NaOH solution and the vinegar.

He put the vinegar in the titrator and the measured amount of NaOH in the beaker. He then added a drop of the phenolphthalein to the solution in the beaker. Does the student need to empty out all of the solutions and start over again or can he go ahead and run the titration? If he runs the titration using the solutions as given above, what should he expect to see happen for results?
Chemistry
1 answer:
Keith_Richards [23]3 years ago
7 0

Answer:

The answer is "Only at the end, a transformation of rose to color is made ".

Explanation:

 In this student puts its vinegar in the titrator, and NaOH throughout the beaker, which implies phenolphthalein has also been poured into water.

Phenolphthalein does have a pH range of 8.3-10 (approx). It's indeed pink in the basic medium therefore, the formulation becomes pink throughout the color by adding phenolphthalein to NaOH.

It is beginning of vinegar was its beginning of neutralization of NaOH from the titrator through full neutralization, a single piece of vinegar is added to a solution as well as the rose solution is lost throughout the beaker.

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. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:

E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ

Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ


2. The wavelength of a photon with this energy would be:

Energy = hc/wavelength

wavelength = hc/energy

wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m

Converting to nanometers gives: 91.16 nm


3. Repeat the calculation in 1, but using n=5.


4. Repeat the calculation in 2 using the energy calculated in 3.

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Describe the properties of alkali metals. Based on their electronic arrangement, explain whether they exist alone in nature.
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Explanation:

Alkali metals are the elements in group 1 of the periodic table. They include Sodium, Lithium, Potassium e.t.c.

Due to the fact they have one atom in their outermost shell, they are very unstable because they easily react with other elements and are therefore don't exist alone in nature but combined with other elements for this same reason.

Since alkali metals don't easily attract other elements due to it's lone pair in the outer most shell, it can be said to have low electro negativity.

Also, they don't need energy to discharge their electrons since they are highly reactive due to their lone pair in the outermost shell and so we say they have low ionization energy.

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Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76
aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.  

The no. of moles can be determined by using the formula,  

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

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The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.  

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,  

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,  

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.  

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