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3 years ago

How many atoms are present in a sample of Potassium (K) weighing 33.49g?

Chemistry

1 answer:

madam [21]3 years ago

3 0

Answer:

5.158 × 10²³ atoms K

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

33.49 g K

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of K - 39.10 g/mol

Step 3: Convert

$33.49 \ g \ K(\frac{1 \ mol \ K}{39.10 \ g \ K} )(\frac{6.022 \cdot 10^{23} \ atoms \ K}{1 \ mol \ K} )$ = 5.15797 × 10²³ atoms K

Step 4: Check

We are given 4 sig figs. Follow sig figs and round.

5.15797 × 10²³ atoms K ≈ 5.158 × 10²³ atoms K

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Suppose a laboratory wants to identify an unknown pure substance. The valence electrons of the substance's atoms feel an effecti

zalisa [80]

Answer:

The answer is the third option in the list: It would have smaller atomic radii than Si and higher ionization energies than Si.

Explanation:

The effective nuclear charge is that portion of the total nuclear charge that a given electron in an atom feels.

Since, the inner electrons repel the outer electrons, the effective nuclear charge of a determined electron is the sum of the positive charge (number of protons or atomic number) that it feels from the nucleus less the number of electrons that are in the shells that are are closer to the nucleus than the own shell of such (determined) electron.

Mathematically, the effective nuclear charge (Zeff) is equal to the atomic number (Z) minus the amount (S) that other electrons in the atom shield the given (determined) atom from the nucleus.

Zeff = Z - S.

Since, the valence electrons are the electrons in the outermost shell of the atom, you can find certain trend for the value Zeff.

Let's look at the group to which Si belongs, which is the group 14. This table summarizes the relevant data:

Element Z Group # valence electrons S Zeff = Z - S

C 6 14 4 6 - 4 = 2 6 - 2 = +4

Si 14 14 4 14 - 4 = 10 14 - 10 = +4

Ge 32 14 4 32 - 4 = 28 32 -28 = +4

Sn 50 14 4 50 - 4 = 46 50 - 46 = +4

Pb 82 14 4 82 - 4 = 78 82 - 78 = +4

With that, you have shown that the valence electrons of the unknown substance's atoms feel an effective nuclear charge of +4 and you have a short list of 4 elements which can be the unknown element: C, Ge, Sn or Pb.

The second known characteristic of the unknown substance's atoms is that it has a higher electronegativity than silicon (Si).

So, you must use the known trend of the electronegativity in a group of the periodic table: the electronegativity decreases as you go down in a group. So, three of the elements (Ge, Sn, and Pb) have lower electronegativity than Si, which has left us with only one possibility: the element C. The valence electrons of carbon (C) atoms feel an effective nuclear charge of +4 and it carbon has a higher electronegativity than silicon.

Other two periodic trends attending the group number are the atomic radii and the ionization energy.

The atomic radii generally increases as you go from top to bottom in a group. This is because you are adding electrons to new higher main energy levels. So, you can conclude that the originally unknwon substance (carbon) has a smaller atomic radii, than Si.

The ionization energies generally decreases as you go from top to bottom in a group. This os due to the shielding effect: as seen, the effective nuclear charge of the atom's valence electrons remains constant, while the distance of the electrons from the nucleus increases (the valence electrons are farther away from the nucleus), which means the upper the element in a given group, the larger the ionization energy of the atoms.

With this, our conclusions about the unnkown substance are:

Since it has a higher electronegativity value than silicon (Si), it is right up of Si, and there is on only element possible element than can be (C).

Since, it is upper than silicon (Si), it would have smaller atomic radii.

Due to the shielding effect, it would have larger ionization energies.

The answer is the third option in the list: It would have smaller atomic radii than Si and higher ionization energies than Si.

6 0

2 years ago

When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

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2 years ago

For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

SVETLANKA909090 [29]

Answer:

$\frac{1}{24}$

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = $\frac{1}{6}$

So, fraction of the whole athletic field reserved for each fifth class = $\frac{1}{4}(\frac{1}{6})=\frac{1}{24}$

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3 years ago

Explain in words or using your diagrams how beryllium atoms would react with fluorine atoms.

Marysya12 [62]

Explanation:

To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;

Be = 2, 2

F = 2, 7

Beryllium is a metal with two valence electrons whereas fluorine is a halogen with seven valence electrons.

When Be loses two electrons it becomes isoelectronic with He;

Be → Be²⁺ + 2e⁻

Also, when fluorine gains an electron, it becomes isoelectronic with Ne;

F + e⁻ → F⁻

This loss and gain of electrons between the two elements creates an electrostatic attraction them and they enter into an electrovalent bond.

Hence;

Be²⁺ + 2F⁻ → BeF₂

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2 years ago

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LiRa [457]

Answer:

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Explanation:

i hope this will help you :)

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3 years ago