One plus one hundred and thirty five

PolarNik [594]

Organic chemical compounds as recommended by the (IUPAC)

8 0

3 years ago

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

3 years ago

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

3 years ago

Why was Rutherford's Gold Foil Experiment so important?

Oksanka [162]

Answer:

Answer is 'B' I think

Explanation:

Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Through previous experiments of shooting alpha particles, Rutherford knew they had considerable mass and speed.

4 0

3 years ago

Write Lewis structures for the following molecules: (a) ICl, (b) PH3, (c) P4 (each P is bonded to three other P atoms), (d) H2S,

spin [16.1K]

Answer : The Lewis-dot structure for the following molecules are shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

Now we have to determine the Lewis-dot structure for the following molecules.

(a) The given molecule is, $ICl$

As we know that iodine and chlorine have '7' valence electrons.

Therefore, the total number of valence electrons in $ICl$ = 7 + 7 = 14

According to Lewis-dot structure, there are 2 number of bonding electrons and 12 number of non-bonding electrons.

(b) The given molecule is, $PH_3$

As we know that phosphorous has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $PH_3$ = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

(c) The given molecule is, $P_4$

As we know that phosphorous has '5' valence electrons.

Therefore, the total number of valence electrons in $P_4$ = 4(5) = 20

According to Lewis-dot structure, there are 6 number of bonding electrons and 14 number of non-bonding electrons.

(d) The given molecule is, $H_2S$

As we know that sulfur has '6' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $H_2S$ = 6 + 2(1) = 8

According to Lewis-dot structure, there are 4 number of bonding electrons and 4 number of non-bonding electrons.

(e) The given molecule is, $N_2H_4$

As we know that nitrogen has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $N_2H_4$ = 2(5) + 4(1) = 14

According to Lewis-dot structure, there are 10 number of bonding electrons and 4 number of non-bonding electrons.

(f) The given molecule is, $HClO_3$

As we know that chlorine has '7' valence electrons, oxygen has '6' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $HClO_3$ = 1 + 7 + 3(6) = 26

According to Lewis-dot structure, there are 12 number of bonding electrons and 14 number of non-bonding electrons.

(g) The given molecule is, $COBr_2$

As we know that bromine has '7' valence electrons, oxygen has '6' valence electrons and carbon has '4' valence electrons.

Therefore, the total number of valence electrons in $COBr_2$ = 4 + 6 + 2(7) = 24

According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.

3 0

3 years ago