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swat32
4 years ago
10

An exception to this trend occurs in Group 13, and gallium is smaller than aluminum. Justify this exception to the expected peri

odic trend by invoking the electron configurations of each element.
Chemistry
1 answer:
kumpel [21]4 years ago
7 0

Answer:

Reduction in effective nuclear charge due to the filling of 3d sub-shell

Explanation:

Although, according to the trend in the periodic table, the size of shell go on increasing but an anomaly in Group 13 s observed in case of Aluminium and Gallium. Gallium should have a bigger shell configuration according to the trend but it is small as compared due to the first filling of 3d sub-shell by electron as we can see from their electronic configurations as:

Aluminium = [Ne] 3s² 3p¹

Gallium = [Ar] 3d10 4s2 4p1

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A research team has just discovered a new element called Likhitium. Now they need to determine the average atomic mass in order
earnstyle [38]

The question is incomplete, the complete question is shown in the image attached to this answer.

Answer:

139.13

Explanation:

The average atomic mass of the element Likhitium is the sum of the relative abundance of all the isotopes of Likhitium.

We obtain the relative atomic mass of Likhitium as follows;

(44.7/100 * 138) + (52.3/100 * 139) + (0.5/100 * 140) + (2.5/100 * 141)

61.7 + 73.2 + 0.7 + 3.53 = 139.13

Hence the relative abundance of Likhitium is 139.13

6 0
3 years ago
Lab report calorimetry and specific heat <br> HELP PLZ
dolphi86 [110]

Explanation:

Specific heat capacity can be calculated using the following equation: q = mc∆T In the equation q represents the amount of heat energy gained or lost (in joules ), m is the mass of the substance (in grams), c is the specific heat capacity of the substance (in J/g °C), and ∆T is the temperature change of the substance 

8 0
3 years ago
Working on-board a research vessel somewhere at sea, you have (carefully) isolated 12.5 micrograms (12.5 ×10–6 g) of what you ho
german

Answer:

The value is Z  =  311.33 \ g/mol

Explanation:

From the question we are told that

The mass of saxitoxin is m  =  12.5 mg = 12.5 * 10^{-6} g

The volume of water is V  =  3.10  mL  =  3.10 *10^{-3} L

The osmotic pressure is P =  0.236 =  \frac{0.236}{760}  =  3.105 * 10^{-4} atm

The temperature is T  =  19^oC  =  19 + 273 =  292 \  K

Generally the osmotic pressure is mathematically represented as

P  =  C  *  T  * R

Here R is the gas constant with value

R =  0.0821 ( L .atm /mol. K)

and C is the concentration of saxitoxin

So

3.105 * 10^{-4}  =  C * 0.0821   *  292

C = 1.295 *10^{-5} mol/L

Generally the number of moles of saxitoxin is mathematically represented as

n = C  *  V

=> n = 1.295 *10^{-5}   *3.10 *10^{-3}

=> n = 4.015 *10^{-8} \  mol

Generally the molar mass of saxitoxin is mathematically represented as

Z  =  \frac{m}{n}

=> Z  =  \frac{12.5 * 10^{-6}}{ 4.015 *10^{-8}}

=> Z  =  311.33 \ g/mol

5 0
4 years ago
A closed, nonreactive system contains species Aand Bin vapor/liquid equilibrium. Species Ais a very light gas, essentially insol
musickatia [10]

Answer:

Remain unchanged.

Explanation:

The total number of moles of liquid remain unchanged as the some moles of species B are added to the system because specie B that is added in the liquid phase is again restored after addition. If the specie B did not restored after addition to the liquid phase so the total number of moles increases in the liquid phase so that's why we can say that the liquid phase remain unchanged.

5 0
3 years ago
The chemical formula for the initial sample is mnso4·h2o and the chemical formula for the final sample is mnso4. use their molec
sammy [17]
The initial sample has a molecular formula of MnSO₄·H₂O. This molecule is a hydrate as it has a unit of water within its structure for every molecule of MnSO₄. This sample is being dehydrated to remove the water to give.

MnSO₄·H₂O → MnSO₄ + H₂O

MnSO₄·H₂O has a molecular mass of 169.02 g/mol. While MnSO₄ has a molecular mass of 151 g/mol. Water has a molecular mass of 18.02 g/mol. We now can use the ratio of the mass of water to the mass of the initial sample to determine the percentage of each component by mass.

% water by mass:

18.02/169.02 x 100% = 10.7% Water by mass.

% MnO₄ by mass:

151/169.02 x 100% = 89.3% MnSO₄ by mass.

Water makes up 10.7% of the initial mass of MnSO₄·H₂O.
4 0
3 years ago
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