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anyanavicka [17]
3 years ago
11

Groups are an important method of classifying because

Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
4 0
<h3>Answer:</h3>

                Option-D, "their elements have very similar properties and exhibit a clear trend" is the correct answer.

<h3>Explanation:</h3>

                         In periodic table the elements are arranged in tabular form with respect to their atomic masses, atomic numbers, electronic configurations and chemical properties. It is called periodic because the properties of elements repeats periodically.

                         Furthermore, the rows from left to right also called Periods basically shows the metallic and non metallic characters of elements. \those on the left are metals and those at the rights are non metals.

                        The columns from top to bottom are called as Groups. Groups contain elements of same chemical properties.

<h3>Example:</h3>

Group I elements are called as <em>Alkali Metals</em>. They all have one valence electron and hence, loose one electron to form the corresponding cation. i.e.

                                           M    →    M⁺¹  +  1 e⁻

Group VII elements are called <em>Halogens</em>, they all are non metals and tend to gain one electron to form the corresponding halide ion. i.e.

                                           X  +  1 e⁻    →    X⁻¹

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Using the word bank explain why increasing the temperature increases the rate of reaction​
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How many atoms of phosphorus are in 2.50 million of copper (II) phosphate
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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
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