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Ede4ka [16]
3 years ago
12

Describe the range of em radiation

Chemistry
1 answer:
natima [27]3 years ago
3 0
The range for em radiation should be a range of 50 yards hope this help's. <span />
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Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the nu
Alja [10]

Answer:

(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹: Number of unpaired electron is 1 and belongs to group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8 of the periodic table.

Explanation:

In the Periodic table, the chemical elements are arranged in 7 rows, called periods and 18 columns, called groups. They are organized in increasing order of atomic numbers.

(a) [Noble gas] ns² np⁵

As the total number of electrons in the p-orbital is 5. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 5 electrons in np orbital. <u>So there are 7 valence electrons.</u>

Therefore, this element belongs to the group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²

As the total number of electrons in the d-orbital is 2. Therefore, the number of unpaired electrons is 2.

This element has 2 electrons in ns orbital and 2 electrons in (n-1)d orbital. So there are <u>4 valence electrons.</u>

Therefore, this element belongs to the group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹

As the total number of electrons in the p-orbital is 1. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 1 electron in np orbital. So there are <u>3 valence electrons</u>.

Therefore, this element belongs to the group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶

As the total number of electrons in the f-orbital is 6. Therefore, the number of unpaired electron is 6.

This element has 2 electrons in ns orbital and 6 electrons in (n-2)f orbital. So there are <u>8 valence electrons.</u>

Therefore, this element belongs to the group 8 of the periodic table.

3 0
3 years ago
The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his
QveST [7]

Explanation :

As we know that Mendeleev arranged the elements in horizontal rows and vertical columns of a table in order of their increasing relative atomic weights.

He placed the elements with similar nature in the same group.

According to the question, the atomic weight of iodine is less than the atomic weight of tellurium. So according to this, iodine should be placed before tellurium in Mendeleev's tables. But Mendeleev placed iodine after tellurium in his original periodic table.

However, iodine has similar chemical properties to chlorine and bromine. So, in order to make iodine queue up with chlorine and bromine in his periodic table, Mendeleev exchanged the positions of iodine and tellurium.

As we know that the positions of iodine and tellurium were reversed in Mendeleev's table because iodine has one naturally occurring isotope that is iodine-127  and tellurium isotopes are tellurium-128 and tellurium-130.

Due to high relative abundance of tellurium isotopes gives tellurium the greater relative atomic mass.

3 0
3 years ago
A large pot (10 kg) of beef stew is left out on the kitchen counter to cool. The rate of cooling, in kJ/min, equals 1.955 (T-25)
olga_2 [115]

Answer:

16 minutes

Explanation:

First, we need to calculate the amount of heat needed to cool the beef stew:

Q = mcΔT

Where <em>m</em> is the mass, <em>c</em> is the heat capacity and <em>ΔT</em> is the variation of the temperature.

Q = 10x4x(40 - 90)

Q = -2000 kJ

So, the beef stew needs to lost 2000 kJ to cool.

With the initial temperature at 90ºC, the rate of cooling(r) will be:

r = 1.955x(90 - 25)

r = 127.075 kJ/min

So, to lose 2000 kJ, will be necessary:

t = Q/r

t = 2000/127.075

t = 16 minutes

5 0
3 years ago
The solubility of potassium chloride is 34 g KCl
kenny6666 [7]

Answer: When 20 grams of potassium chlorate, KClO3, is dissolved in 100 grams of water at 80 ºC, the solution can be correctly described as:, unsaturated

At approximately what temperature does the solubility of sodium chloride, NaCl, match the solubility of potassium dichromate, K2Cr2O7?, 60 ºC

Explanation:

8 0
3 years ago
What is the amount of matter in an object.<br><br> A mass<br> B volume<br> C Density
Sever21 [200]
Mass mass mass mass mass mass
7 0
2 years ago
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