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3 years ago

What is the abbreviated electron configuration of iodine (I)?

Chemistry

1 answer:

castortr0y [4]3 years ago

5 0

Answer:

b) [Kr] 5s² 4d¹⁰ 5p⁵

Explanation:

Properties of iodine:

Iodine is present in group seventeen.

It is halogen element.

Its atomic mass is 127 amu.

Its atomic number is 53.

It is present in solid form.

It is crystalline in nature.

It is very corrosive and has pungent odor.

It can not react with oxygen and nitrogen.

Electronic configuration:

I₅₅ = 1s² 2s² 2p⁶ 3s³ 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁵

Abbreviated electronic configuration:

I₅₅ = [Kr] 5s² 4d¹⁰ 5p⁵

Abbreviated electronic configuration is shortest electronic configuration by using the noble gases.

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3 years ago

Using the van der Waals equation, calculate the pressure for a 1.25 mol sample of xenon contained in a volume of 1.000L at 75Â°C

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3 years ago

What is the difference between atomic mass, relative atomic mass and average atomic mass

vesna_86 [32]

$\bold{\huge{\underline{\purple{ Answer }}}}$

Difference between Atomic mass, relative atomic mass and average atomic mass :-

<h3>Atomic Mass :-</h3>

Atomic mass is the mass of neutrons and protons present in the nucleus of an atom .
It is always calculated for a single element and having direct value
For isotopes also, the atomic mass is calculated separately . Example :- Carbon 12 , carbon 13 and carbon 14 have different atomic mass. 
The SI unit of Atomic mass is " u" and "amu"

<h3>Relative Atomic mass :-</h3>

Relative atomic mass is mean mass of the atoms of an element which is compared to the 1/12th mass of carbon - 12 .
Carbon - 12 is taken as a relative when we calculate the relative atomic mass of any element
For calculating relative atomic mass, we need to know the masses, percentage and abundance of all types of elements
Relative atomic mass is a dimension less quantity

<h3>Average Atomic Mass :-</h3>

Average atomic mass is the average mass of an atoms of a particular element by considering it's isotopes
While we calculate average atomic mass is a standardized number. Whereas, Average atomic mass sometimes varies geologically .
It also includes percentage, abundance and masses of given element .
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2 years ago

An excess of Ba(No3)2 reacts with 250ml of H2SO4 solution to give 0.55g of BaSo4.determine The concentration in moles per litre

Wewaii [24]

Chemical reaction: Ba(NO₃)₂ + H₂SO₄ → BaSO₄ + 2HNO₃.
V(H₂SO₄) = 250 mL ÷ 1000 mL/L = 0,25 L.
m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.

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3 years ago

If 190dm of hydrogen gas collected at 20°c and 760mmHg .Calculate it's volume at stp (standard pressure=760mmHg

Andrei [34K]

Answer:

177.1 L

Explanation:

The excersise can be solved, by the Ideal Gases Law.

P . V = n . R . T

In first step we need to determine the moles of gas:

We convert T° from, C° to K → 20°C + 273 = 293K

We convert P from mmHg to atm → 760 mmHg = 1atm

1Dm³ = 1L → 190L

We replace: 190 L . 1 atm = n . 0.082 . 293K

(190L.atm) / 0.082 . 293K = 7.91 moles.

We replace equation at STP conditions (1 atm and 273K)

V = (n . R .T) / P

V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L

We can also make a rule of three:

At STP conditions 1 mol of gas occupies 22.4L

Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L

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3 years ago