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lubasha [3.4K]
2 years ago
9

what is the molarity of liters of an aqueous solution that contains 0.50 mole of potassium iodide, KI

Chemistry
1 answer:
Marizza181 [45]2 years ago
5 0

Answer:

The molarity of 2.0 liters of an aqueous solution that contains 0.50 mol of potassium iodide is 0.25M.HOW TO CALCULATE MOLARITY:The molarity of a solution can be calculated by dividing the number of moles by its volume. That is;Molarity = no. of moles ÷ volumeAccording to this question, 2.0 liters of an aqueous solution that contains 0.50 mol of potassium iodide. The molarity is calculated as follows:Molarity = 0.50mol ÷ 2LMolarity = 0.25MTherefore, the molarity of 2.0 liters of an aqueous solution that contains 0.50 mol of potassium iodide is 0.25M.Learn more about molarity at: brainly.com/question/2817451

Explanation:

Mark me brainliest please!!! I spent a lot of time on this!!

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Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the
NikAS [45]

The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

Hf, Am, In, Ta, As, Se, Rn

<u>Answer:</u>

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

<u>Explanation:</u>

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is ns^{1-2}

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is np^{1-6}

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is [(n-1)d^{1-10}ns^{0-2}]

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is [(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]. They are also known as lanthanide and actinide series.

For the given elements:

  • <u>Option 1:</u> Hf

Hafnium is the 72nd element of the periodic table having electronic configuration of [Xe]4f^{14}5d^26s^2

As, the last electron is entering the d subshell, it is a transition element.

  • <u>Option 2:</u> Am

Americium is the 95th element of the periodic table having electronic configuration of [Rn]5f^{7}6d^07s^2

As, the last electron is entering the (f subshell), it is a inner transition element.

  • <u>Option 3:</u> In

Indium is the 49th element of the periodic table having electronic configuration of [Kr]5s^25p^1

As, the last electron is entering the p subshell, it is a main group element.

  • <u>Option 4:</u> Ta

Tantalum is the 73rd element of the periodic table having electronic configuration of [Xe]4f^{14}5d^56s^2

As, the last electron is entering the d subshell, it is a transition element.

  • <u>Option 5:</u> As

Arsenic is the 33rd element of the periodic table having electronic configuration of [Ar]4s^24p^3

As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

  • <u>Option 6:</u> Se

Selenium is the 34th element of the periodic table having electronic configuration of [Ar]4s^24p^4

As, the last electron is entering the p subshell, it is a main group element.

  • <u>Option 7:</u> Rn

Radon is the 86th element of the periodic table having electronic configuration of [Xe]4f^{14}5d^{10}6s^26p^6

As, the last electron is entering the p subshell, it is a main group element.

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\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here's the balanced equation for given Double displacement reaction ~

\sf Pb(NO_3)_2 +2 \:  KI = PbI_2  +2 \:  KNO_3

The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )

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