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3 years ago

7

What name should be used for the ionic compound LiI?

2 answers:

Gelneren [198K]3 years ago

8 0

<u>Answer: </u>The given compound is named as lithium iodide.

<u>Explanation:</u>

$LiI$ is an ionic compound because lithium is a metal and Iodine element is a non-metal.

We know that, a bond which is formed between a metal and a non-metal is always ionic in nature.

Ionic bond is formed when there is a complete transfer of electron from an electropositive element (a metal) to an electronegative element (non-metal).

The nomenclature of ionic compounds is given by:

Positive ion is always written first.
The negative ion is written next with a suffix added at the end of the negative ion. The suffix used is '-ide'.

Hence, the name given to $LiI$ will be lithium iodide.

timurjin [86]3 years ago

5 0

What name should be used for the ionic compound LiI?

The name that should be used for the ionic compound LiI is Lithium iodide. When naming ionic compounds, you have to take into consideration pairing of two components.

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Lewis dot structure for N3-

labwork [276]

Answer :

The steps involved in the electron dot structure of $N^{3-}$ are :

First we have to determine the total number of valence electron in $N^{3-}$ .

Number of valence electrons in N = 5

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The image is shown below.

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3 years ago

Read 2 more answers

Determine the percentage composition of Ptl2

Hitman42 [59]

Answer:

Percent Composition

1. Find the molar mass of all the elements in the compound in grams per mole.

2. Find the molecular mass of the entire compound.

3. Divide the component's molar mass by the entire molecular mass.

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1 year ago

How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide

Marat540 [252]

Answer:

$\boxed{\text{2.00 mol}}$

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r: 58.32

Mg(OH)₂ + … ⟶ … + 2HOH

m/g: 58.3

(a) Moles of Mg(OH)₂

$\text{Moles of Mg(OH)$_{2}$} =\text{58.3 g Mg(OH)$_{2}$} \times \dfrac{\text{1 mol Mg(OH)$_{2}$}}{\text{58.32 g Mg(OH)$_{2}$}}\\\\=\text{0.9997 mol Mg(OH)$_{2}$}$

(b) Moles of H₂O

The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

$\text{Moles of H$_{2}$O}= \text{0.9995 mol Mg(OH)$_{2}$} \times \dfrac{\text{2 mol {H$_{2}$O}}}{ \text{1 mol Mg(OH)$_{2}$}}\\\\= \textbf{2.00 mol H$_{2}$O}$

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3 years ago

Calculate the volume of a balloon that can hold 113.4 g of nitrogen dioxide, NO2 gas at STP-

Karolina [17]

Answer:

55.18 L

Explanation:

First we convert 113.4 g of NO₂ into moles, using its molar mass:

113.4 g ÷ 46 g/mol = 2.465 mol

Then we<u> use the PV=nRT formula</u>, where:

P = 1atm & T = 273K (This means STP)

n = 2.465 mol

R = 0.082 atm·L·mol⁻¹·K⁻¹

V = ?

Input the data:

1 atm * V = 2.465 mol * 0.082atm·L·mol⁻¹·K⁻¹ * 273 K

And <u>solve for V</u>:

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Diano4ka-milaya [45]

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3 years ago