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viktelen [127]
1 year ago
10

When do you use the roman numerals when naming ionic compounds

Chemistry
1 answer:
Ganezh [65]1 year ago
6 0
<h2>Answer:</h2>

Oxidation State

In chemistry, when naming ionic compounds there will sometimes be a roman numeral after the first element. This number represents the oxidation state of the element. An oxidation state is defined as the hypothetical charge of an atom, assuming that all of its bonds are fully ionic. In other words, the oxidation state shows how many electrons an atom gives or receives after ionization.

In ionic compounds, there will only ever be a roman numeral after the first element. This element will always be the cation (positively charged ion). So, the roman numeral will show how many electrons each cation gave.

When Roman Numerals are Necessary

This is necessary when you deal with transition metals or any other element that has multiple oxidation states. For example, iron can give 2 or 3 electrons. So, when naming a compound with iron you need to use a (II) or (III) to show how many electrons each iron atom gave.

When Roman Numerals are Unnecessary

On the other hand, roman numerals are unnecessary when using an element that only has one possible oxidation state. For example, Na can only give 1 electron ever, so it does not need a roman numeral. Additionally, elements in the second group like Mg can only ever give 2 electrons, so they also don't need a roman numeral to follow their name.

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Aqueous potassium sulfide and aqueous cobalt (II) chloride are mixed, and a double replacement reaction occurs. What is the corr
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<h3>What are Carbonates?</h3>

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brainly.com/question/704297

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2 years ago
The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
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Answer:

\large \boxed{\text{21.6 L}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

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

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}

(b) Moles of CO₂

\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}

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