What is the formal charge of carbon in CO?|C≡O|

Chemistry

1 answer:

Anastasy [175]3 years ago

7 0

Answer:

FC of C = -1

General Formulas and Concepts:

Lewis Dot Structures
Resonance Structures
Formal Charge: FC = # of valence e⁻ - # of lone e⁻ - # of bonds

Explanation:

<u>Step 1: Define</u>

Carbon (C)

4 valence e⁻
2 lone e⁻
3 bonds

Oxygen (O)

6 valence e⁻
2 lone e⁻
3 bonds

Substitute: FC = 4e⁻ - 2e⁻ - 3
Subtract: FC = -1

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Un tecnico di laboratorio deve preparare una soluzione di carbonato di sodio decaidrato, Na2CO3⋅ 10 H2O per eseguire alcune anal

fiasKO [112]

Answer:

1. 3.70 g Na₂CO₃·10H₂O

2. 50.0 mL of the first solution

Explanation:

1. Prepare the solution

(a) Calculate the molar mass of Na₂CO₃·10H₂O

$\begin{array}{rrr}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 2\times22.99 & 45.98\\\text{1C} & 1\times 12.01 & 12.01\\\text{13O}&13 \times16.00 & 208.00\\\text{20H}&20 \times 1.008 & 20.16\\&\text{TOTAL =} & \mathbf{286.15}\\\end{array}$

The molar mass of Na₂CO₃·10H₂O is 286.15 g/mol.

(b) Calculate the moles of Na₂CO₃·10H₂O

$\text{Moles of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\= \text{0.250 L solution} \times \dfrac{\text{0.0500 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 L solution}}\\\\= \text{0.0125 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

(c) Calculate the mass of Na₂CO₃·10H₂O

$\text{Mass of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O }\\= \text{0.012 50 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O } \times \dfrac{\text{296.15 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}\\\\= \text{3.70 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\\text{You need $\large \boxed{\textbf{3.70 g}}$ of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

2. Dilute the solution

We can use the dilution formula to calculate the volume needed.

V₁c₁ = V₂c₂

Data:

V₁ = ?; c₁ = 0.0500 mol·L⁻¹

V₂ = 100 mL; c₂ = 0.0250 mol·L⁻¹

Calculation:

$\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\V_{1}\times \text{0.0500 mol/L} & = & \text{100 mL} \times\text{0.0250 mol/L}\\0.0500V_{1}& = & \text{2.500 mL}\\V_{1}&=& \dfrac{\text{2.500 mL}}{0.0500}\\\\& = & \text{50.0 mL}\\\end{array}\\\text{You need $\large \boxed{\textbf{50.0 mL}}$ of the concentrated solution.}$