According to the following electron structure, how many valence electrons do these fluorine have?
2 answers:
Answer : Fluorine has Seven valence electrons.
Explanation :
Element = Fluorine
Atomic number = 9
Number of electrons = 9
Electronic configuration =
Electron structure is also known as electronic configuration. It is defined as the arrangement of electrons of an atom in an atomic orbitals.
According to Electron configuration, the 2s and 2p are the outermost energy shell and the number of electrons in 2s and 2p are 2 & 5 respectively. That means Fluorine has '7' valence electrons.
The electron structure of fluorine is shown below.
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When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams
<h3>Stoichiometric calculation</h3>From the equation of the reaction:
The mole ratio of iron(III) oxide to produced iron is 1:2.
Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles
Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles
Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams
More on stoichiometric calculations can be found here; brainly.com/question/27287858
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Answer:
Explanation:
Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:
- an equilibrium constant is, first of all, a fraction;
- in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
- in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
- each concentration should be raised to the power of the coefficient in the balanced chemical equation;
- only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.
Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid: