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3 years ago

The Lewis dot notation for two atoms is shown.

Chemistry

2 answers:

NISA [10]3 years ago

7 0

To help, I drew a diagram. This represents an ionic bond between Na and Cl. Na is giving his single electron to Cl, which is indicated by the arrow, to make Cl full with 8 electrons.

yaroslaw [1]3 years ago

3 0

Explanation:

Atomic number of sodium is 11 and its electronic distribution is 2, 8, 1. So, it means there is only one valence electron present in a sodium atom. This valence electron is represented by a dot in the Lewis structure.

Atomic number of chlorine is 17 and its electronic distribution is 2, 8, 7. So, it means there are seven valence electrons present in a chlorine atom. These valence electrons are represented by seven dots in the Lewis structure.

Also, in order to attain stability sodium will lose one valence electron and this valence electron will be gained by chlorine atom.

Therefore, an arrow pointed from dot near Na to Cl shows that an electron is being donated from Na to the Cl atom.

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1. empirical formula for Cu2o4?

Wittaler [7]

We know that

1) empirical formula of a compound is the representation of a molecules with the elements in simplest mole ratio.

2) the molar mass is the actual proportion of moles of elements present in a molecule

3) there is a simple ratio between the molar mass and empirical mass a molecule

Now let us solve each problem

1) empirical formula of $Cu_{2}O_{4}$ will be $CuO_{2}$

2) empirical formula of $Zn_{3}(NO_{3})_{4}$ will be $Zn_{3}(NO_{3})_{4}$

3) empirical formula of $Ti_{2}S_{2.5}$ will be $Ti_{4}S_{5}$

4) the molecular formula will be :

$Ratio=\frac{molarmass}{empiricalformula}=\frac{112.04}{28.01}=4$

the molecular formula = 4 X empirical formula = 4 X CO = $C_{4}O_{4}$

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3 years ago

Why are some rocks more easily weathered than other rocks

dmitriy555 [2]

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4 years ago

36. Write a balanced chemical equation for each chemical reaction. (a) Solid copper reacts with solid sulfur to form solid coppe

stiks02 [169]

Answer:

a) 2Cu(s) + S(s) → Cu₂S(s)

b) 2SO₂(g) + O₂(g) → 2SO₃(g)

c) 4HCl(aq) + MnO₂(s) → MnCl₂(aq) + 2H₂O(l) + Cl₂(g)

d) C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3H₂O(l)

Explanation:

First, we have to write the chemical formulas for reactants and products. Then, to identify which is the state of matter of each compound (liquid: l, solid: s, aqueous: aq, gas: g). Finally, we write the equation and balance the atoms.

a) Solid copper reacts with solid sulfur to form solid copper(I) sulfide.

Reactants: solid copper (Cu(s)) and solid sulfur (S(s)). Product: copper(I) sulfide (Cu₂S) because sulfur anion has two negative charges (S⁻²) and we need two copper ions with one positive charge (Cu⁺) to obtain the neutral compound.

The balanced requires a coefficient of 2 for Cu(s) to balance the 2 Cu atoms on the products side:

2Cu(s) + S(s) → Cu₂S(s)

(b) Sulfur dioxide gas reacts with oxygen gas to form sulfur trioxide gas.

Reactants: sulfur dioxide (SO₂) and oxygen gas (which is a diatomic molecule: O₂). Products: sulfur trioxide gas (SO₃(s)).

The balanced equation is:

2SO₂(g) + O₂(g) → 2SO₃(g)

We need a coefficient of 2 for SO₂ and SO₃ to balance O and S atoms.

(c) Aqueous hydrochloric acid reacts with solid manganese(IV) oxide to form aqueous manganese(II) chloride, liquid water, and chlorine gas.

Reactants: aqueous hydrochloric acid (HCl(aq)) and solid manganese (IV) oxide (MnO₂(s), because we need two oxygen anions O²⁻ to neutralize 4 positive charges in Mn IV). Products: aqueous manganese(II) chloride (MnCl₂, because we need two chloride ions Cl⁻ to neutralize two positive charges in Mn(II)), liquid water (H₂O), and chlorine gas (Cl₂, because it is a diatomic molecule).

The balanced chemical equation is:

4HCl(aq) + MnO₂(s) → MnCl₂(aq) + 2H₂O(l) + Cl₂(g)

(d) Liquid benzene (C₆H₆) reacts with gaseous oxygen to form carbon dioxide and liquid water.

Reactants: benzene (C₆H₆) and gaseous oxygen (O₂). Products: carbon dioxide (CO₂, because C has valence IV) and liquid water (H₂O).

The balanced chemical equation is:

C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3H₂O(l)

In this case, we need a coefficient of 15/2 for O₂ because on the product side there are 15 atoms of O (12 in CO₂ and 3 in H₂O).

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