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Paraphin [41]
3 years ago
11

Will Rb and I form an iconic or covalent bond?

Chemistry
1 answer:
sergeinik [125]3 years ago
8 0

Answer:

Most compounds of Rubidium are formed by ionic bonds.

Explanation:

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Use the chart to determine which type of bond is formed between potassium (K) and chlorine (Cl).
iris [78.8K]
Potassium and Chloride forms an ionic bond.
(K+) + (Cl-) = KCl

Potassium is under Group IA (Alkali Metal), wherein elements under this group can easily lose electrons.

Chlorine is under Group VII (Halogens), in which these elements can gain electrons easily.

The inner shell electrons on potassium will merge with the outer shell of electrons of chlorine to make potassium chloride.
6 0
4 years ago
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castortr0y [4]

Answer:

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7 0
3 years ago
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In _, Solon was chosen as an Athenian statesman with reformation powers.
NeX [460]
594 B.C.E.

In 594 B.C.E., Solon was chosen as an Athenian statesman with reformation powers.
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4 years ago
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Which of the following has the greatest mass, in grams?
Elina [12.6K]
Mole represents the a huge Avogadro number (way larger than  atom), so we can eliminate option A and D.

And we know that Mercury is a solid , which mean it has greater mass than chlorine

So the answer is  : B. 1.0 mol mercury (Hg) Atom


7 0
3 years ago
90 POINTSSSS!!!
Katarina [22]

Answer:

\large \boxed{\text{-2043.96 kJ/mol}}

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn:    C₃H₈(g) +   O₂(g) ⟶   CO₂(g) +   H₂O(g)

Balanced eqn:    C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}

(b)Total enthalpies of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\=  \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}

ΔᵣH° is negative, so the reaction is exothermic.

5 0
3 years ago
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