Why do the copper square, penny, and the statue of Liberty all soluble in water?

Chemistry

1 answer:

Romashka-Z-Leto [24]2 years ago

5 0

Answer:

<em>It’s because of the green substance on the copper square, pennies and the Statue of Liberty giving those objects the ability to all be soluble in water. </em>

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1 . a chemical bond formed by the electrostatic attraction between ions covalent bond

liberstina [14]

Explanation:

When there occurs sharing of electrons between two chemically combining atoms then it forms a covalent bond. Generally, a covalent bond is formed between two non-metals.

An ionic bond is defined as the bond formed due to transfer of one or more number of electrons from one atom to another. An ionic bond is always formed between a metal and a non-metal.

Every atom of an element will have orbitals in which electrons are found. These orbitals are known as energy level.

A molecule is defined as the smallest particle present in a substance or atom.

A metallic bond is formed due to mobile valence electrons shared by positive nuclei in a metallic crystal.

Thus, we can conclude that given statements are correctly matched as follows.

1). a chemical bond formed by the electrostatic attraction between ions - ionic bond

2). a chemical bond formed by two electrons that are shared between two atoms - covalent bond

3). the orbitals of an atom where electrons are found - energy level

4). the smallest particle of a covalently bonded substance - molecule

5). a bond characteristic of metals in which mobile valence electrons are shared among positive nuclei in the metallic crystal - metallic bond

7 0

3 years ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$