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uysha [10]
2 years ago
6

Name the covalent compound: I8P

Chemistry
1 answer:
SOVA2 [1]2 years ago
6 0

There are certain rules to follow when naming covalent compounds. But first, let us look at the definition of Covalent Compounds.

<h3>What are Covalent Compounds?</h3>

When covalent bonds aid the creation of a molecule, in which the atoms have at least one similar pair of valence electrons, a covalent compound is said to have been formed.

A very common example is water (H₂O)

<h3>How are Covalent Compounds named?</h3>

To name a covalent compound, simply list the first element in the formula using the name of the element, then name the second element by adding the suffix "ide" to the stem of the second element's name.

If there is only one atom in the molecule of the first element, then no prefix should be added.

It is to be noted that if the second element in the compound is oxygen, then we should say:

  • monox<em>ide</em> instead of monoox<em>ide</em> and
  • triox<em>ide</em> instead of trox<em>ide</em>, all depending on how many atoms that are involved.

See the attached for the prefixes related to the various number of atoms in the compounds.

It is to be noted that the covalent compound to be named here is not stated hence the general answer.

Learn more about naming covalent compounds at:
brainly.com/question/9841865

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Answer: A reversible reaction is a reaction that takes place in back and front directions. If the reaction were to reach equilibrium, the rate of forward direction would be equal to that of the reverse reaction.

Explanation:

Reversible reactions :

These are the reaction in which reactants reacts to give product and products reacts to give reactants as a product in return.

        A+B\leftrightharpoons C+D

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3 0
3 years ago
1) The heat of combustion for the gases hydrogen, methane and ethane are −285.8, −890.4 and −1559.9 kJ/mol respectively at 298K.
Morgarella [4.7K]

Answer:

The enthalpy of the reaction is 64.9 kJ/mol.

Explanation:

H_2 + \frac{1}{2}O_2\rightarrow H_2O,\Delta H_1 =-285.8 kJ..[1]

CH_4 + 2O_2\rightarrow CO_2 + 2H_2O,\Delta H_2 =-890.4 kJ..[2]

C_2H_6 + \frac{7}{2}O_2\rightarrow 2CO_2 + 3H_2O,\Delta H_3= -1559.9 kJ..[3]

2CH_4(g)\rightarrow C_2H_6(g) + H_2(g),\Delta H_4=?..[4]

2 × [2] - [1]- [3] = [4]  (Using Hess's law)

\Delta H_4=2\times \Delta H_2 -\Delta H_1 -\Delta H_3

\Delta H_4=2\times (-890.4 kJ)-(-285.8 kJ) -(-1559.9 kJ)

\Delta H_4=64.9 kJ/mol

The enthalpy of the reaction is 64.9 kJ/mol.

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