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3 years ago

Major classification of crude product

Chemistry

2 answers:

Mnenie [13.5K]3 years ago

7 0

pls mark as brainlist

Explanation:

Crude is classified, based on density, as light, medium, heavy, or extra heavy. It can also be classified, based on sulfur content, into a sour and sweet category.

MakcuM [25]3 years ago

3 0

<h2>Answer::</h2>

Crude Oil are:: Very light oils – these include:: Jet Fuel, Gasoline, Kerosene, Petroleum Ether, Petroleum Spirit, and Petroleum Naphtha. Light oils – These include Grade 1 and Grade 2 Fuel Oils, Diesel Fuel Oils as well as Most Domestic Fuel Oils.','.

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WHAT IS THE UNBALCNED EQUATION OF Methane burns in oxygen to produce carbon dioxide and water

deff fn [24]

Answer:

$\boxed{CH_{4}+O_2 --> CO_2 + H_2O}$

Explanation:

<h2>Part 1: Naming compound formulas given the names</h2>

<u>Step 1.</u> Methane's formula is $CH_4$ .

<u>Step 2:</u> Oxygen is a diatomic molecule (it exists bonded to itself for stability purposes), so by itself in chemical equations, it is written as $O_2$ .

<u>Step 3:</u> Carbon dioxide is the molecular compound of one atom of carbon and two atoms of oxygen → $CO_2$ .

<u>Step 4:</u> Water is the common name of the compound of two hydrogen atoms and one oxygen atom → $H_2O$ .

<h2>Part 2: Writing the skeleton equation</h2>

<u>Step 1:</u><em> </em>Use the determined formulas for the reactants and plug them into the equation. We are told that methane burns in oxygen -- hinting at a combustion reaction. Therefore, we may infer that these are the reactants that yield the products.

Skeleton equations are written with the reactant(s) on the left -- if there are several, they are separated by an addition symbol (+).

With this information, we may begin our equation: $CH_4 + O_2$ , where $CH_4$ is methane and $O_2$ is the diatomic molecule of oxygen.

<u>Step 2:</u> Use the determined formulas for the products and plug them into the equation. We are told that the methane burns in oxygen to produce carbon dioxide and water. Hence, we can separate these two as we did with the reactants.

Now, our products side of the reaction will look like this: $--> CO_2 + H_2O$ , where $CO_2$ is carbon dioxide and $H_2O$ is water.

<u>Step 3:</u> Write the final equation. All you must do after determining both sides of the equation is simply push them together. Place the reactant side of the equation on the left and the product side of the equation on the right.

This gives us our final equation, $\boxed{CH_4 + O_2 --> CO_2 + H_2O}$ .