Would 1 pound of peanut butter occupy more or less space than 1 pound of water?

Is there a way to tell whether or not to put two double bonds or not. For example SeS2 seems to work if it has two double bonds

Nina [5.8K]

Your first step should be to analyse the compound. For example, if the compound is carbon, you know it always has a valence of four, so, if it has a formula C2H4 (ethylene) it obviously has a double bond. There are difficulties here because benzene C6H6 can be considered to have 6 1.5 C-C bonds, being aromatic.

A second step is to look at its structure. Double bonds are traditionally shorter than single bonds; triple bonds shorter still. Covalent bonds do have typical lengths, nevertheless you can still have problems.

<span>A third step is to consider reactivity. For example, if you have a C=C double bond, you can add, say, bromine to it Thus C2H4 gives C2H4Br2, and by adding two bromine atoms you know you have one double bond. Again, benzene becomes an awkward molecule, but because of this, you know benzene does not have double bonds in the traditional sense</span>

8 0

3 years ago

For each compound below, identify any polar covalent bonds and indicate the direction of the dipole moment using the symbols 84

Lostsunrise [7]

Answer:

H+ ----- Br-

H+ ----- Cl-

O₋₋ -----2H++

CH3 ----- O- ------ H+

Explanation:

Dipole moment occurs when there is bonding between a very strong electronegative element and hydrogen atom.

Electronegative elements are the element which attract electrons towards themselves, (that is they have strong affinity for electrons).

Generally, group 7 elements (Fluorine, Chlorine, Bromine, Iodine) of the periodic table are highly electronegative

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3 0

3 years ago

What is the concentration of OH− in a solution with a pOH value of 6.09

djyliett [7]

Answer:

I'm not sure

Explanation:

I'm just not sure

5 0

2 years ago

How are the helium atoms in a glass model different from real atoms

djyliett [7]

The helium in a glass model is a gas

5 0

3 years ago

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, $\boxed{CH_4 + O_2 --> CO_2 + H_2O}$ .

Because the problem asks for the unbalanced equation, we do not need to take any further steps of balancing the equation.

5 0

3 years ago

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