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professor190 [17]
3 years ago
14

Calculate the empirical formula of a compound that is 56.27% phosphorus and the remainder is oxygen.

Chemistry
1 answer:
Inessa [10]3 years ago
5 0

Answer:

ikjnfjngvfnjfdkmldskmofevivrtbhurt3gdkjbfongidfngherghihierihgdifngodfnvgndfugiripegndbjgindgndfvcl vncf 'g

Explanation:

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What is the total number of orbitals found in the second energy level? openstudy?
Setler79 [48]
On the second shell there are two individual subshells:
The "s" subshell has only 1 orbital with max. two electrons spinning around; and the so-called "p" subshell has 3 orbitals with max. 6 electrons (2 on each!)
In total, there are four orbitals with 8 revolving electrons on the second shell.
Hope could help :)
3 0
3 years ago
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I know how to do electron configuration, but I think I’m doing the rest wrong. Answers and explanations would be much appreciate
Natasha_Volkova [10]

Your answers seem great so far, except for a tiny issue: With the ionic symbols, try to get into the habit of using "+", with metals, like sodium, and try to use the integer first. So, for example, a potassium ion would be K^+, while an oxide ion would be O^2-


Let's take aluminium as an example I'll work through:

Aluminium, with it's atomic number of 13, will have an electronic configuration of 1s2 2s2 2p6 1s2 2p1. So it would have 2, 8, 3 electrons in the first three energy levels, respectively.

Usually, if an elemental atom has a valence electron (highest energy level electron) count less than 4, it almost always will lose electrons. Since aluminium has 3, it will also lose the electrons.

It loses the 3 valence electrons, and so will end up with 10 electrons.

Since the atomic number also tells how many protons it has, we know that an aluminium atom has 13 protons, which doesn't change.

Since the size of the charges of a proton and an electron are the same, with protons being positive and the electrons being negative, an aluminium ion would have a charge of +3, and the Ionic symbol would be Al^3+



Hope I helped! xx


4 0
3 years ago
Which product is used on the natural nail prior to product application to assist in adhesion and serves to chemically bond the e
ira [324]

The product that is used on the natural nail prior to application to assist in adhesion and serves to chemically bond the enhancement product to the natural nail is known as nail primer.

<h3>What is a nail primer?</h3>

A nail primer is a chemical agent used in esthetic centers before applying a colored polish to the nails and serves as an adhesive product.

The nail primers are also very useful for improving the cleaning efficiency of the product before its application.

Nail care products include different types of chemical formulations such as, for example, creams that reinvigorate the cuticle.

In conclusion, the chemical formulation employed on the natural nail that is capable of enhancing and also assisting adhesion is called the nail primer.

Learn more about nail esthetic products here:

brainly.com/question/14498053

#SPJ1

8 0
2 years ago
What would a substance that could undergo condensation polymerization most likely have? two double bonds a triple bond two aldeh
blagie [28]
The fourth option on Edgen, "two alcohol functional groups". You're welcome :)
5 0
3 years ago
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"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"
Salsk061 [2.6K]

Answer:- The Ka for the acid is 1.53*10^-^5 .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

Ka = [H^+][A^-]\frac{1}{HA}

Where, Ka is the acid ionization constant. Let's plug in the values.

Ka = \frac{X^2}{0.25-X}

Let's calculate the value of X first using the equation:

pH = -log[H^+][/tex]

on taking antilog ob above equation we get:

[H^+]=10^-^p^H

[H^+]=10^-^2^.^7^1

[H^+] = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

Ka=\frac{(0.00195)^2}{0.25-0.00195}

Ka=1.53*10^-^5

So, the value of Ka for butyric acid is 1.53*10^-^5 .

8 0
3 years ago
Read 2 more answers
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