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Semenov [28]
2 years ago
10

The common constituent in all acid solutions is

Chemistry
1 answer:
Oksi-84 [34.3K]2 years ago
8 0

Answer:

H+/H3O , H2O

Explanation:

The ability to be a proton donor is the Bronsted-Lowry definition of acids. The Lewis definition of an acid is an electron pair acceptor, which covers molecules liKE BF3

The ability to accept a pair of electrons is what is common to all acids, not the ability to be a proton donor.

All acid solutions contain hydronium ions (H3O+), hydroxide ions (OH-) and water molecules. Each different acid solution will then have an anion that is exclusive to that acid. For example, hydrochloric acid solution will contain all of the above and chloride ions (Cl-).

All acids contain the acidic substance dissolved in water. Water naturally dissociates to a small amount, creating hydronium and hydroxide ions. But most of the water remains as water molecules.

Then when we add an acid, like HCl, the oxygen on the water attracts the hydrogen from the HCl. The electrons in the covalent bond remain with the chlorine, giving it a negative charge and thus it becomes the chloride ion (Cl-). The hydrogen now has a positive charge and as said before, is attracted to the water (specifically the lone pair of electrons on the oxygen) to create hydronium ions.

This creates extra hydronium ions, making the solution acidic. But remember, there are still water molecules, hydroxide ions and the negative ion all in solution for all acids.

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Which of the following phrases describes valence electrons?
kykrilka [37]

Answer:

C

Explanation:

because valence electrons are located at the last energy level

7 0
2 years ago
Add electron dots and charges as necessary to show the reaction of potassium and bromine to form an ionic compound
S_A_V [24]

Explanation: Electron dot structures are the lewis dot structures which represent the number of valence electrons around an atom in a molecule.

The electronic configuration of potassium is [Ar]4s^1

Valence electrons of potassium are 1.

The electronic configuration of Bromine is [Ar]4s^24p^5

Valence electrons of bromine are 7.

These two elements form ionic compound.

Ionic compound is defined as the compound which is formed from the complete transfer of electrons from one element to another element.

Here, one electron is released by potassium which is accepted by bromine element. In this process, Potassium becomes cation having +1 charge and Bromine become anion having (-1) charge.

The ionic equation follows:

K^++Br^-\rightarrow KBr

The electron dot structure is provided in the image below.

8 0
1 year ago
Read 2 more answers
After a reaction, a new compound contains 0.73 g Mg and 0.28 g N. What is the empirical formula of this compound?
FromTheMoon [43]

Answer:

Mg₃N₂

Explanation:

The empirical formula of a chemical compound is defined as the simplest positive integer ratio of atoms present in a compound. Using molecular mass of Mg (24,305g/mol) and mass of nitrogen (14,006g/mol), moles of each element are:

0,73g × (1mol / 24,305g) = 0,03 moles of Mg

0,28g × (1mol / 14,006g) = 0,02 moles of N

Dividing each value in 0,01 to obtain natural numbers:

0,03 moles of Mg / 0,01 = 3

0,02 moles of N / 0,01 = 2.

Thus, empirical formula is: <em>Mg₃N₂</em>

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I hope it helps!

5 0
1 year ago
Fill in the blanks to complete each statement about the heating of earths surface. Earth takes in thermal energy from the sun in
VLD [36.1K]

Answer: absorption and insulation

8 0
1 year ago
Read 2 more answers
Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying t
-Dominant- [34]

Answer:

The description including its given problem is outlined in the following section on the clarification.

Explanation:

The given values are:

RBCC = 0.12584 nm

RFCC = 0.12894 nm

The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:

√3 ABCC = 4RBCC

⇒  ABCC = \frac{4RBCC}{\sqrt{3} }

⇒             = \frac{4\times 0.12584}{\sqrt{3}}

⇒             = 0.29062 \ nm

Likewise AFCC as well as RFCC are interconnected by  

√2AFCC = 4RFCC

⇒  AFCC = \frac{4RFCC}{\sqrt{2}}

⇒             = \frac{4\times 0.12894}{\sqrt{2} }

⇒             = 0.36470 \ nm

Now,

The Change in Percent Volume,

= \frac{V \ final-V \ initial}{V \ initial}\times 100 \ percent

= \frac{(VFCC)unit \ cell-(VBCC)unit \ cell}{(VBCC)unit \ cell}\times 100 \ percent

= \frac{(aFCC)^3-(aBCC)^3}{(aBCC)^3}\times 100 \ percent

= \frac{(0.36470)^3-(0.29062)^3}{(0.29062)^3}\times 100 \ percent

= 97.62 \ percent (approximately)

Note: percent = %

7 0
1 year ago
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