When a substance dissociates into OH- ions, they are classified as _______

An ionic bond occurs between what particles?

atroni [7]

Answer:

A positive and negative ion.

Explanation:

Ionic bond:

It is the bond which is formed by the transfer of electron from one atom to the atom of another element.

Both bonded atoms have very large electronegativity difference.

The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.

For example:

Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion. Sodium have one valance electron while chlorine have 7 valance electrons. In order to complete the octet chlorine require one electron while sodium need to lose its one electrons. That's why when both atom combine sodium lose its electron and becomes positive ion i.e cation while chlorine accept its electron and becomes negative ion called anion and bond between them is ionic bond.

4 0

3 years ago

Read 2 more answers

PLS HELP QUICK ALOTTT OF POINTS

timofeeve [1]

Answer:

$\boxed {\boxed {\sf 0.80 \ mol\ F}}$

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

$\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}$

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

$4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}$

Flip the ratio so the units of atoms of fluorine cancel each other out.

$4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}$

$4.8 \times 10^{23} *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }$

Condense into 1 fraction.

$\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F$

Divide.

$0.7970773829 \ mol \ F$

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

$0.80 \ mol \ F$

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>

6 0

3 years ago

What type of reaction is shown below? <br>a) Addition reaction <br>b) Esterification

Gemiola [76]

Answer:

a) Addition reaction, is your answer

6 0

3 years ago

Explain why a nuclear chain reaction is self-sustaining.

skelet666 [1.2K]

answer is written in word file attached below

Download docx

6 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

When a substance dissociates into OH- ions, they are classified as ________ ?