All categories

3 years ago

3. Ions in the same group tend to form ions with the same

Chemistry

1 answer:

kolezko [41]3 years ago

4 0

Answer:

Same charge.

Explanation:

Ions in the same group tend to form ions with the same charge. Elements belong to the same group on the periodic table form ions with the same charge because they have the same number of valence electrons. For example, all members of alkali metals have different number of shells but same number of electrons in their outermost shell so due to same number of outermost shell these alkali metals will make bonding with halogens which is a seven group that have seven electrons and needs one electron for attaining stability and forming ions on both atoms.

You might be interested in

What is the atomic mass of<br> an atom?

Alborosie

Answer:

1 unit

Explanation:

It is expressed as a multiple of one-twelfth the mass of the carbon-12 atom, 1.992646547 × 10−23 gram, which is assigned an atomic mass of 12 units.

1/12 * 12 = 1. I think so.

4 0

3 years ago

Classify methanal according to the position of the c=o group

seraphim [82]

Answer : Methanal also known as Formaldehyde $CH_{2}O$ is a chemical Aldehyde which contain ( -CHO) group.

Explanation :

In organic chemistry, a carbonyl group is a functional group which contain a carbon atom double-bonded to an oxygen atom i.e, ( C=O).

If carbonyl group is present in a compound then it can be a carboxylic (RCOOH), aldehyde (RCHO), ketone (RCOR'), ester ((RCOOR') or amide (RCONR'R") group.

Here are some functional groups naming according to the<em> IUPAC</em> rules and image also attached,

Carboxylic acid → (RCOOH) → ( name end in 'OIC ACID' )

Aldehyde → (RCOH) → ( name end in 'AL' )

Ketone → (RCOR') → ( name end in 'ONE' )

Ester → (RCOOR') → ( name end in 'ATE' )

Amide → (RCONR'R") → ( name end in 'AMIDE' )

In an aldehyde, atleast one hydrogen atom must be attached to the carbonyl carbon. For an aldehyde, remove ( -e) from alkane name and add ( -al) at the end of the compound.

Methanal is the IUPAC name for Formaldehyde.

3 0

3 years ago

What is the pH of a 0.0042 M hydrochloric acid solution?

babymother [125]

Answer:

E) 2.38

Explanation:

The pH of any solution , helps to determine the acidic strength of the solution ,

i.e. ,

Lower the value of pH , higher is its acidic strength

and ,

Higher the value of pH , lower is its acidic strength .

pH is given as the negative log of the concentration of H⁺ ions ,

hence ,

pH = - log H⁺

From the question ,

the concentration of the solution is 0.0042 M , and being it a strong acid , dissociates completely to its respective ions ,

Therefore , the concentration of H⁺ = 0.0042 M .

Hence , using the above equation , the value of pH can be calculated as follows -

pH = - log H⁺

pH = - log ( 0.0042 M )

pH = 2.38 .

4 0

4 years ago

1. Name the compound present in a sample of dry air.

aev [14]

Answer:

not air

Explanation:

I think so

hahaha

7 0

2 years ago

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell = 0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0

3 years ago

3. Ions in the same group tend to form ions with the same​

3. Ions in the same group tend to form ions with the same