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3 years ago

To which third period element do these ionization values belong?

Chemistry

2 answers:

Arlecino [84]3 years ago

7 0

Answer:

Aluminum, Al

Explanation:

aleksklad [387]3 years ago

4 0

IE1 = 578 kJ/mol

IE2 = 1820 kJ/mol

IE3 = 2750 kJ/mol

IE4 = 11600 kJ/mol

If the following set of successive ionization energies are your ionization values this would likely belong to Aluminum. Since there is a huge point between the third and fourth ionization energies, which designates that the atom reached noble gas configuration after the third electron was removed. The element which has 3 valence electrons in the third period is aluminum.

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4-hydroxypentanal reacts with one equivalent of methanol to form a cyclic acetal.

quester [9]

Answer:

Explanation:

The equation is given as:

CH3CHOHC2H4CHO + CH3OH --> CYCLIC ACETAL + H2O

This above equation is carried out in the presence of a strong acid. There are five mechanisms employed and they are:

Step 1:

Initial formation of the hemiacetal which takes several steps

Step 2:

Addition of a proton. The hemicetal is protonated on the hydroxyl group (-OH group)

Step 3:

As seen a bond is broken to give the H2O molecule and a resonance stabilized cation.

The carbonyl group on the cation is enriched with the oxygen-18 got from the H2O molecule as seen in the mechanism.

Step 4:

An attraction occurs between electrophile and nucleophile i.e the stabilised cation and the lone paids of the methanol.

Step 5:

Finally, a proton (+) is removed from the molecule by a lone pair of electron on the methanol.

Attached are the Steps 1 - 5 mechanism below

4 0

3 years ago

When an alkane reacts with an element from group 7a, the reaction is referred to as?

Elena-2011 [213]

Alkanes are saturated hydrocarbon, that is they contain hydrogen and carbon without a double or triple bond between the carbon atoms, e.g. ethane, propane. Group 7a in the periodic table are called halogens e.g chlorine, bromine. Alkanes react with halogens in a reaction called substitution, where halogens replace hydrogen atoms in alkanes.

7 0

3 years ago

Read 2 more answers

4. A student was not sure that she accurately recorded the final temperature of her melting point sample. Therefore, she decided

Mumz [18]

Answer:

AWF DSF

Explanation:

W AFSDDDDDDDDD

3 0

2 years ago

1. Matter can be measured in many ways for specific purposes. Which measurement depends on the force of gravity it is not reliab

Ganezh [65]

Answer:

1. B- weight

2. B- 2.0g/cm3

3. C- 22.8g/cm3

Explanation:

6 0

3 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

3 years ago