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2 years ago

Why does iron not fit the pattern in column 7

1 answer:

3 0

Iron doesn't fit because it doesn't have enough atoms or protons in its nucleus so there for it belongs in column 2. <span />

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A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2

4vir4ik [10]

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

$C_3H_7+5O_2\rightarrow 3CO_2+4H_2O$

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

$\frac{3}{1}\times 3.85 mol=11.55 mol$ of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

$\frac{4}{1}\times 3.85 mol=15.4 mol$ of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

$\frac{4}{5}\times 0.647 mol=0.5176 mol$ of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

Then 0.647 moles of oxygen will give:

$\frac{1}{5}\times 0.647 mol=0.1294 mol$ of propane.

0.1294 moles of propane are consumed.

5 0

1 year ago

2CH4(g)⟶C2H4(g)+2H2(g)

Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of $CH_4$ will be,

$2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)$ $\Delta H^o=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890.3kJ$

(2) $C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$ $\Delta H_2=-136.3kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=-571.6kJ$

(4) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$ $\Delta H_4=-3120.8kJ$

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)$ $\Delta H_1=2\times (-890.3kJ)=-1780.6kJ$

(2) $C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)$ $\Delta H_2=-(-136.3kJ)=136.3kJ$

(3) $H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)$ $\Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ$

(4) $2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)$ $\Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ$

The expression for enthalpy of the reaction will be,

$\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)$

$\Delta H=201.9kJ$

Therefore, the enthalpy change for the reaction is, 201.9 kJ

5 0

1 year ago

Which of the following concerning second ionization energies is true?

Dennis_Churaev [7]

Answer:

Explanation:

The fact is that the both elements belong to different groups in the periodic table. Mg is in group 12 while Al is in group 13. The outer most electron configuration for Mg2+=[Ne]3s0

Al+= [Ne]3s13p0

It is evident that in Al+, the second electron is to be removed from a filled 3s subshell which is energetically unfavourable. Unlike In Mg2+ where the two electrons are removed. There is always a tendency towards a quick loss of all the valence electrons in the valence shell. It will be more difficult to remove an electron from a filled shell.

5 0

1 year ago

6. From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 25ºC: Reaction A: ΔH = 10.5 kJ

qwelly [4]

Answer:

Both reaction A and reaction B are non spontaneous.

Explanation:

For a spontaneous reaction, change in gibbs free energy ( $\Delta G$ ) should be negative.

We know, $\Delta G=\Delta H-T\Delta S$ , where T is temperature in Kelvin scale.

Reaction A: $\Delta G=(10.5\times 10^{3})-(298\times 30)J/mol=1560J/mol$

As $\Delta G$ is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

$\Delta H-T\Delta S< 0$

or, $T> \frac{\Delta H}{\Delta S}$

or, $T> \frac{10.5\times 10^{3}}{30}$

or, $T> 350$

So at a temperature greater than 350 K, the reaction is spontaneous.

Reaction B: $\Delta G=(1.8\times 10^{3})-(-113\times 298)J/mol=35474J/mol$

As $\Delta G$ is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

$\Delta H-T\Delta S< 0$

or, $T> \frac{\Delta H}{\Delta S}$

or, $T> \frac{1.8\times 10^{3}}{-113}$

or, $T> -16$

So at a temperature greater than -16 K, the reaction is spontaneous.

3 0

2 years ago

Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.

Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0

2 years ago