Chemistry: Post your doubts here!

[hela]

AlCl3 ... its that concept about the charge density of Al being too high that it attains a covalent character by attracting the electrons more towards itself.... if u don't get it them reply and i'll explain in detail

the answer in ms is MgCl2 I DO NOT GET WHY

 

[littlecloud11]

the answer in ms is MgCl2 I DO NOT GET WHY

SiCl4 is a covalent compound. The bonding in AlCl3 is more covalent than it is ionic. When the question asks which is ionic which SOME covalent character it is asking about compound where the main bond is ionic.

The question says ionic with some covalent character. this only happens when

- the charge on the cation is large ( Mg > Na )
- the cation is small ( Mg < Na )
- the size of the anion is large

since both anion are the same, and Mg is smaller in size and has a greater charge, it has a higher covalent character then Na.​

The main bonding in MgCl2 and NaCl are ionic bond and MgCl2 has more covalent tendency so the answer is B.​

 

[Ahmedraza73]

Ethanol is much more soluble in water than EthylEthanoate.
Which statement help to account for this?
A: A hydrogen bond forms between the hydrogen of the -OH group in ethanol and the hydrogen of a water molecule.
B: A hydrogen bond forms between the hydrogen of the -OH group in ethanol and the Oxygen of a water molecule.
C: Ethanol is a polar molecule ,but ethylethanoate is non-polar.
D: Ethanol is able to dissocaite into hydrogen ions and ethoxide ions,but ethylethanoate is not able to dissociate

Please some one answer this question Briefly?

 

[MONJUR HASAN NISHAD]

1) The answer is one and two because the value can only be calculated theoretically and cannot be carried out practically.
1- because both are solid under standard conditions
2- because methane is in it's gaseous state under standard conditions.

since both experiments cannot be carried out, Hess' law is use to calculate the enthalpy changes.
The combustion enthalpy can easily be calculated through a laboratory experiment.

I can't understand the concept of option 1 and option 2. 1- If both are solid, what's the problem? Why can't we calculate the enthalpy change practically if both are solids? 2- If methane is a gas, what's the problem? Why can't we calculate the enthalpy change practically if methane is a gas? 3- I understood.

 

[Ahmedraza73]

I2(g) +3Cl2(g) _________ 2ICl3(s) delta H= -214 kJ/mole
I2(s)_____________I2(g) deltqa H=+38 kJ/mole
What is the standard entahply change of the Formation of Iodine trichloride,ICl3?
A: +176 kJ/mole
B: +138 kJ/mole
C: -88 kJ/mole
D: -138 kJ/mole
E: -214 kJ/mole
Please can any one solve this

 

[littlecloud11]

I can't understand the concept of option 1 and option 2. 1- If both are solid, what's the problem? Why can't we calculate the enthalpy change practically if both are solids? 2- If methane is a gas, what's the problem? Why can't we calculate the enthalpy change practically if methane is a gas? 3- I understood.

Well it is difficult to measure to measure the enthalpy change of hydration for solids accurately in a direct way because the hydration process cant be controlled directly. It involves the formation of attractive bonds between the ion and the water molecules and can be very spontaneous.
As for methane, H2 and C simply does nor react together under standard conditions, so how can you calculate the formation of methane directly when the very definition requires the elements to react under standard conditions?

 

[MONJUR HASAN NISHAD]

Well it is difficult to measure to measure the enthalpy change of hydration for solids accurately in a direct way because the hydration process cant be controlled directly. It involves the formation of attractive bonds between the ion and the water molecules and can be very spontaneous.
As for methane, H2 and C simply does nor react together under standard conditions, so how can you calculate the formation of methane directly when the very definition requires the elements to react under standard conditions?

Ok, I understood option 1. But in option 2 the question only asked about the formation of methane from its elements. This implies that it is not mandatory for the reactants or products to be at standard conditions. So, why can't we calculate the enthalpy change practically?

 

[hela]

I2(g) +3Cl2(g) _________ 2ICl3(s) delta H= -214 kJ/mole
I2(s)_____________I2(g) deltqa H=+38 kJ/mole
What is the standard entahply change of the Formation of Iodine trichloride,ICl3?
A: +176 kJ/mole
B: +138 kJ/mole
C: -88 kJ/mole
D: -138 kJ/mole
E: -214 kJ/mole
Please can any one solve this

I2(s)_____________I2(g) deltqa H=+38 kJ/mole
+
I2(g) +3Cl2(g) _________ 2ICl3(s) delta H= -214 kJ/mole
ADD OBTEIN _176 FOR 2 MOLES THEN DIVIDE BY 2 OBTAIN _88

 

[hela]

can you solve

 

[hela]

can you solve

 

[darknessinme]

can you solveView attachment 22101

q4. delta H= sigma(bonds broken) - sigma(bonds formed)
180=(994+496)-(2E(N-O))
So E(N-O)=655
Answer is A

q5. From the definition of standard enthalpy of atomisation, it involves only the breaking of bonds, which is endothermic, so only positive values.
For standard enthalpy of formation, you should have seen enough examples to know that it can be either positive or negative.
So it leaves B as the only option available, where solution can be either positive or negative.

 

[Soldier313]

Aoa wr wb
This is an A2 qn, if someone could please provide a detailed explanation for it.....





JazakAllah khair

littlecloud11 help please?

 

[Alice123]

The lattice energies(enthalpies) of rubidium fluoride(RbF) and caesium chloride(CsCl) are -760 KJ/mol and -650kJ/mol respectively. What is the lattice energy of caesium Fluoride(CsF) likely to be? (Rb-37;Cs-55) ANS B
A.-620kJ/mol
B.-720 kJ/mol
C.-800kJ/mol
D.-900kJ/mol
can any1 explain n is this an AS question?

 

[littlecloud11]

Ok, I understood option 1. But in option 2 the question only asked about the formation of methane from its elements. This implies that it is not mandatory for the reactants or products to be at standard conditions. So, why can't we calculate the enthalpy change practically?

Ok, do you remember how we manufacture methane? By reacting hydrogen with CO2. Not carbon. There is a reason for it. H2 does not react with C directly below temperatures of 900 degree Celsius. It is not feasible to carry out such reactions to find the enthalpy of formation of methane. Standard or not, the enthalpy of formation requires the compound to be formed directly from it's element, and in the case of methane, the reaction is just not feasible to carry out experimentally.

 

[daredevil]

In options 1 and 2, the products are formed by electrophilic addition of DBr and alkene. Option 3 is not possible because when DBr reacts with alcohol in a nucleophilic substitution reaction the hydrogen in the OH is not replaced by D. So the answer is option 1 and 2, right? Correct me if I am wrong. Thanks for your help.

yup u're ryt

 

[daredevil]

the answer in ms is MgCl2 I DO NOT GET WHY

ohhh i'mm sorry!! it will be MgCl2 because ALCl3 is covalent with some ionic character because of the EXTREMELY high charge density. Mg2+ however has a high charge density but is still lower than Al3+ so it is actually an ionic compound but due to the high charge density of Mg2+ the electron favours it to Cl- just a tad bit more and thus a litttllleeeee teeeny tiny bit of covalent character is attained.
if u don't get what charge density is then ask me again and i'll explain in another post otherwise it might get confusing over here.

 

[daredevil]

Ethanol is much more soluble in water than EthylEthanoate.
Which statement help to account for this?
A: A hydrogen bond forms between the hydrogen of the -OH group in ethanol and the hydrogen of a water molecule.
B: A hydrogen bond forms between the hydrogen of the -OH group in ethanol and the Oxygen of a water molecule.
C: Ethanol is a polar molecule ,but ethylethanoate is non-polar.
D: Ethanol is able to dissocaite into hydrogen ions and ethoxide ions,but ethylethanoate is not able to dissociate

Please some one answer this question Briefly?

what is the answer?? is it A?

 

[daredevil]

Q. Find the amount of Fe formed when 96 dm3 of CO are made to react with 320 g of Fe2O3
3CO(g) + Fe2O3(s) --> 2Fe(s) + 3CO2 (g)

please solve this question for me... i did solve it but then i cudn't get the right answer ._.

 

[littlecloud11]

Q. Find the amount of Fe formed when 96 dm3 of CO are made to react with 320 g of Fe2O3
3CO(g) + Fe2O3(s) --> 2Fe(s) + 3CO2 (g)

please solve this question for me... i did solve it but then i cudn't get the right answer ._.

No of moles of CO = 96/24 = 4mols
Fe2O3 Mr =160
No of moles of Fe2O3 = 320/160= 2
2 mols Fe2O3 requires 6 moles CO for complete reaction but only 4 moles of CO are present, and is therefore limiting.
So, 3 mols CO = 112 g Fe
4 mols CO =149.3 g Fe (ans)

 

[daredevil]

No of moles of CO = 96/24 = 4mols
Fe2O3 Mr =160
No of moles of Fe2O3 = 320/160= 2
2 mols Fe2O3 requires 6 moles CO for complete reaction but only 4 moles of CO are present, and is therefore limiting.
So, 3 mols CO = 112 g Fe
4 mols CO =149.3 g Fe (ans)

thanx but 112g of Fe ?? where did that come from?? the Mr??