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Serga [27]
3 years ago
12

Which has a higher melting point and why? LiF or Rbl?

Chemistry
1 answer:
ExtremeBDS [4]3 years ago
7 0

Explanation:

To get a better understanding of the answer,let's take use of Fajan's rule which says that no bond is neither purely covalent or purely ionic.

For ionic compounds MP and BP are higher than covalent compounds. So in order to find the one with higher MP,it's enough to find the most ionic compound among the given options.

In LiF,LiCl,LiBr and LiI,The cation is same and anion only differs.According to Fajans rule Smaller anion and larger cation favours ionic bonding and vice versa. Since Fluoride is the smallest anion,LiF is the most ionic compound and hence it is the one with highest MP from the options.Hope it helps

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compute the mass-specific enthalpy change associated with Nz that is undergoing a change in state from 400 k to 800 k
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Answer:

The correct answer is "430 kJ/kg". A further explanation is given below.

Explanation:

The given values are:

T₁ = 400 k

T₂ = 800 k

The average temperature will be:

= \frac{T_1+T_2}{2}

= \frac{400+800}{2}

= 600 \ k

From table,

At 600 k the C p will be = 1.075

Now,

⇒ The specific enthalpy = Cp(T_2-T_1)

⇒                                \Delta h=1.075 (800-700)

⇒                                      =430 \ kJ/kg

8 0
3 years ago
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Answer:

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Given the equation:
Citrus2011 [14]

Answer: a)  Fe is the limiting reagent

b) 3.59 g

c) 11.6g

Explanation:

4Al_2O_3+9Fe\rightarrow 3Fe_3O_4+8Al  

To calculate the moles :

   

\text{Moles of} Al_2O_3=\frac{27.5g}{102g/mol}=0.27moles

\text{Moles of} Fe=\frac{8.4g}{56g/mol}=0.15moles

According to stoichiometry :

a) 9 moles of Fe  require= 4 moles of Al_2O_3

Thus 0.15 moles of Fe will require=\frac{4}{9}\times 0.15=0.067moles  of Al

Thus Fe is the limiting reagent as it limits the formation of product and Al is the excess reagent.

b) As 9 moles of Fe give = 8 moles of Al

Thus 0.15 moles of Fe give =\frac{8}{9}\times 0.15=0.133moles  of Al

Mass of Al=moles\times {\text {Molar mass}}=0.133moles\times 27g/mol=3.59g

c) As 9 moles of Fe give = 3 moles of Fe_3O_4

Thus 0.15 moles of Fe give =\frac{3}{9}\times 0.15=0.05moles  of Fe_3O_4

Mass of Fe_3O_4=moles\times {\text {Molar mass}}=0..05moles\times 231.5g/mol=11.6g

8 0
3 years ago
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