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Nastasia [14]
3 years ago
8

For the pair of compound Fe(s) or CCl4(s) the one with the higher melting point is For the pair of compound F e ( s ) or C C l 4

( s ) the one with the higher melting point is blank.. For the pair of compound KCl(s) or HCl(s) the one with the higher melting point is For the pair of compound K C l ( s ) or H C l ( s ) the one with the higher melting point is blank.. For the pair of compound H2O(s) or H2S(s) the one with the higher melting point is For the pair of compound H 2 O ( s ) or H 2 S ( s ) the one with the higher melting point is blank.. For the pair of compound Ti(s) or Ne(s) the one with the higher melting point is___________.
Chemistry
1 answer:
guajiro [1.7K]3 years ago
7 0

Answer:

Fe has a higher melting point than CCl4.

KCl has a higher melting point than HCl

H20 has a higher melting point than H2S

Ti has a higher melting point than Ne

Explanation:

The bonding in molecules or compounds will determine its boiling or melting point. The stronger the bonds between the compound and  molecules, the higher the energy needed to break the bonds which leads to an higher boiling or melting point.

  1. Fe (Iron) has an higher melting point than CCl4 (Tetrachloromethane). The bonding in Fe is metallic while CCl4 has van der waal force. Metallic bonding is stronger than van der waal force. Much energy is needed to break metallic bonding compared to van der waal force
  2. KCl (potassium chloride)has ionic bonding while HCl (hydrochloric acid) has permanent dipole (covalent bonding). Ionic bonding is stronger than any covalent bonds. Therefore KCl will have a higher melting point than HCl
  3. H20 (water) has hydrogen bonding while H2S (hydrogen sulpide) has permanent dipole. Hydrogen bonding is stronger than permanent dipole.
  4. Ti (Titanium) has a metal having metallic bonding while Ne (neon) has van der waal forces. metallic bonding is stronger than van der waal forces.  
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Here is your answer:


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<em>~Nonportrit</em>

6 0
3 years ago
Read 2 more answers
From the following reaction and data, find (a) S o of SOCl2 (b) T at which the reaction becomes nonspontaneous SO3(g) + SCl2(l)
disa [49]

Answer:

618 J/Kmol

T > 1.36 x 10³ K

Explanation:

The  balanced reaction of interest is:

                           SO₃ (g) + SCl₂ (l) ⇒    SOCl₂ (l) +     SO₂ (g)

with the data:

ΔHºf (kJ/mol)      -396          -50.0          -245.6         -296.8

Sº(J/mol·K)             256.7       184               ?               -248.1

ΔGº=  -75.2 kJ

We know, we can find the standard  change inGibb´s free energy from the equation:

ΔGºrxn =  ΔHºrxn - TΔSºrxn

So we can calculate ΔHºrxn  = ∑ ΔHºf prod  -  ΔHºreact, and substitute into this equation to solve Sº SOCl₂.

ΔHºrxn = ( -245.6 + (-296.8) ) - ( -396 - 50) kJ = - 96.4 kJ

Similarly  for ΔSºrxn

 ΔSºrxn = (-0.248.1 +Sº SOCl₂) - (0.256.7 +0.184) kJ/K

= -0.689 kJ /K -+ Sº SOCl₂

Plugging the values for the expression for  ΔGºrxn:

-75.2 kJ = -96.4 kJ - 298 K  x  ( -0.689 kJ /K + Sº SCl₂ )

-75.2 kJ = -96.4 kJ + 205.3 Kj - 298 Sº SCl₂

-184  kJ = -298 K  x Sº SCl₂

0.618 kJ/molK = Sº SCl₂

= 0.618 kJ/K x 1000 J = 618 J/Kmol

For the second part we will still be using the Gibb´s free energy change  equation as above , but now we will solve for T when the reaction becomes  non-spontaneous.

For the reaction to become non-spontaneous  ΔGº is positive, and this happens when the term  TΔSº becomes greater tha ΔHº:

ΔGºrxn =  ΔHºrxn - TΔSºrxn

0 =   ΔHºrxn - TΔSºrxn ⇒  TΔSºrxn  =  ΔHºrxn

                                           T= ΔHºrxn / ΔSºrxn

ΔSºrxn  = -0.689 J/Kmol + 0.618 J/Kmol = -0.0710 kJ/Kmol

( using the value  the value just calculated from above )

T =  - 96.4 kJ / -0.071 kJ/K = 1.36 x 10³ K

For temperatures greater than 1.36 x 10³ K the reaction becomes non-spontaneous.

4 0
4 years ago
From the dissolution of a gas sample, two more gases were obtained, namely 150 cm3 of N2 and 450 cm3 of H2 .What is the formula
natali 33 [55]
450 X 250 X + 25m this is correct
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kvasek [131]
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3 years ago
Which set of work should be used to solve the following problem?
VladimirAG [237]

Answer:

C. 65.5 g PF3 x (1 mole / 87.97 g PF3)

Explanation:

Dimensional analysis can be used to find the amount of moles in 65.5 grams of PF3 gas.

First we would need to determine relevant conversions.

We would need to find the molar mass of PF3 which is already given ( 87.97 g / mol )

So 1 mole of PF3 = 87.97g

From there we can create the dimensional analysis table which will look something like the attached image.

This table correspond with answer choice C.

<u>Reasons </u><u>it's </u><u>not </u><u>the </u><u>other </u><u>answer </u><u>choices</u><u>.</u>

It wouldn't be A because the units must cancel out and they don't as grams is being multiplied by grams

It would be B because it doesn't use correct units.

There are not 87.97 moles in 1 gram of PF3. There are 87.97 grams in 1 mole of PF3

7 0
2 years ago
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