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scoundrel [369]
2 years ago
12

The melting points for the second series transition elements increase from 1526˚C for yttrium to 2623˚C for molybdenum and then

decrease again to 321˚C for cadmium. Account for this trend in terms of band theory.
Chemistry
1 answer:
boyakko [2]2 years ago
6 0

Answer:

The melting points of the second series of transition elements are electrons are entering to the type of orbital.

Explanation:

Transition elements have "d" band that can overlap with the "s' band to give a composite band. The composite band have six molecular orbitals per metal atom.

Half of the molecular orbitals are the anti bonding.So, maximum bonding occurs when the molecular orbitals contains six electrons i.e. the metal has six valence electrons per each metal atom.

When the maximum bonding occurs then it shows maximum melting point.

So, molybdenum shows melting point at 2623^{o}C and after that the electrons goes into the antibonding and reaches minimum for cadmium when the anti boning orbital is completely filled. So, the melting point of the cadmium is tex]321^{o}C[/tex].

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If a mixture looks smooth and the same throughout it is probably what
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Answer:
             <span>If a mixture looks smooth and the same throughout it is probably <u>Homogeneous</u>.

Explanation:
                   Mixture is the combination of different compounds which are unreactive to each other.

Mixture are classified as ...

Solutions; in which the mixed compounds are thoroughly mixed and cannot be distinguished from each other and are said to be homogeneous. In solutions the size of solute is very small (i.e. Less than 1 nm).

Colloids; in which the solute is homogeneous visually but heterogeneous microscopically. The size of particles in this case is between 1 nm to 1 </span>μm.

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4 0
3 years ago
What are some of the forms of energy you used today? How did you use them?
klio [65]

Answer:

Heat or Thermal energy, Solar Energy, Chemical energy, electrical energy, mechanical energy

Explanation:

8 0
2 years ago
When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem
masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

<u>Step 1:</u> Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

<u>Step 2</u>: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

<u>Step 3:</u> Define the formula

Pressure is constant.  → the heat evolved from the reaction is equivalent to the enthalpy of reaction.  

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the  calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for  water (1g/mL and 4.184 J/g°C)

7 0
3 years ago
The name for the following formula C9H18 should be
Elza [17]

Answer:

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7 0
3 years ago
A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp
vichka [17]

Answer:

The balanced chemical equation:

C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = T_1= 21.36^oC

Final temperature of the calorimeter = T_2= 26.37^oC

Heat absorbed by calorimeter = Q

Q=C\times \Delta T

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol

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2 years ago
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