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AVprozaik [17]
3 years ago
10

What are transitional elements in chemistry?

Chemistry
2 answers:
GuDViN [60]3 years ago
8 0
They are the elements from scandium to copernicium. The elements between group 2 and group 3
CaHeK987 [17]3 years ago
7 0
Transitional Elements are d-block elements which when they form ions have incomplete "d" suborbital of electrons. 
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B. Due to Earth's rotation

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What state of matter has the most energy in its molecules
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Explanation:

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a. Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g
Tanya [424]

Answer:

(A) \Delta H^{\circ }_{r}= -144 kJ

(B) \Delta H^{\circ }_{r}= - 2552kJ

Explanation:

(A) 2NO(g) + O₂(g) → 2NO₂(g)

1/2 N_{2}(g)+O_{2}(g)\rightarrow NO_{2}(g), \Delta H^{\circ }_{a}=33.2 kJ....equation (a)

1/2N_{2}(g)+1/2O_{2}(g)\rightarrow NO(g), \Delta H^{\circ }_{b}=90.2 kJ ....equation (b)  

Now, multiplying equation (a) with 2:

⇒ N_{2}(g)+2 O_{2}(g)\rightarrow 2 NO_{2}(g)....equation (a)

Then equation b is reversed and multiplied with 2:

2 NO(g)\rightarrow N_{2}(g)+ O_{2}(g)....equation (b)

Now by adding the equation (a) and equation (b), we get:

⇒  2 NO(g)+ \bcancel N_{2}(g)+\bcancel 2 O_{2}(g)\rightarrow 2 NO_{2}(g) +\bcancel N_{2}(g)+ \bcancel O_{2}(g)

⇒  2NO(g) + O₂(g) → 2NO₂(g)

<u>Therefore, the enthalpy of the reaction:</u>

\Delta H^{\circ }_{r}= 2\times \Delta H^{\circ }_{a} - 2\times \Delta H^{\circ }_{b}

= (2\times33.2)- (2\times90.2)=66.4 - 180.4= -144 kJ

(B) 4B(s)+3O₂(g) → 2B₂O₃(s)    

B_{2}O_{3}(s)+3H_{2}O(g)\rightarrow 3O_{2}(g)+B_{2}H_{6}(g), \Delta H_{a }^{\circ }=+2035 kJ...equation (a)

2B(s)+3H_{2}(g)\rightarrow B_{2}H_{6}(g), \Delta H_{b }^{\circ }= +36 kJ...equation (b)

H_{2}(g)+1/2O_{2}(g)\rightarrow H_{2}O(l), \Delta H_{c }^{\circ }= -285 kJ...equation (c)

H_{2}O(l)\rightarrow H_{2}O(g), \Delta H_{d }^{\circ }=+44 kJ...equation (d)

Now multiplying equation (b) with 2, reversing equation (a) and multiplying with 2. Reversing equation (c) and (d) and multiplying both with 6.

6O_{2}(g)+2B_{2}H_{6}(g)\rightarrow 2B_{2}O_{3}(s)+6H_{2}O(g)...equation (a)  

4B(s)+6H_{2}(g)\rightarrow 2B_{2}H_{6}(g)...equation (b)

6H_{2}O(l)\rightarrow 6H_{2}(g)+3O_{2}(g)...equation (c)

6H_{2}O(g)\rightarrow 6H_{2}O(l)...equation (d)

Now by adding the equations (a), (b), (c), (d); we get:

4B(s)+3O₂(g) → 2B₂O₃(s)

<u>Therefore, the enthalpy of the reaction: </u>

\Delta H^{\circ }_{r}= -2\times \Delta H^{\circ }_{a} + 2\times \Delta H^{\circ }_{b} - 6 \times \Delta H_{c }^{\circ } - 6 \times \Delta H_{d }^{\circ }

= -2\times (+2035 kJ)+ 2\times (+36 kJ) - 6 \times (-285 kJ)- 6 \times (+44 kJ) = -4070 + 72 + 1710 - 264 = - 2552kJ

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