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3 years ago

Three biologically important diatomic species, either because they promote or inhibit life, are

Chemistry

1 answer:

Naddik [55]3 years ago

3 0

Answer:

See Explanation

Explanation:

We can write the molecular orbital configuration of molecules in the same way as we write the orbital electron configuration of atoms. The valence electrons in the molecule are filled into molecular orbitals of appropriate energy in accordance to the Aufbau principle.

For CO;

σ2s2, σ*2s2, Π2py2, Π2pz2, σ2px2

For NO;

σ2s2, σ*2s2, Π2px2, Π2py2, σ2pz2, Π*2px1

For CN-;

σ2s2, σ*2s2, Π2px2, Π2py2, σ2pz2

These are the ground state electron configurations of these molecules.

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Please help me. ? :)

BaLLatris [955]

Answers and explanation:

To answer these questions you use a periodic table.

A is In because it has 49 protons and In is element number 49.

B is 27 because Co is element number 27.

C is 73 because Ta is element number 73.

D is 49 because the number of electrons in an atoms is always equal to the number of protons.

E is 56 because the atomic number of Ba is 56.

F is 54 because 56 - 54 = 2. The charge is equal to the number of protons minus the number of electrons.

G is 66. The number of neutrons is equal to the atomic mass minus the number of protons. 115 - 49 = 66.

H is 108 because 181 - 73 = 108.

I is 32. The atomic mass of an element is equal to the amount of protons plus the amount of neutrons. So 16 + 16 = 32.

3 0

4 years ago

How many valence electrons do elements in the same family as Aluminum contain?

Nady [450]

Aluminium family has 3 electrons in their valence shell....i hope its correct..｡◕‿◕｡

5 0

3 years ago

Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is

mote1985 [20]

Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(5\times -1271.94)+(9\times -285.83)]-[(2\times 73.2)+(12\times 0)]=-9078.57kJ/mol$

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

From the reaction we conclude that,

As, 2 moles of compound released heat = -9078.57 kJ

So, 1 moles of compound released heat = $\frac{-9078.57}{2}=-4539.28kJ$

For per gram of compound:

Molar mass of $B_5H_9$ = 63.12 g/mole

$\Delta H^o_{rxn}=\frac{-4539.28}{63.12}=-71.915kJ/g$

Therefore, the heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

7 0

3 years ago

Please help and match them

alina1380 [7]

The answers from top to bottom right to left are
3,1,2,4

7 0

3 years ago

MOLE TO ATOM CONVERSIONS

Advocard [28]

Answer:

Answers are given below.

Explanation:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1) 0.680 moles of H₂S

0.680 mol × 6.022 × 10²³ molecules / 1mol

4.09× 10²³ molecules

2) 8.96 moles of KI

8.96 mol × 6.022 × 10²³ molecules / 1mol

53.96 × 10²³ molecules

3) 6.500 moles of KClO

6.500 mol × 6.022 × 10²³ molecules / 1mol

39.143 × 10²³ molecules

4) 0.950 moles of NaOH

0.950 mol × 6.022 × 10²³ molecules / 1mol

5.721 × 10²³ molecules

5) 4.73 moles of Na₂CO₃

4.73 mol × 6.022 × 10²³ molecules / 1mol

28.48 × 10²³ molecules

7 0

3 years ago