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mash [69]
3 years ago
7

Determine the electron geometry, molecular geometry, and idealized bond angles for each of the following molecules. CF4CF4 NF3NF

3 OF2OF2 H2SH2S In which cases do you expect deviations from the idealized bond angle
Chemistry
1 answer:
vfiekz [6]3 years ago
5 0

Answer:

CF4

Molecular geometry- tetrahedral

Electron geometry- tetrahedral

NF3

-molecular geometry - trigonal pyramidal

Electron geometry - tetrahedral

OF2

Molecular geometry - bent

Molecular geometry - tetrahedral

H2S

Molecular geometry- bent

Electron geometry - tetrahedral

Explanation:

According to Valence Shell Electron Pair Repulsion Theory, the shape of a molecule depends on the number of electron pairs on the valence shell of the central atom in the molecule.

For all the compounds listed, the central atom has four points of electron density. This correspond to a tetrahedra electron pair geometry. The presence of lone pairs on the central atom of OF2,NF3 and H2S accounts for the departure of the observed molecular geometry from the geometry and idealized bond angle predicted on the basis of the VSEPR theory.

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Explain why the air in Earth's atmosphere is a mixture, not a compound. Brainiest offered need it by 6:45!!!
Georgia [21]

Answer:

air is not a mixture because of scientists freezing it and finding different liquids, it is a mixture because the compounds that make up air e.g. oxygen (o2), Carbon dioxide (co2) and the most important Nitrogen which is an element and makes up 78.09% of air are not chemically bound in the way that compounds are

Explanation:

8 0
3 years ago
Burning a candle is a combustion reaction that uses wax and oxygen to produce carbon dioxide, water, and energy. When you burn a
ss7ja [257]

Answer:

The wax

Explanation:

The limiting reactant is the reactant in short supply in a particular reaction. This reactant determines the extent of the reaction. Oxygen in the reaction is the one in excess. As we know that oxygen is free in the atmosphere and it is in excess supply.

After the wax finishes burning the reaction stops and no more production of carbon dioxide, water and energy but oxygen still continues to abound all around.

In order to determine the amount of carbon dioxide and water that would be produce, we simply use the amount of wax to relate with them.

4 0
3 years ago
Calculate the density of a sample if the mass is 2.98g and the volume is 2.12l
sineoko [7]
Hey there!:

Mass = 2.98 g

Volume = 2.12 L

Therefore:

Density = mass / volume

Density = 2.98 / 2.12

Density = 1.405 g/L
6 0
4 years ago
The electrode that contains the item to be electroplated<br><br> a) anode<br> b)cathode
cricket20 [7]
The answer is A, anode
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
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