Answer:
The electron geometry, molecular geometry and idealized bond angles for these molecules respectively are:
a. CF4: tetrahedral, tetrahedral and 109.5 degrees
b. NF3 tetrahedral, trigonal pyramidal and 102.5 degrees
c. OF2 tetrahedral, angular and 103 degrees
d. H2S tetrahedral, angular and 92.1 degrees
Explanation:
The electron geometry considers the bound atoms and unbound electron pairs to determine the geometry. The four molecules have four bound atoms and/or unbound electrons pairs, thus they have a tetrahedral geometry. On the other hand, the molecular geometry only considers the position of bound atoms to determine the geometry.
Between H3O and H2O, H2O has a smaller bond angle due to the two unbound electron pairs. The bond angle decrease as the number of unbound electron pairs increases in every molecule.
CO2 and CCl4 are both nonpolar because of the 3D geometry of the molecule. Each individual bond is polar but both molecules have symmetrical geometry so the dipole bonds are canceled.
CH3F is a polar molecule because the dipole between the C-H and C-F bonds are differents thus, besides the symmetrical geometry the dipole bonds are not canceled.
In the bookstore Avanti gave four books so it would be 1
Answer:
Option A, The Rutherford experiment proved the Thomson "plum-pudding" model of the atom to be essentially correct.
Explanation:
Thomson's plum pudding model:
Plum pudding model was proposed by J.J Thomson. In Thomson's model, atoms are proposed as sea of positively charge in which electrons are distributed through out.
Result of Rutherford experiment:
As per Rutherford's experiment:
Most of the space inside the atom is empty.
Positively charge of the atom are concentrated in the centre of the atom known as nucleus.
Electrons are present outside the nucleus and revolve around it.
As it is clear that, result of Rutherford experiment did not supported the Thomson model.
This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and and are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
Finally we convert this result to kJ:
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0.424209104545485 is the answer my friend lol or at least what I got