Answer:
The strength of a bond depends on the amount of overlap between the two orbitals of the bonding atoms
Orbitals bond in the directions in which they protrude or point to obtain maximum overlap
Explanation:
The valence bond theory was proposed by Linus Pauling. Compounds are firmed by overlap of atomic orbitals to attain a favourable overlap integral. The better the overlap integral (extent of overlap) the better or stringer the covalent bond.
Orbitals overlap in directions which ensure a maximum overlap of atomic orbitals in the covalent bond.
Answer:
THE STRENGTH OF THE BOND DEPENDS ON THE AMOUNT OF OVERLAP BETWEEN THE TWO ORBITALS OF THE BONDING ATOMS
ORBITALS BOND IN THE DIRECTION OR POINT IN WHICH THEY PROTRUDE OR POINT TO OBTAIN MAXIMUM OVERLAP.
Explanation:
Valence bond theory describes the covalent bond as the overlap of half-filled atomic orbital yields a pair of electrons shared between the two bonded atoms. Overlapping of orbitals occurs when a portion of one orbital and the other occur in the same region of space. The strength of a bond is determined by the amount of overlap between the two orbitals of the bonding atoms. In other words, orbitals that overlap more and in the right orientation of maximum overlapping form stronger bonds that those with less overlap and right orientation for maximum overlap. The bonding occurs at a varying distance in different atoms from which it obtains its stable energy caused by the increase in the attraction of nuclei for the electrons.
Orbitals also bond in the direction to obtain maximum overlap as orientation of the atoms also affect overlap. The greater overlap occurs when atoms are oriented on a direct line mostly end to end or side by side between the two nuclei depending on the type of bond formed. A sigma bond is formed when atoms overlap end to end in which a straight line exists between the two atoms that is the internuclear axis indicating the concentrated energy density in that region. Pi bond exits in when overlap occurs in the side -to -side orientation and the energy density is concentrated opposite the internuclear axis.
Answer:
333.6 atm
Explanation:
The following data were obtained from the question:
Mole of O2 (nO2) = 3.96 moles
Mole of N2 (nN2) = 7.49 moles
Mole of CO2 (nCO2) = 1.19 moles
Total pressure = 563 mmHg
Partial pressure of N2 =..?
Next, we shall determine the total number of mole in the container.
This can be obtained as follow:
Mole of O2 (nO2) = 3.96 moles
Mole of N2 (nN2) = 7.49 moles
Mole of CO2 (nCO2) = 1.19 moles
Total mole =?
Total mole = nO2 + nN2 + nCO2
Total mole = 3.96 + 7.49 + 1.19
Total mole = 12.64 moles
Next, we shall determine the mole fraction of N2.
This can be obtained as follow:
Mole fraction = mole of substance/total mole
Mole of N2 (nN2) = 7.49 moles
Total mole = 12.64 moles
Mole fraction of N2 =?
Mole fraction of N2 = 7.49/12.64
Finally, we shall determine the partial pressure of N2.
This can be obtained as follow:
Mole fraction of N2 = 7.49/12.64
Total pressure = 563 mmHg
Partial pressure of N2 =..?
Partial pressure = mole fraction x total pressure
Partial pressure of N2 = 7.49/12.64 x 563
Partial pressure of N2 = 333.6 atm
The, the partial pressure of nitrogen, N2 is 333.6 atm
coefficients of the reactants equal the coefficients of the products,
products and reactants are the same chemicals,
same number of each kind of atom appears in the reactants and in the products,
same number of each kind of atom appears in the reactants and in the products
Answer:
C₆H₆
Explanation:
Each border of the figure represents 1 atom of carbon. We have 6 borders = 6 atoms of carbon.
Each atom of carbon form 4 bonds. All the carbons are doing a double bond and a single bond with other carbons. That means are bonded 3 times. The other bond (That is not represented in the figure. See the image) comes from hydrogens. As we have 6 carbons that are bonded each 1 with one hydrogen. There are six hydrogens and the molecular formula is:
This structure is: Benzene
The given reaction will shift towards cis-2-butene once placed in equilibrium. This can be determined by calculating the reaction quotient and comparing it with the equilibrium constant.
The reaction could either shift towards the cis-2-butene or trans-2-butene depending on whether the reaction quotient, Q, is lesser or greater than the equilibrium constant, Kp.
Bear in mind that Kp = Ptrans/Pcis. Let's say that Pt is the partial pressure of trans-2-butene and Pc is the partial pressure of cis-2-butene at equilibrium. If we start with 5 atm of each gas, the change in Pc is -x and the change in Pt is +x.
So, Kp = (5+x)/(5-x). We are given that Kp = 3.4. Solving these two equations will show that x is a negative value, which means that the system shifts towards cis-2-butene.
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For the isomerization reaction cis-2-butene ⇌ trans-2-butene, with an initial pressure of 5.00 atm for both gases and a Kp of 3.40, the system will shift towards the product, trans-2-butene, as Kp > Qp (1). This reflects the principle that a chemical system at equilibrium will shift to counteract any change.
In terms of the equilibrium constant (K), for gas-phase reactions, Kp represents equilibrium in terms of partial pressures, while Kc represents it in molar concentrations. For instance, in the isomerization reaction given cis-2-butene ⇌ trans-2-butene, Kp is given as 3.40. To determine the behavior of the system, we need to compare it to reaction quotient (Q). Given that the flask initially contains 5.00 atm of each gas, Qp is 1 (since Qp = partial pressure of trans-2-butene / partial pressure of cis-2-butene). Since Kp > Qp, the reaction will shift towards the products, hence the system will shift towards trans-2-butene. From this, it is clear that the equilibrium constant and reaction quotient play vital roles in determining the direction of shift in a chemical equilibrium.
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