The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol
<h3>Further explanation</h3>
Given
Ratio of the concentrations of the products to the concentrations of the reactants is 22.3
Temperature = 37 C = 310 K
ΔG°=-16.7 kJ/mol
Required
the free energy change
Solution
Ratio of the concentration : equilbrium constant = K = 22.3
We can use Gibbs free energy :
ΔG = ΔG°+ RT ln K
R=8.314 .10⁻³ kJ/mol K
They are very stable (not reactive).
This is because their outer shell has a total of 8 valence electrons. All elements strive to get towards a full outer shell, but since these elements already have a full outer shell, they are fairly unreactive.
1. P = F/A; weight is a force (the force of gravity on an object), so divide the weight by the area given. P = 768 pounds/75.0 in² = 10.2 pounds/in².
2. Using the same equation from question 1, rearrange it to solve for A: A = F/P. We're given the force (the weight) and the pressure, so A = 125 pounds/3.25 pounds/in² = 38.5 in².
3. Again, using the same equation from question 1, rearrange it this time to solve for F: F = PA = (4.33 pounds/in²)(35.6 in²) = 154 pounds.
4. We can set up a proportion given that 14.7 PSI = 101 KPa. This ratio should hold for 23.6 PSI. In other words, 14.7/101 = 23.6/x; to solve for x, which would be your answer, we compute 23.6 PSI × 101 kPa ÷ 14.7 PSI = 162 kPa.
5. We are told that 1.00 atm = 760. mmHg, and we want to know how many atm are equal to 854 mmHg. As we did with question 4, we set up a proportion: 1/760. = x/854, and solve for x. 854 mmHg × 1.00 atm ÷ 760. mmHg = 1.12 atm.
6. The total pressure of the three gases in this container is just the sum of the partial pressures of each individual gas. Since our answer must be given in PSI, we should convert all our partial pressures that are not given in PSI into PSI for the sake of convenience. Fortunately, we only need to do that for one of the gases: oxygen, whose partial pressure is given as 324 mmHg. Given that 14.7 PSI = 760. mmHg, we can set up a proportion to find the partial pressure of oxygen gas in PSI: 14.7/760. = x/324; solving for x gives us 6.27 PSI oxygen. Now, we add up the partial pressures of all the gases: 11.2 PSI nitrogen + 6.27 PSI oxygen + 4.27 PSI carbon dioxide = 21.7 PSI, which is our total pressure.
The hybridization for C in acetylene, HCCH, or C₂H₂ is 'sp'.
Discussion:
There are three different forms of hybridization -
- sp- The first occurs when two carbon atoms are triple linked.
- sp₂- When two carbon atoms are double-bonded to one another, this is known as sp₂.
- sp₃- When a single bond joins two carbon atoms, this is known as sp₃.
In the case of acetylene(HCCH or C₂H₂):
- The carbon atom requires additional electrons to establish four bonds with hydrogen and other carbon atoms in the synthesis of C₂H₂. One 2s₂ pair is consequently transferred to the vacant 2pz orbital. Each carbon has two sp hybrid orbitals after the 2s orbital in each atom combines with one of the 2p orbitals.
- As a result of the atoms' symmetrical alignment in a single plane, C₂H₂ possesses a linear molecular structure. Due to their lower electronegative nature than Hydrogen atoms, all Carbon atoms are situated near the center of the Lewis structure of C₂H₂.
H-C≡C-H
Therefore, it is concluded from the above discussion that the hybridization type of acetylene is 'sp'.
Learn more about hybridization here:
brainly.com/question/14140731
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Answer:
Zn2+ is colourless
Explanation:
We know that transition metal salts are usually coloured due to the possibility of d-d transition.
This d-d transition can only occur when there are vacant d-orbitals. The electronic configuration [Ar] 4s23d8 suggests the presence of vacant d-orbitals and the possibility of the compounds of Zn2+ being coloured.
However, the absence of colours in Zn2+ compounds shows that there is no d-d transition(electronic) spectra observed for Zn2+ because the d orbitals are completely filled. This means that the correct electronic configuration of the ion is [Ar] 3d10.
