Most molecules do not have a "lower energy triplet state". For most molecules, the ground state is a singlet, in which all the electrons are paired. An exception is dioxygen, which has two degenerate antibonding orbitals and two electrons to go in them. In this case, the ground state is the...
Consistent ones. I would convert everything to SI units.
The one thing you must never do is blindly plug numbers into a formula without consideration of units, e.g. mixing atm, L, kg etc. (some SI and some not).
How big is your dewar? A rough calculation suggests your resistor would boil off ca. 500 mL liquid Ar per hour. Is that significant compared to the volume of your dewar? But how good is your dewar anyway, if it boils off in an hour?
That still gives no help to answering the question. Presumably a) and b) are alternative answers to the question, but what was the question? Can you quote it - as exactly as possible please.
The clue is in the name. Surface characterisation techniques characterise the surface of a material. Other techniques may characterise the bulk material as a whole.
Nothing you say makes any sense. What are a and b? "The ratio of x = y" is meaningless. The meaning of pKa - 1 is the value of pKa minus one; where's the problem?
I would say it means the position of reaction at equilibrium - whatever that means! The trouble is that "position of reaction" and "position of equilibrium" are not well-defined terms.
The equilibrium constant K is constant at constant temperature, but changes with change of temperature. At...
The reaction shifts to the right, i.e. there is a net conversion of A and B to C and D. The position of equilibrium as defined by K does not change, Q = K at equilibrium, but the individual concentrations will be different from before.
No, it will not. The rates of both forward and backward reactions decrease with decreasing temperature, but the back reaction more so.
Q is not a measure of the position of equilibrium. K is. Q = K when the reaction is at equilibrium. Reducing the quotient (by reducing [C]) does not move the...
Some compounds whose molecules are roughly spherical in shape can form a plastic crystal, in which the molecules are fixed in their crystal positions (can't translate), but can freely rotate - kind of the opposite of a liquid crystal. E.g. sulfolane between 16 and 28°C.
Just come across the term "kronoseismology, or the ability to see what's happening inside Saturn":
https://www.nationalgeographic.com/science/2019/01/how-long-is-day-on-saturn-astronomers-just-found-out/
The Washburn equation is used in mercury porosimetry:
ΔPr = -2γcosθ
where ΔP is the pressure difference (usually just the pressure, as the sample is evacuated); r is the pore radius, γ is the surface tension of mercury and θ is the contact angle of the mercury with the solid. Using typical...
What are you taking as G and G0 here?
Are you assuming that the blast furnace is at equilibrium? Is that true? Do they add extra CO to force the reaction to the right?
There's an important point you omitted. "There's no difference in internal energy between a polymer that has been stretched into a straight chain, or one that's in a curled state" - at the same temperature. If you stretch or relax the band slowly, so that it remains in thermal equilibrium with...
I'm no expert on relativity, but I think you're confusing things. Time dilation isn't a frequency. A rate isn't a frequency. QM doesn't say that the "frequencies of things" are linearly proportional to their energy. E = hν relates the energy and frequency of a photon, specifically.
Consider for...
Your answer looks correct. The field is discontinuous at the negative charge; it tends to positive infinity as you approach the negative charge from below, and negative infinity as you approach it from above. So you follow the black curve for z<d, and the red curve for z>d.
As I understand, ordinary stars don't form iron. They stop around oxygen, I think. Heavier elements like iron are formed in supernova explosions. If a star contains iron, the iron must have been present in the gas cloud from which it condensed, as the result of earlier supernova explosions. As...
Your calculation is correct, but where did you get your numbers from? Standard cylinders I use, about 40L, at 230 bar, contain about 400 moles of gas. 112 L looks high, if the other numbers are right.
That's because the sun is at a focus of the ellipse, not the centre. In this case, the semi-major axis is the average of the closest and farthest distances. You can't get the semi-minor axis from these (it is independently variable), but you should be able to find it easily, or the value of the...
There seem to be a few mistakes in your formula. If I've got it right, "sqrta^2b^2" should be sqrt(a^2 - b^2). (Maybe just a typo on your part.)
{[(sqrta^2b^2)/a]/3}^4 should be {[(sqrt(a^2-b^2))/a]^4}/3
And I don't know where you're getting your a and b from. In this formula, a is the...
It might be more helpful if you think about it the other way round, that energy depends on temperature (though it comes to the same thing in the end). Have you come across the principle of equipartition of energy? It says that the energy is (on average) equally distributed between all the...
The probability is 1/4 if both parents carry the affected allele. But what is the probability of that? You know they don't have the disorder, but each has a sibling who does. What does that tell you about the probability of their being carriers?
Not really, without computer number-crunching. Practically, I think it would work the other way round - measure the ionisation energy, calculate the quantum defect, compare periodic trends and see what that tells you about the orbitals.
You have ignored the initial angular velocity. ω = ω0 + ∫a(t)dt.
That still doesn't give the given answer. Are you sure the initial tangential velocity wasn't -0.5 m/s? Either you or the book seems to have made a typo.
It looks like you are using x and y for different things. There is the equation of the semicircle: x2 + y2 = a2. Then there are the coordinates of a point in space (x, y, z), for which you want to find the potential and its derivative. But you can't use the same x and y for both, and try to...