Stack Exchange Network

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange

New answers tagged

1

You should use the ideal gas law to convert between $n_i$ (amount of gas) and $p/T$. For each temperature and pressure pair you can write $$\begin{align}n_{g,1}=\frac{p_1V}{RT_1}\\n_{g,2}=\frac{p_2V}{RT_2}\end{align}$$ where $$\begin{align}n_{g,1}&=n_{g,H_2O}+n_{Ar}\\n_{H_2O}&=n_{g,H_2O}+n_{l,H_2O}\\n_{g,2}&=n_{H_2O}+n_{Ar}\end{align}$$ You also ...


0

I have been informed that glass is a super-cooled liquid and is also considered to be an amorphous solid. Can it be both and, if not, what category does it fall into? This is dependent on temperature, but before I go into that, we need to understand what a solid and liquid are as that is where mush of the misinformation comes from. A liquid is defined ...


3

After a lot of help, I have the following to suggest as an answer: Imagine a reaction with $\Delta_\mathrm{r} G^{\circ} = -1000Jmol^{-1}$ at 298K. Using the following equations: $$\Delta_\mathrm{r} G^{\circ} = - RT\ln K$$ $$e^{-\frac{\Delta_\mathrm{r} G^{\circ}}{RT}} = K$$ This would give a K value of 1.50, which indicates the Q value on the curve where G ...


4

The scheme gives $rate=k_2\ce{[O3][O]}$ and $\displaystyle \frac{d\mathrm{[O]}}{dt} =k_1\mathrm{[O_3]}-k_{-1}\mathrm{[O_2][O] } -k_2\mathrm{[O_3][O] }=0$ for steady state on the oxygen atom. This gives the rate expression as: rate=$\displaystyle \frac{k_1k_2\mathrm{[O_3]^2} }{k_{-1}\mathrm{[O_2]}+k_2\mathrm{[O_3]}}$. Thus the problem when the oxygen ...


3

What does this $[\ce{O2}]^{-1}$ part mean? It means the higher the concentration of oxygen, the lower the rate. The two-step mechanism you show relies on oxygen atoms (radicals) to react with ozone in the second step. Dioxygen acts as a radical scavenger (first reaction, reverse direction). So if dioxygen concentrations are high, the reaction will not ...


5

Logically, that would mean that absolutely pure reactants have infinite G (which doesn't seem right). It's the slope (rate of change) that is infinite, not the actual property (Gibbs free energy). $\Delta_r G$ (expressed sometimes as $\Delta G'$, sometimes you'll see just $\Delta G_m$ without subscripts indicating explicitly that this is the molar Gibbs ...


5

G is a finite quantity for ξ = 0 or max. The slope of the graph is vertical on the extremes, though. This is because the chemical potential for reagents approaches negative infinity on one side, and that of the product positive infinity on the other side. (A vertical or infinite slope does not mean that the function value has to be infinite - a half-circle ...


6

The spontaneous emission of light by a substance not resulting from heat is luminescence (a of cold-body radiation). $\ce{^1}$ A few types of luminescence are: Bioluminescence: Made by living creatures such as fireflies, glow-worms, and many marine creatures. Chemoluminescence: made by a chemical reaction. Glow sticks work this way. Electroluminescence: ...


1

1) How do we know the particles in a solution are more separated? When a solution forms, intermolecular bonds in the solid solute are broken and replaced by interactions with the solvent. The solute is then free to diffuse in the solvent. If the solute molecules still interacted in a highly favorable way they would tend to bond and precipitate out. This is ...


1

Chemisorption involves chemical attachment (electrostatic or covalent bonding) with the surface. Since there are a finite number of such bonding sites on the surface, the maximum occupancy is one monolayer. Physisorption involves weaker (dipolar or dispersion) interactions, and additional layers might be able to attach to underlying layers via similar weak ...


1

Here is how the y-axis works :


3

You can work from the principle that at a phase transition the partial molar Gibbs free energy (chemical potential) of the substance in each phase is equal, so that $$\Delta_\mathrm{sub} G = 0$$ from which $$\Delta_\mathrm{sub} S_\mathrm{m} = \frac{\Delta_\mathrm{sub} H_\mathrm{m}}{T_\mathrm{sub}}$$ Note however that the value of $\Delta_\mathrm{sub} H_\...


0

This is my first answer in stack exchange if some mistake comment on. $$E^\circ_\ce{Fe^3+/Fe} \neq E^\circ_\ce{Fe^3+/Fe^2+} + E^\circ_\ce{Fe^2+/Fe}$$ but $$\Delta_r G^\circ_1 = \Delta_r G^\circ_2 + \Delta_r G^\circ_3$$ for the three reactions: $$\ce{Fe^3+ + 3/2 H2 -> Fe + 3H+}\tag{1}$$ $$\ce{Fe^3+ + 1/2 H2 -> Fe^2+ + H+}\tag{2}$$ $$\ce{Fe^2+ + ...


3

boiling point is the temperature at which vapour pressure of the liquid becomes equal to the external pressure. But in case of closed vessel on boiling volume remains the same so pressure increases due to increase in temperature,which also cause the pressure of the air molecules in the container to rise, along with that of the liquid present so in that case ...


1

If we consider the classical Daniel cell $$\ce{Zn|Zn^2+||Cu^2+|Cu|}\text{(wiring to Zn)}$$ then due higher tendency of zinc ions to leave the metal, getting hydrated in solution, the zinc electrode obtains more negative electrostatic potential due extra electrons, compared to cupper electrode. As electrodes are galvanically connected, electrons flow ...


0

The constant $a$ only describes attractive forces while the constant $b$ describes repulsive forces. So by definition $a$ cannot be negative. Can anyone tell me if this is correct? It is correct. Neither parameter $a$ or $b$ is regarded as negative, because a negative parameter would be opposite to its original meaning within the van der Waals theory. In ...


6

You can define the size of atoms and molecules in various ways. You can, for instance, derive a measure of the size of a hydrogen molecule from the density of solid hydrogen: Solid hydrogen has a density of 0.086 g/cm3 making it one of the lowest-density solids. From the above you can derive an effective radius for $\ce{H2}$ of $\pu{2.10 Å}$, which is ...


2

The definition of generations of physicochemical properties can be attributed to Hugh et al. [1] as recapitulated by e.g. Ferraz et al. [2]: The search for new and different ILs has led to the progressive development and application of three generations of ILs: 1) The focus of the first generation was mainly on their unique intrinsic physical and ...


0

The mechanism is complicated by quantum chemistry aspects, but the principle can be tracked down to the low level classical electrostatic energy analysis, or even to the classical gravitational orbit energy analysis. An object $O$ is on a circular orbit around a massive central object $C$. This $C$ puts the attractive force $F=A/r^2$ on the object $O$ (this ...


0

how is the potential energy of the system reduced? To be honest, there is no specific mechanism to decrease in potential energy. Potential energy is merely a mathematical function that scientists came up with to describe energy changes in physical systems after rigorous measurements and calculated through equations backed by experimental results. These ...


1

The potential energy that changes during reactions is electrostatic energy (+/- attraction, +/+ or -/- repulsion) from the charged particles (electrons and protons) within the atoms. The theory is described by Coulomb's Law (involving the potential energy of charged particles): http://guweb2.gonzaga.edu/faculty/cronk/CHEM101pub/energy.html Generally, this ...


4

Even if the final temperature of the products is the same as the initial temperature of the reactants, you can still have a change in entropy, enthalpy, and Gibbs free energy of products relative to the reactants. This is the result of making and breaking chemical bonds and not the result of a temperature change. Entropy and enthalpy are not just functions ...


7

One reason we write equations this way is because all of the parameters on the right-hand-side can be measured or computed independently. Consider an analogous (but more intuitive) equation, the Stokes-Einstein relation for the diffusion coefficient $D$ of a particle in a liquid: $$D = \frac{k_\mathrm{B} T}{6\pi\eta r}$$ where $\eta$ is the dynamic ...


4

If you rotate the molecule by $45^\circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$. Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. ...


1

To answer this kind of problem, it'd be better starting with elimination approach. It is given the solubility of sodium carbonate in $\pu{100 g}$ of water as $\pu{25.0 g}$ at $\pu{22 ^{\circ}C}$. By that, you know, solubility of its hydrate should be more than $\pu{25.0 g}$ at $\pu{22 ^{\circ}C}$ because the mass of hydrate is including additional mass of ...


0

One should be cautious about attempting to directly compare partial atomic charges with formal oxidation states (see this Nature Materials article for a nice discussion on the topic). As a trivial example, I just ran an M06/def2-TZVP calculation of the HO2 molecule (see Gaussian 16 input file below), which formally should assign charges of H+ and O2-. Based ...


0

Your calculation would be good, if you did not forget that hydrates contain water. The solubility of the anhydrous carbonate is 25 g in 100 g of water. Both carbonate forms, once diluted, are the same compound, forming the same ions plus eventually releasing water. So in case of the hydrate, the ratio carbonate:water must be 1:4 as well. $M_{\ce{Na2CO3}}=...


2

This is a confusing question because, while solubilities can be reported in mL/L, there can be ambiguity when choosing a pressure during conversion to this unit, for instance using the following equation to convert from molarity $c$ to volume/volume units: $$ \rho = \frac{cRT}{p}$$ In this online data page, for instance , in some columns the solubility is ...


5

As noted by Wikipedia, "Henry's Law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid." So you are absolutely right. 4 times the pressure of oxygen should result in 4 times as much oxygen being dissolved. The answer key for the problem must be wrong. User Delta_G made an ...


1

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$ $$\ce{NaOH + H3O+ -> Na+ + 2H2O}$$ The base removes the hydrogen ions from solution, and the acid further dissociates to maintain LeChatelier's equilibrium. The $\ce{H3O+}$ ion concentration will decrease as it is continually removed whereas the $\ce{A-}$ concentration will increase until the acid is fully ...


2

The ideal gas law (or other equation of state) is valid only when a gas is in thermodynamic equilibrium. In a rapid irreversible expansion or compression, the pressure depends not only on the volume (and temperature) but also on the rate of change of volume. This is because a gas has a viscous property which acts to dissipate mechanical energy and convert ...


4

If you have two fragments that are very far apart, certain quantum chemistry methods can cause spurious effects from including both fragments in the same calculation. A well known example of this is dissociation curves for $\ce{H2}$ with Restricted Hartree Fock. By using a density based method, we are less likely to localise the electrons on any one atoms ...


1

$$\Delta H=\Delta E + p \cdot \Delta V$$ $\Delta E$ is energy change of a system at constant volume. $\Delta H$ as is energy change of a system at constant pressure. As the molar amount of gases is decreased by the reaction, $p \cdot \Delta V \lt 0$ at constant $T,p$. Therefore, $\Delta H \lt \Delta E$ Neither that $\Delta E<0$ even at constant $T$, ...


1

By definition, $$\Delta H=\Delta E+\Delta(PV)$$For a reacting ideal gas mixture at constant temperature, $$\Delta(PV)=(\Delta n)RT$$where $\Delta n$ is the change in the number of moles between reactants and products. Therefore, $$\Delta H=\Delta E+(\Delta n)RT$$For the reaction under consideration, $$\Delta n=-\frac{1}{2}$$


1

You have to assume your unknown hydroxide is completely dissociate in aqueous solution. If so, then you can calculate $\mathrm{pOH}$ of the solution as $\mathrm{pOH} = -\log \ce{[OH-]}$. Then, $\mathrm{pH} = \mathrm{p}K_\mathrm{w} - \mathrm{pOH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}$. Now, you see, since $\mathrm{p}K_\mathrm{w} = 14$, only unknown is $\...


0

In this case, the absolute configuration remains same(as R) for SN1 and chages(into S) for SN2. This is because the priority order of non substituted groups(wrt C2) remains the same before and after substitution of Br by OH. Let the initial mole fraction of (R)-2bromobutane be 1. After the reaction, let the mole fractions of R and S 2-bromobutane be x and 1-...


6

Even with single-particle systems, analytical solutions are only possible for a small subset of quantum mechanical problems: https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions For example, if you were to change the external potential of the particle in a box into something even a tad weirder, like a polynomial function,...


1

Your doubt is legitimate because you are assuming that $K_{(a)}$ is a parameter that you can plug in every reaction. Actually $K_{(a)}$ is calculated using this equation in a solution of the acid with water: $$ \ce{K_{(a)} = \frac{[H+][CN-]}{[HCN]}} $$ That is derived from this ideal equilibrium: $$\ce{HCN ⇌ CN- + H+}$$ Even though it reflects the ...


3

The answer should be C. One way would be to write the thermochemical equations, however since we just want to compare, we can do it without going into that. The formation of the H-Cl bond, (like all bonds) is exothermic. And the breaking of the C-H, N-H and O-H bonds is endothermic (since their formation is exothermic too). Since the contribution of the H-...


1

The energy and speed do not affect particles themselves, but their interactions with other diatomic objects. And, of course, they affect their kinematics in electromagnetic field. Kinetic energy of a particle is $$E_\mathrm{k}=\frac 12 \cdot m \cdot v^2$$ The force on a charged particle in an electrostatic field of the strength $E$ is $$\vec F=q\...


1

Sucrose is an covalent compound and it does not ionize. $\ce{KNO3}$, $\ce{NaOH}$ are good ionic compounds hence the can form electolytic solutions. Ammonium acetate also can ionize. Hence the answer is 2


1

For an exothermic reaction $\Delta H \lt 0$ and $K_p$ must decrease as the temperature increases. (It is the opposite way round for an endothermic reaction, i.e. more dissociation at higher temperature which makes sense if the reaction is endothermic: 'more energy more product'.) Using the integrated Van Hoff equation you can calculate what happens; $\...


1

You can understand this as follows : Le Chatelier's principle says that on changing some factor ( like temp.,pressure,find.,etc ) the equilibrium will shift in a direction so as to undo the effect of change . Now in this case on increasing the temperature the reaction equilibrium is shifting in backward direction (descrease in value of K ) so reverse ...


0

Option (a) is more appropriate as in case of (c) the temperature or the condition is not mentioned. As in case of 698 K, since $K_\mathrm{eq}$ is reasonably less, the equilibrium will shift in the backward direction, so HI will be stable.


0

Physical method is an old school terminology for what we would call as instrumental method of chemical analysis. There is a pretty famous textbook on "Physical Methods for Chemists." The book by Drago is quite popular. These physical methods, as the name indicates were borrowed from physicists. This included NMR, Group Theoretical Applications for IR & ...


2

I'm going to continue to resist calling C the heat capacity because, as used here, it is not path-independent. Instead, I will take it to merely represent the derivative of the heat added per mole of ideal gas with respect to temperature for a polytropic process path. With this caveat, here is my interpretation of what is happening mathematically when C ...


1

You are right. Only when the process is reversible we have: $dS = \frac{dQ} {T}$ For an irreversible process we have instead: $dS > \frac{dQ} {T}$ This is the Clausius inequality and when replacing $dQ$ with its expression derived from the first law ($dU = dQ - PdV $; assuming only PV work) we obtain a general criterion for spontaneous change: $dU + ...


1

Researchers were of the opinion that, chemiluminescence could be a prime suspect to justify why it was so hard to observe Marcus inverted region in electron transfer reactions. To explain why so, let us assume that the equilibrium reactant and product curves have a large free energy gap relative to the reorganization energy, which would mean that the ET ...


5

What definition, where? The short answer is given in the IUPAC Gold Book: chemical element A species of atoms; all atoms with the same number of protons in the atomic nucleus. A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. Sometimes this concept is called the elementary substance as distinct ...


1

I don't know how you call it in English but you can apply this approximation $\frac{d[D]}{dt} = 0$ because (3) is fast and $D$ have a really short lifetime, so you can say that $$k_2[B]^2[C]=k_3[D][E]$$ This approximation is why the teacher is always saying that only the slow reaction is important Fast equilibrium means that the equilibrium is quickly ...


Top 50 recent answers are included