# Tag Info

2

What I understood: Let's say $\ce{H2O}$ has a (saturated) vapour pressure of $x$ atm at a certain temperature. Now Relative humidity (RH) is the ratio of the partial pressure of water vapour to the equilibrium vapour pressure of water at a given temperature. So what this means is that if RH = 100% then the maximum amount of water vapours is present in the ...

2

As gas particles expand into a vacuum there's no transfer of their kinetic energy into anything else, they do not collide with anything but with themselves. Since the temperature of a gas is a measure of the average kinetic energy of its particles, the temperature remains constant. (This is different than in the case when the expansion would involve for ...

6

The strength of permanganic acid that you quote, combined with that of potassium hydroxide as a base, would guarantee that pure potassium permanganate is neutral in aqueous solution. But commercially prepared potassium permanganate is made in the presence of alkali, the use of potassium instead of sodium arising from the fact that the reaction scheme does ...

0

This is an adiabatic irreversible expansion so $q=0$. The change in internal energy is $\Delta U=\int_{T_1}^{T_2}C_V dT$ where $C_V$ is the heat capacity at constant volume for the monoatomic gas (=3R/2). The work done is $w=-\int_{V_1}^{V_2} pdV$ at the expense of the internal energy. As the temperature drops as no heat added or lost and then $\Delta U =w$...

4

Michaelis-Menten kinetics is given by the equation: $$V = V_\mathrm{max}\frac{S}{K_M+S} \tag1$$ Where $V_\mathrm{max} = k_\mathrm{cat}\cdot [\ce{E_T}]$ and $K_M = \dfrac{k_\mathrm{on} + k_\mathrm{cat}}{k_\mathrm{off}}$ (using the conventional description of Briggs and Haldane's derivation of the Michaelis-Menten equation; Ref.1): $$\ce{E + S <=>[k_\... 1 Reason lies in definition of enthalpy of reaction. Enthalpy of reaction is heat exchanged between our system in which reaction happens and surroundings when reaction is carried at constant temperature and pressure. If reaction is exothermic, it releases heat and increases temperature of our system and so to keep it at the same temperature you need to give ... 4 Michaelis-Menten kinetics is given by$$V = V_{\max} \frac{S}{K_M + S}$$Taking the inverse of both sides to linearise the equation,$$\frac{1}{V} = \frac{1}{V_\max} +\frac{K_M}{V_{\max} S}$$Taking y = 1/V and x = 1/S in y = A + Bx leads by linear regression to A = 0.4767, B = 4.4754 (not including units). Therefore,$$V_\max = \frac{1}{A} = \pu{2....

3

One contribution is the number of identical conformations of a molecule that can be generated by symmetry operations, which is described by a symmetry number $\sigma$. In the high temperature limit $$S_\mathrm{rot} /R = 1 + \ln\left( \frac{8\pi^2IkT}{h^2\sigma} \right)$$ A larger symmetry number decreases the entropy. This number is 2 for nitrogen and 1 for ...

3

The bond dissociation enthalpy* $D$ is the enthalpy change for conversion of bonded into dissociated molecules following homolytic cleavage of the bond in question: $$\ce{A-B (g)->A(g) + B(g)},\qquad D(\ce{A-B})=\Delta H^\circ_f (\ce{A(g)})+\Delta H^\circ_f (\ce{B(g)})-\Delta H^\circ_f (\ce{A-B(g)})$$ This is the property usually discussed as representing ...

1

The issue as I can see it here is a simple misunderstanding that can rectified: $K_\mathrm a$ is a constant even if $\ce{[A]}$ is not. (which is how it was defined) The definition of $k_\mathrm a$ is as follows: $K_\mathrm a$ , the acid ionization constant, is the equilibrium constant for chemical reactions involving weak acids in aqueous solution. Nowhere ...

4

To answer the question, over and above the errors noted in the other answer the integral is just not well defined - it is infinite. There are two ways to see this The "common sense" method. Consider $$\int^\infty_{-\infty} \left[ A\sin(kx) + B \cos(kx)\right]^2dx$$ The quantity under the integral is positive semi-definite - it is always greater ...

0

Thank you very much for your idea and help. I think that your calculations are correct. But I think you'd better to try other simpler choices to make oxygen. One way is air separation. If you separate nitrogen from air, you gain oxygen. This can be done by "Zeolite oxygen generators". You can read these site about them. 1- https://www.zeochem.com/...

1

Your attempt has multiple issues, most of which are very basic. The first issue that I noticed was how you expanded and simplified your integral. \begin{align} \int^\infty_{-\infty}[A^2\sin^2(kx) &+ AB\sin(kx)\cos(kx) \\ &+ AB\cos(kx)\sin(kx) +B^2\cos^2(kx)]\mathrm dx = 1 \tag{1} \end{align} Since, $\sin^2 x + \cos^2 x = 1$, $$\implies \int^\... 2 The curves labelled Harmonic/Morse show how the potential energy changes with internuclear separation, and the horizontal lines are the allowed total energies, potential plus kinetic. The potential energy is a maximum at the turning points (end of the horizontal lines in your picture) and zero at the r_e value but because the total energy is constant the ... 0 Considering the example of NaCl and CuCl: In NaCl, the electrons effectively shield (or screen the outermost electron from the attraction of nucleus) or one can say the shielding is 100\%. Whereas in case of CuCl, there are 10 3d electrons present which ineffectively shields the nucleus, so the value of screening decreases. The increasing order of ... 2 All generalities are false (as is this one). Though impurities usually lower the melting point (m.p.) by disrupting crystallization on the atomic order, consider the phase diagram of the binary alloy (amalgam) HgxAg1- x: Pure Hg melts ~-39°C, and adding even a little bit of impurity raises the melting point considerably! And without resorting to entropic ... 6 The kinetic energy is not ignored. Each quantized state (labeled by the vibrational quantum number) has a total energy given by the expectation value of the Hamiltonian operator for the wavefunction. The contribution of the kinetic energy can be computed as the expectation value of the QM kinetic energy operator, or as the difference between the expectation ... 4 I am not sure you even need the density in this case. Say you have \ce{1kg} or \ce{1000g} water (the solvent). That would mean that there is \ce{4.16 mol} of \ce{KNO_3}. Since we can figure out the molar mass of \ce{KNO_3} = \pu{101.11g mol-1}, the mass of \ce{KNO_3} in the sample must be (4.16)(101.11)= \pu{420.62 g}. We already assumed that ... 9 Because the acid and conjugate bases are equimolar - and because the equilibrium constant is small, obviating the need to solve a quadratic if you approached this in another way - the Henderson-Hasselbalch equation is:$$\pu{pH} = {\rm p}K_\mathrm{a} + {\rm log}_{10}\left({[A^-]\over[HA]}\right)\pu{pH} = 4.74 + {\rm log}_{10}\left({0.2\ {\rm M}\over 0.2\...

2

Out of the four equations that you have, you need to find $\Delta H^\circ_\mathrm f$ of $\ce{MgO}$, or rather, the $\Delta H$ value of the reaction $$\ce{Mg(s) + 1/2O2(g) -> MgO(s)} \tag{1} \label{1}$$ Why is this? This is because of the definition of heat of formation is as follows: Standard heat of formation of a compound is the change of enthalpy ...

3

why is dH=dE at constant volume? It isn't, and nowhere in the problem or answer is this implied. First of all some definitions. For a combustion reaction, the enthalpy change can be equated with the heat of combustion at constant pressure, whereas the internal energy is the heat of combustion at constant volume: $$\Delta U = q_V ~~~~~\text{constant volume} \... 13 The ambiguity doesn't lie so much in the definition of the rate, because as long as the constants a and b in ap_\ce{X} and b[\ce{X}] are chosen appropriately, the two expressions are entirely equivalent. So you can define the rate either way, and there is no physical difference, not even a mathematical difference. It is somewhat akin to asking ... 0 Notice, K_\text{f} for water is 1.86 \pu{°C⋅kg⋅mol^{-1}}. This is just a constant to know. Now that you know K_f, you can solve for \Delta T. Assume 1 litre of solvent. That means there will be 7.71 moles of LiCl. Next, the mass of 1L of solution is 1.1 678\pu{cm^{-1}} because \pu{g/cm^3} is is the same as \pu{kg/L}. Next, you can subtract the ... 3 OP has given a good effort to solve the problem using correct path. Only loose point was not considering the diatomic nature of the gases as Safdar Faisal pointed out in a comment. Suppose the amount of \ce{O2} in the gas mixture is n_1 \ \pu{mol} and the amount of \ce{N2} in the gas mixture is n_2 \ \pu{mol} in volume of \pu{1.0 L} container. Thus,... 3 The point of the question is to test your understanding i. of Raoult's law (that increasing the solute concentration decreases the vapor pressure and that the identity of the solute is not important in an ideal solution; the mole fraction of solute determines \Delta p: \Delta p = p-p^\circ = -\chi_\textrm{solute}p^\circ ii. that vapor pressure increases ... 2 Since joule-thomson effect is isoenthalpic process, so everything is done at the cost of internal energy. Taking a as the term involving intermolecular forces of attraction and b as the excluded volume, we get:$$T_i = \frac{2a}{Rb}$$As per above equation inversion temperature is the function of intermolecular forces of attraction. When the temperature ... 1 The temperature of the milk increases because of work done on the milk molecules which cause , increase the velocities of the moecules and since velocity is directly proportional to the temperature. Velocity α Temperature So the temperature of milk increases due to the shaking process. Although this effect is very small but heat up the molecules. 2 Temperature will increase. If system is adiabatic, it doesn't neccesarily mean that temperature of the system can't change because even though system can't exchange energy with surroundings, processes inside can generate heat. In your example, exothermic reaction. Note that, internal energy in your case stays the same even though temperature increased. This ... 2 For a system in contact with a constant temperature reservoir at the same temperature as the system in its initial state, and also in contact with a constant pressure surrounding at the same pressure as the system in its initial state,$$\Delta U=Q-W=Q-P\Delta V-W_{npv}$$where W_{npv} is the amount of non-PV work done by the system. In addition, from the ... 2 Let$$moles\ CH_4=1-e_1moles\ H_2O=1-(e_1+e_2)moles\ CO=e_1-e_2moles\ CO_2=e_2moles\ H_2=3e_1+e_2$$where the e_1 is the number of moles of CH_4 destroyed and e_2 the number of moles of CO_2 that are produced. So the equilibrium equations are going to read:$$\frac{(e_1-e_2)(3e_1+e_2)^3}{(1-e_1-e_2)(1-e_1)(2+2e_1)^2}=K_1$$and$$\...

0

I would answer only the constant of dissociation $K_a$, because it is a thermodynamical constant and you can calculate it from potentiometric measurements without dealing with activities and so on. If you have the $\Delta G$ of reaction, which is related with the difference of electric potential of the cell, you can calculate the constant as \Delta G = -RT \... 1 The graph is a little bit confused, because there exist different sorts of Gibbs energy. i It can be the Gibbs energy of the rectant. It can be the Gibbs energy of the products. And it can also be the total of these two Gibbs energies. First let's discuss the Gibbs energy of the reactant. In the beginning of the reaction, the Gibbs energy of the reactant is ... 1 Here is a labeled and annotated version of your graph (from OpenStax Chemistry). 2 The starting point to solve the problem is Raoult's law. For each component we can write that the vapour pressure (related by Dalton's law to the total pressure) is equal to the product of the mole fraction in the solution and the vapour pressure of the pure component: $$y_iP_S = \chi _i P^{\circ} _i$$ You can exploit the fact that the molar amounts in the ... 1 I am going to use the physics conventions of work according to which, work done by the gas is positive. So if you want the answers as per chemistry conventions than just flip the sign of the work done, keeping the magnitude same. Part (a): Reversible Isothermal Expansion \begin{align} \Delta U_a &=nC_v\Delta T\\ \Delta U_a &=\pu{0 J}\\ \end{align} \... -1 The statement that it is impossible to calculate the absolute energy at an ambient T seems to be intuitively incorrect, difficult possibly, but not impossible. Since U is a state function the value of U for a given set of conditions is invariable and can be set as an arbitrary zero and changes in U calculated. So all one must do is work in reverse to ... 0 Ivan may be right that the question was designed by morons, but an argument can be made that the strength of an acid governs the pH in a solution with a given concentration, rather than the pH governing the strength. The question apparently asked what properties of the acid can be used to assess its strength and thus the predict the pH you get at a given ... 2 Enthalpy is a fundamental physical property of the material(s) comprising a system. It can often be presented for a given material in the literature in tables (e.g., the steam tables for water), relative to some specific datum state (at which it is taken to be zero). This is analogous to potential energy which is expressed relative to some specified ... 1 I remember my days when I started confusing myself in the exact same way. The issue I believe is that you haven't been explained the details of how all this happens. Evaporation: All molecules in a liquid have various amount of energy. But the ones at the top, the ones that form the surface - they have the maximum energy in them because they are present ... 1 Enthalpy vs ∆Enthalpy We know that enthalpy represents the energy of a system. So, now think about it? Measuring the energy of a system? Consider all the changes and influences a system is under. This makes it really difficult (we could say impossible) to calculate the enthalpy of a system. But we do know one thing, that the enthalpy depends on two factors. ... 1 An ideal gas is defined as one of non-interacting point particles and its internal energy depends on temperature alone, thus the slope of a plot ofU$vs$T$is a straight line with slope$C_V$. A real gas (or liquid or solid) is made up of molecules and these have internal energy levels due to a molecule's ability to rotate and vibrate. At low temperatures ... 1 For a multicomponent reactive system you would rather write the enthalpy differential as $$d(nH) = \left(\frac{\partial (nH)}{\partial T}\right)_{P,n_i}dT + \left(\frac{\partial (nH)}{\partial P}\right)_{T,n_i} dP + \sum_{i = 1}^N \left(\frac{\partial (nH)}{\partial n_i}\right)_{P,T,n_{j\neq i}}dn_i$$ Where$H$is the molar enthalpy. At constant ... 2 The only way that an exothermic reaction can occur with "constant temperature" is if the heat generated is constantly removed in some way. Even if the heat removal is essentially instantaneous, it is incorrect to say that there is no temperature change. Rather there is an infinite number of infinitely small temperature changes as the system heats ... 2 My textbook says that critical temperature is the temperature above which a gas cannot be liquified no matter how much pressure we apply on it. Below the critical temperature, you can compress a gas and a second phase (liquid) will gradually form. In places with gravity, the liquid will collect at the bottom of the container. As you decrease the volume, the ... 2 One useful distinction between a liquid and a gas is that while they are in equilibrium with each other, so at the same temperature and pressure but not the same density, molecules of the liquid need to gain energy to leave the liquid phase and enter the gas phase. A surface, with surface energy, exists between the two phases, and allows them to segregate ... 1 [OP] This gives a net 0 (carboxy) + 1 (amino) + 1 (side chain) = +2 charge. This is approximately correct. See DavePhD's answer for a more accurate treatment. [OP] Why do sites such as this say that at$\mathrm{pH} = 2$, lysine's charge is only +1, not +2? The table you cite is for proteins. When lysine gets incorporated into a protein, it forms peptide ... -3 [OP] But according to the equation$dH=nC_{p}dT$, if temperature$T$is constant then$dT =0$and$dH =0$. Even if this equation would describe a system completely, a constant temperature does not imply that$dH$is zero. Instead, you can have a temperature bath of infinite size (n approaches infinity) so that the average temperature does not increase. Of ... 1 The molar heat capacity of classical ideal gases at constant volume is temperature independent, because there are not considered quantization steps of electronic, vibrational and rotational energy. The molar heat capacity ideal gases in the context of quantum-mechanic-aware gas theory is not temperature independent anymore. Because the electronic, ... 1 Assuming you meant$\ce{MnO^-4}$and not$\ce{MnO^{4-}}$, your reduction half reaction is incorrect. Because according to your equation$\ce{Mn^{2+}}$is being oxidised to$\ce{MnO^-4}$rather than being reduced. The correct half cell reactions would be: $$\ce{Zn -> Zn^{2+} + 2e-}\\ \ce{MnO^-4 + 8H+ + 5e- -> Mn^{2+} + 4H2O}\\$$ And so the overall ... 4$ \begin{align} (n_{\ce{NaOH}})_i= \pu{0.2 mmol}\\ (n_{\ce{HCl}})_i= \pu{0.1 mmol}\\ (n_{\ce{NaH2PO4}})_i= \pu{0.1 mmol}\\ (n_{\ce{Na3PO4}})_i= \pu{0.05 mmol} \end{align} $At first the$\ce{NaOH}$will react with$\ce{HCl}$as per the following reaction:$\$ \begin{align} \begin{array}{ccccc} \ce{NaOH} & + & \ce{HCl} & \ce{->} & \ce{NaCl} ...

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