16
The critical point is a point of convergence of all state properties of the respective liquid and gas. It can be considered as the degeneration point, where there is no difference between gas and liquid and this distinguishing does not make sense any more.
It can be also said the supercritical fluid near the critical point is neither gas neither liquid. It ...
14
It really does liquefy. But it does not do so in exactly the same way as you see below the critical temperature and pressure.
As an example, suppose you heat steam to 400°C and then compress it, isothermally, to 5000 bars pressure*. When you are done, you find that the water has a density and viscosity more or less similar to ordinary liquid water; what ...
5
Small magnesium metal bars are available commercially, packed with a little iron rod for starting fires. You smack the magnesium with the iron rod and get some sparks. If you have some light fluffy kindling, you are supposed to be able to get a fire going. I suppose a little steel wool would help.
If you soak some cotton balls with KNO$_3$, dry them ...
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:
$$
\begin{align}
\begin{array}{ccccc}
\ce{NaOH} & + & \ce{HCl} & \ce{->} & \ce{NaCl} ...
4
An exothermic reaction has a reduced equilibrium constant at higher $T$ because while the contribution of the change in the entropy of the system is a fixed quantity (for a small $T$ change), the effect of transferring heat to the surroundings is reduced at higher $T$ (because it causes a smaller change in the entropy of the surroundings). Mathematically ...
3
[OP] The equilibrium constant would not include the solid $\ce{I2}$, but why is this?
Let me explain this with a different example. If you have a saturated solution (e.g. lemonade with too much sugar in it) it is at equilibrium. If you add more sugar, the lemonade does not get sweeter. That tells you that the amount of solid does not matter (as long as ...
3
We normally assume that the molecules are in contact with a heat bath at a fixed temperature $T$ and then the Maxwell distibution of speeds in range $v\to v+dv$ is
$$ F(v)dv = 4\pi n\left(\frac{m}{2\pi k_BT}\right)^{3/2}v^2e^{-mv^2/2k_BT}dv$$
It rises due to the $v^2$ term and falls due to the exponential giving a curve with a gentle maximum. The average ...
3
No, it does not depend directly on the number of molecules. The expression is
$$c_{rms}=\sqrt{\frac{3RT}{M}}$$
where $T$ is the temperature in K, $M$ is the molar mass in kg/mol and $R$ is of course the gas constant in J mol-1 K-1.
However, it is possible to have a situation where the number of molecule affects the temperature of the gas, which would change ...
3
Apply an electric current to a suitable metal powder mix where, at time of employment, one adds sulfur or potassium nitrate to the open mix (that is, never placed in a confined container). This a possible option and follows the common recipe for flash powder.
However, given the desired applications of an electric current as a fire starting mechanism, ...
2
I have come up with a solution to use steel wool with electric power. Steel wool can generate sparks when electricity is passed through. I hope some fuel can take over from there to start a fire. This might have become off-topic here but I hope someone would find it useful.
Anyways, now I know it's not safe to use $\ce{KMnO4 + H2SO4}$
2
The actual volume occupied by the gas is less than the measured volume(volume of the container).
In reality gaseous particles are not point objects and don't occupy zero space, which is against one of the postulates of the Kinetic Theory of Gases.
Thus while dealing with real gases we have to account for the volume which has been occupied by the gaseous ...
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 ...
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
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 ...
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$$
You can exploit the fact that the molar amounts in the ...
1
I remember my days when I started confusing myself in the exact same way. The issue I believe is that you haven't been explained the details of how all this happens.
Evaporation:
All molecules in a liquid have various amount of energy. But the ones at the top, the ones that form the surface - they have the maximum energy in them because they are present ...
1
Enthalpy vs ∆Enthalpy
We know that enthalpy represents the energy of a system. So, now think about it? Measuring the energy of a system? Consider all the changes and influences a system is under. This makes it really difficult (we could say impossible) to calculate the enthalpy of a system.
But we do know one thing, that the enthalpy depends on two factors.
...
1
For a multicomponent reactive system you would rather write the enthalpy differential as
$$
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
$$
Where $H$ is the molar enthalpy. At constant ...
1
An ideal gas is defined as one of non-interacting point particles and its internal energy depends on temperature alone, thus the slope of a plot of $U$ vs $T$ is a straight line with slope $C_V$.
A real gas (or liquid or solid) is made up of molecules and these have internal energy levels due to a molecule's ability to rotate and vibrate. At low temperatures ...
1
The molar heat capacity of classical ideal gases at constant volume is temperature independent, because there are not considered quantization steps of electronic, vibrational and rotational energy.
The molar heat capacity ideal gases in the context of quantum-mechanic-aware gas theory is not temperature independent anymore. Because the electronic, ...
1
Assuming you meant $\ce{MnO^-4}$ and not $\ce{MnO^{4-}}$, your reduction half reaction is incorrect. Because according to your equation $\ce{Mn^{2+}}$ is being oxidised to $\ce{MnO^-4}$ rather than being reduced.
The correct half cell reactions would be:
$$
\ce{Zn -> Zn^{2+} + 2e-}\\
\ce{MnO^-4 + 8H+ + 5e- -> Mn^{2+} + 4H2O}\\
$$
And so the overall ...
1
For a container with a constant initial amount of gas at constant T, the rate of effusion decreases with time, since
$$\phi_N = \frac{\Delta P A N_A}{\sqrt{2\pi MRT}}$$
where A is the aperture, $\Delta P$ the pressure drop at the exit of the container, M is the molar mass of the gas, and T is the temperature. You can also consult a physical chemistry ...
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 ...
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