CHEMISTRY COMMUNITY

Victor Qiu 1C

K=Q only when the reaction is at equilibrium. K is the constant at equilibrium, while Q can be the constant at any time. I think this website can help: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/E...

Victor Qiu 1C

Yes, it depends on the units.

There are 4 values that are used more often:
R = 0.0821 L·atm/(mol·K) = 8.3145 J/(mol·K) = 8.2057 m3·atm/(mol·K) = 62.3637 L·Torr/(mol·K)

Victor Qiu 1C

I think there are several characteristics that are assumed for ideal gas: (1) the collisions between gas molecules are elastic and conservation of energy occurs among the molecules; (2) the total volume of the individual molecules is negligible comparing with the volume the gas occupies; (3) no inte...

Victor Qiu 1C

Because it is ClO₄^- in KClO₄ but it is Cl^- in KCl.

Victor Qiu 1C

For group 1 elements, the oxidation state is always +1. For group 2 elements, the oxidation state is always +2. Oxygen is usually -2, except in examples like H 2 O 2 and F 2 O. I think you can find more information on this website: https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplem...

Victor Qiu 1C

Yes, the activation energy is very high, so it would take very long for diamond to become graphite. There is more information about this on this website:
https://chemistry.stackexchange.com/questions/34193/are-diamonds-really-forever

Victor Qiu 1C

I think this website can help:
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Comparing_Strengths_of_Oxidants_and_Reductants

Victor Qiu 1C

Yes. I think most of data are collected at STP, which is at 101.325kPa (1atm) and 0°C (273.15K), and when we try to calculate E or G at different temperatures or pressures, we can use the compare them with the E° and G° to get the answer.

Victor Qiu 1C

I think the values are:

1 atm = 101325 Pa = 101.325 kPa = 1.013 bar = 760 torr

Victor Qiu 1C

I think the values are:
Standard Temperature and Pressure (STP): 101.325kPa (1atm), 0°C (273.15K);
Standard Ambient Temperature and Pressure (STP): 101.325kPa (1atm), 25°C (298.15K).

Victor Qiu 1C

I think if you are dealing with a chemistry problem, you can assume that the gas is an idea gas, and you can apply ideal gas law for most of the times. Nevertheless, the real characteristics of ideal gas should include: (1) the collisions between gas molecules are elastic and conservation of energy ...

Victor Qiu 1C

You can still use the equation ΔG = ΔG^。 + RTlnQ to derive.

Victor Qiu 1C

I think because combustion usually releases lots of heat, which means ΔH<0, and the reaction is very exothermic, ΔG would very likely be G<0.

Victor Qiu 1C

ΔH<0, exothermic, releases heat;
ΔH>0, endothermic, absorbs heat;
ΔG<0, exergonic, spontaneous;
ΔG>0, endergonic, not spontaneous.

Victor Qiu 1C

1 atm = 101325 Pa = 101.325 kPa = 1.013 bar = 760 torr

Victor Qiu 1C

It is not true that if there are more products than reactants, the formation of reactants would be favored. You can just think of an example of dissociation of a strong acid.
HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl^-(aq)
The products are favored even when there are more products than reactants.

Victor Qiu 1C

Kb = 1.38065 × 10⁻²³J/K

Victor Qiu 1C

There are several characteristics that are assumed for ideal gas: (1) the collisions between gas molecules are elastic and conservation of energy occurs among the molecules; (2) the total volume of the individual molecules is negligible comparing with the volume the gas occupies; (3) no intermolecul...

Victor Qiu 1C

Standard Temperature and Pressure (STP): 101.325kPa (1atm), 0°C (273.15K);
Standard Ambient Temperature and Pressure (STP): 101.325kPa (1atm), 25°C (298.15K).

Victor Qiu 1C

You can just treat other variables as constants to do the integral.
$-\int p dV=-pV+C$

Victor Qiu 1C

I think the equation is not correct. I think you should use
ΔH_f^o(Zn²⁺)+2ΔH_f^o(Cl^-)-2ΔH_f^o(HCl)
which gives
-153.89 kJ/mol + 2 × (-167.16 kJ/mol) - 2 × (-167.16 kJ/mol)
and the result is
ΔH_r = -153.89 kJ/mol

Victor Qiu 1C

ATP → ADP + Pi
ATP: adenosine triphosphate
ADP: adenosine diphosphate
Pi: phosphate ion
This reaction releases energy.

Victor Qiu 1C

0K = -273.15°C
This, if we are considering a change in temperature (ΔT), 80 units in °C equal to 80 units in K.

Victor Qiu 1C

ΔG is free energy at a moment, while ΔG^o is standard-state free energy.
ΔG = ΔG^o + RTlnQ

Victor Qiu 1C

There are several characteristics of ideal gases: (1) the collisions between gas molecules are elastic and conservation of energy occurs among the molecules; (2) the total volume of the individual molecules is negligible comparing with the volume the gas occupies; (3) no intermolecular forces exist ...

Victor Qiu 1C

It depends on the units in the problem. I think we should use the value of R whose units are the same as the units in the problem.

There are 4 values that are used more often:
R = 0.0821 L·atm/(mol·K) = 8.3145 J/(mol·K) = 8.2057 m3·atm/(mol·K) = 62.3637 L·Torr/(mol·K)

Victor Qiu 1C

Strong Acids: HCl, HBr, HI, (3 halogens)
HClO₃, HClO₄, (2 HClO_n)
HNO₃, H₂SO₄ (appears in problems very often)
Strong Bases: Group 1 hydroxides, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂, Group 1 and 2 oxides

Victor Qiu 1C

Take H₃PO₄ as an example, K_a1=7.1 × 10^-3, K_a2=6.3 × 10^-8, K_a3=4.5 × 10^-13. The difference between Ka1 and Ka2 is at the scale of 10⁵. Thus, in these cases, the second deprotonation can often be ignored.

Victor Qiu 1C

log₁₀(ab) = log₁₀(a) + log₁₀(b) is correct

Also,
if pKw = pKa + pKb
10^-pKw = 10^-(pKa+pKb)
10^-pKw = 10^-(pKa) × 10^-(pKb)
Kw = Ka × Kb

Victor Qiu 1C

I think there are 4 values that are used more often:
R = 0.0821 L·atm/(mol·K) = 8.3145 J/(mol·K) = 8.2057 m3·atm/(mol·K) = 62.3637 L·Torr/(mol·K)

Victor Qiu 1C

I don’t think the units can cancel out all the time.
If A_(aq) + B_(aq) → 2C_(aq) + D_(aq)
K_c = [C]² [D] / ([A] [B])
Because there is a square, one unit of [C] would not cancel out, and the unit of K_c should be the unit of [C].

Victor Qiu 1C

ΔH_forward=-ΔH_reverse
Thus, if you are given the ΔH for forward reaction and you wish to use the ΔH for reverse reaction, you should add a negative sign to the value.

Victor Qiu 1C

Yes, partial pressure only applies to gases. Just imagine adding a solid, a liquid, and a gas into a closed box. The solid and liquid would just occupy some volume of that box, but the gas would occupied the entire volume of that box. Therefore, only the gas would have partial pressure.

Victor Qiu 1C

It depends on the units you are using.
R = 0.0821 L·atm/(mol·K) = 8.3145 J/(mol·K) = 8.2057 m³·atm/(mol·K) = 62.3637 L·Torr/(mol·K)

Victor Qiu 1C

To give an example, heat capacity is an extensive property, while specific heat capacity is an intensive property. Heat capacity is the amount of heat given to a material with certain mass to make 1°C change in its temperature, while specific heat capacity is amount of heat needed to add to 1g of th...

Victor Qiu 1C

R = 0.0821 L·atm/(mol·K)

Victor Qiu 1C

Hello! I think posts above have already explained the first question very well, and I would just add a little bit information on the second question. The first law of thermodynamics states that energy cannot be destroyed or created (conservation of energy). Thus, it is very reasonable to think that ...

Victor Qiu 1C

I think that means that we can easily see if we can apply PV=nRT in the question.

Victor Qiu 1C

If the environment has higher temperature and lower pressure, the gas would act more ideal.

Victor Qiu 1C

K_w=1×10^-14, so that pK_w=14=pH+pOH at 25°C.

Victor Qiu 1C

Strong acids and bases would dissociate fully in water, but weak acids and bases would only dissociate partially.

Victor Qiu 1C

I think that means the equilibrium favors one direction very much. The reaction would favor where the arrow points very much.

Victor Qiu 1C

I think H₂SO₄ is an exception because its K_a2 = 1.0×10⁻², which is close to other acids' K_a1, so it should not be ignored.

Victor Qiu 1C

I think in most of the times, yes. This is because the second Ka is often much smaller than the first Ka. Take H 3 PO 4 as an example, K a1 =7.1 × 10 -3 , K a2 =6.3 × 10 -8 , K a3 =4.5 × 10 -13 . The difference between K a1 and K a2 is at the scale of 10 5 . Thus, I think the second Ka can often be ...

Victor Qiu 1C

Take H₃PO₄ as an example, K_a1=7.1 × 10^-3, K_a2=6.3 × 10^-8, K_a3=4.5 × 10^-13. The difference between K_a1 and K_a2 is at the scale of 10⁵. Thus, I think if it is not clarified specifically, the second deprotonation can often be ignored.

Victor Qiu 1C

The way to calculate K c and Q c is very similar. Both are the relative ratio of products to reactants. The only difference is that K c is the relative ratio of products to reactants when the reaction is at equilibrium, while Q c is the relative ratio of products to reactants at a given instant (the...

Victor Qiu 1C

Basically, K is the relative ratio of products to reactants when the reaction is at equilibrium, while Q is the relative ratio of products to reactants at a given instant (the reaction might not be at equilibrium).

Victor Qiu 1C

Nan_Guan_1L wrote:I think you can still draw the structure horizontally it doesn't really matter. as long as you represent the bond angle is roughly 180 degree I think it works either way you draw it.

I don't think you need to consider the bond angle when drawing a lewis structure.

Victor Qiu 1C

ΔG^o=-RTlnK

Victor Qiu 1C

I think just using same units in calculation would be fine. You can always convert units.
R=8.314 J mol^-1 K^-1=0.08205 L atm mol^-1 K^-1=62.363 mmHg L mol^-1 K^-1

Victor Qiu 1C

According to the Le Chatelier's principle, the reaction would shift to counteract the change. Therefore, if more products are added, the reaction should shift towards the reactants because in this way, there will be less products and more reactants and the reaction counteracts the change of adding m...

Victor Qiu 1C

The formula should be easy to derive using PV=nRT:
P=nRT/V, because n/V is concentration,
P=cRT,
and because when calculating equilibrium constants, the stoichiometric coefficients should be the exponents, so take that into consideration,
K_P=(RT)^ΔnK_c, Δn=n_Products-n_Reactants

Victor Qiu 1C

I don't think the unit of pressure should must be in atm. I think just using same units in calculation would be fine.
R=8.314 J mol^-1 K^-1=0.08205 L atm mol^-1 K^-1=62.363 mmHg L mol^-1 K^-1

Victor Qiu 1C

I think you should clarify if the K is K c or K P . I think if you are referring to K c , you should not assume that P/R is directly proportional to K c because when calculating K c , the concentrations are the bases while the stoichiometric coefficients are the exponents. The stoichiometric coeffic...

Victor Qiu 1C

I think for d orbitals, there are d_xy, d_yz, d_xz, d_x²-y², and d_z².

Victor Qiu 1C

Because it should follow the order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 5f 6d...

Victor Qiu 1C

It should follow the order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 5f 6d... However, I think [Ar] 3d¹4s² and [Ar] 4s²3d¹ are the same, and it does matter whether 4s or 3d is written first. They both represent the electron configuration of Sc.

Victor Qiu 1C

Valence shell configuration refers to the electrons in the outermost shell. The alkali earth metals are beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra), and the respective electron configurations are [He] 2s 2 , [Ne] 3s 2 , [Ar] 4s 2 , [Kr] 5s 2 , [Xe] 6s 2...

Victor Qiu 1C

The electron configuration of Zn is [Ar] 3d¹⁰ 4s². Therefore, we can infer that Zn would likely to lose the 2 electrons in 4s² and form Zn²⁺. Therefore, as 2+2×(-1)=0, we should write Zn(OH)₂ because it's Zn²⁺ and 2OH^-.

Victor Qiu 1C

I think Br₃^- is tribromide, while Br^- is bromide.

Victor Qiu 1C

I think Kevin Wright’s answer is correct. Just to add on it, Xe forms 3 double bonds with 3 O, respectively, and has a lone pair electrons. Also, each O has 2 pairs of lone pair electrons.

Victor Qiu 1C

I think its H — C: — H, and it’s called methylene.

Victor Qiu 1C

C has 4 valences electrons, O has 6 valence electrons. If there is a double bond in CO, C still needs 2 pairs of lone pair electrons to satisfy octet rule, and O also needs 2 pairs of lone pair electrons to satisfy the octet rule. Therefore, in total there should be 2×2+2×2+2×2=12 electrons. However...

Victor Qiu 1C

Valence electrons are the electrons in the outer shell. If we use element carbon as an example. Carbon has the electron configuration [He]2s²2p². We can see that its outer shell is n=2, which is 2s and 2p. Thus, its valence electrons are 2+2=4.

Victor Qiu 1C

I do not think there is a reason. When drawing the Lewis structure, the same bond can be drawn in various ways (the bond angle does not matter). I think this is because Lewis structure is 2-D while the molecule is 3-D, therefore, there has to be some kind of distortion when trying to reflect the rea...

Victor Qiu 1C

To add on NeelPatelLec4's answer, for seesaw, the bond angle between axial and axial is 180°, between equatorial and equatorial is 120°, and between axial and equatorial is 90°. For trigonal pyramidal, the bond angle should be less than 109.5° due to the repulsion from the lone pair electrons.

Victor Qiu 1C

For seesaw structure, the bond angle between axial and axial is 180°, between equatorial and equatorial is 120°, and between axial and equatorial is 90°. For square planar structure, the bond angles should be 90°. These are the bond angles in ideal situations.

Victor Qiu 1C

Generally, lone pair electrons would make the bond angle to be smaller, because the lone pair electrons would have a repulsion force. For example, in a water molecule, there are 2 pairs of lone pair electrons, which makes the bond angle H-O-H to become 104.5°.

Victor Qiu 1C

I think molecules that are AX₂E₁ or AX₂E₂ have a bent shape.

Victor Qiu 1C

Yes, that means valence shell electron pair repulsion theory. Thus VSEPR theory.

Victor Qiu 1C

Yes, the amount is determined, like for tetrahedral, there are always 4 atoms bonding to central atom and no lone pair electrons at central atom. However, I think it requires much work to memorize these. You can just draw the Lewis structure and determine the shape.

Victor Qiu 1C

I think we also have square pyramidal, pentagonal bipyramidal, pentagonal pyramidal, and others.

Victor Qiu 1C

Glycine is an amino acid. Amino acid always has an amine group, a carboxyl group, and a side chain (R group). Also, There is an α carbon at the center that is connected to the amine group, the carboxyl group, and the side chain. I think if we know that glycine follows this structure, it would be pre...

Victor Qiu 1C

The Cl atom in the middle forms 2 single bonds, and it has 3 lone pair electrons. Thus, the answer should be 10 valence electrons.

Victor Qiu 1C

I always use the order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p 8s...

Victor Qiu 1C

Just as what is said in the Dr. lavelle's lecture, London Dispersion Force always exists because electrons always exist, and electrons always fluctuate.

Victor Qiu 1C

OH^- is not radical, •OH is radical.

Victor Qiu 1C

We can consider how the element would lose or gain electrons in its outer shell. For example, Mg is [Ne] 3s², so it has the tendency to lose 2 electrons, which makes Mg ion Mg²⁺, and Al is [Ne] 3s²3p¹, so it has the tendency to lose 3 electrons, which makes Al ion Al³⁺.

Victor Qiu 1C

We can consider the angular momentum quantum number. Electrons in s-, p-, d-, and f- orbitals have the angular momentum quantum number l = 0, 1, 2, and 3.

Victor Qiu 1C

Yes, there is an easy method. You can simply distinguish the s-,p-, d-, and f- blocks by thinking about the graph. You can try to think like you want to cut the periodic table into rectangles. First you move He to the left to make a thin rectangle, making a 2×7 rectangle that represents s-block. The...

Victor Qiu 1C

I would say yes, but one thing to be noticed is that, for example, NaCl(s) is a salt, but when it is dissolved in water, I would not call it a salt, because in water there are Na⁺ and Cl^-, I would just call them sodium ion and chloride ion.

Victor Qiu 1C

This is because when we count valence electrons, we only count the electrons that are in the outer shell. The electrons whose principle quantum number is the greatest are in the outer shell. What does that mean? The electron configuration for Sb is [Kr]4d 10 5s 2 5p 3 , 5>4, that means the electrons...

Victor Qiu 1C

It follows the same principle, but there are some tips. Carbon is always on the backbone, and for -COOH, it is always like C forms a single bond with an O, and for this O, it forms another single bond with H and has 2 pairs of electrons surrounding it, and for another O, it forms a double bond with ...

Victor Qiu 1C

Br has 7 valence electrons while O has 6, and because it is BrO - , there should be 7+6+1=14 electrons in total. If there is a double bond, there should be 10 electrons that do not form bonds surrounding Br and O. When there is a double bond, Br allows 2 pairs of electrons around it, and O also allo...

Victor Qiu 1C

I think you can just add "Because electron shells are integers, the number is ...".

Victor Qiu 1C

Examples should include
Cu [Ar]3d¹⁰4s¹
and the elements around 4d and 5s
Y [Kr] 4d¹5s²
Zr [Kr] 4d²5s²
Nb [Kr] 4d⁴5s¹
Mo [Kr] 4d⁵5s¹
Tc [Kr] 4d⁵5s²
Ru [Kr] 4d⁷5s¹
Rh [Kr] 4d⁸5s¹
Pd [Kr] 4d¹⁰
Ag [Kr] 4d¹⁰5s¹
Cd [Kr] 4d¹⁰5s²

Victor Qiu 1C

It is always helpful to memorize the elements right after the noble gas elements.
Li [He]2s¹
Na [Ne]3s¹
K [Ar]4s¹
Rb [Kr]5s¹
Cs [Xe]6s¹
Fr [Rn]7s¹

Victor Qiu 1C

Yes, I agree with Sydney_Lam3G. For example, we can just write 1s²2s²2p⁶ for Ne.

Victor Qiu 1C

For example, if 1s²2s¹2p³ is the excited state, 1s²2s²2p² would be the ground state.

Victor Qiu 1C

If an electron jumps from a high state of energy to a low energy state, the change in energy is negative because this process release energy. However, when we use the energy later to calculate either the frequency or the wavelength, we usually do not include the negative sign, or you can understand ...

Victor Qiu 1C

It would go as 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p

Victor Qiu 1C

I think Jordan Young's answer is correct. To make an example, we would write [Ar]3d¹⁰4s¹ for Cu instead of 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

Victor Qiu 1C

I think Megan Chan's answer is correct. Just to add on it, neon's electron configuration can also be represented as [He]2s²2p⁶

Victor Qiu 1C

Because it's one dimensional, there will only be lines. Imagine that there is only an x-axis. The box will only have a line segment that represents its diameter.

Victor Qiu 1C

Yes, the Rydberg Equation works for Hydrogen only, because it is based on Bohr's model which is specifically about the electrons in hydrogen atoms. There is a more complex equation 1/λ = RZ 2 ((1/n 1 2 ) - (1/n 2 2 )) that can be used for other one-electron (meaning one electron affected by nuclear ...

Victor Qiu 1C

If an electron jumps from a high state of energy to a low energy state, the change in energy is negative because this process release energy. However, when we use the energy later to calculate either the frequency or the wavelength, we usually do not include the negative sign, or you can understand ...

Victor Qiu 1C

What Ashley said is correct. In short, energy per photon needs to be great enough so that it can "take away" an electron. Therefore, if a light cannot eject electrons, that means per photon energy is not great enough. Lower wavelength means greater energy. If a light with lower wavelength ...

Victor Qiu 1C

So our eyes work like this: after light hits the retina, photoreceptors turn the light into electrical signals, and our brain tells us that "there is light". Therefore, when at excited state, there is no photon emitted, which means there is no light hitting the retina. Therefore, it's not ...

Victor Qiu 1C

1/λ = R_H (1/n₁² - 1/n₂²)
1/λ × 1/R_H = (1/n₁² - 1/n₂²)
(1/n₁² - 1/n₂²) = 1/(4.34 × 10^-7) × 1/(1.096776 × 10⁷) = 0.21
0.21 = 1/4 - 1/25
4 = 2², 25 = 5²
Therefore, n₂ = 5, n₁ = 2

Victor Qiu 1C

Victor, that's a novel way of thinking about confidence intervals and I definitely never had that thought occur to me. It took me a couple tries, but I think I understood what you're trying to get at in general. which makes the former more precise while the latter more accurate I'm confused about t...

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