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## How does radiation change across the electromagnetic spectrum?

Daniel W.
Featured 14 hours ago

Three things change: frequency, wavelength and method of production.

#### Explanation:

The electromagnetic spectrum is a family of waves that have the same essential nature (they are vibrations in electric and magnetic fields) but different wavelength/frequency and in some cases a different method of production.

The spectrum is often represented by a continuous sinusoidal wave that has a shortening wavelength toward one end – see the diagram below for this.

Note speed is the same for all electromagnetic waves and as frequency increases the wavelength decreases.

As you can see radio waves have the longest wavelengths and gamma-rays have the shortest. The spectrum is divided into seven major regions according to wavelength: radio, microwaves, infrared, visible light, ultraviolet, x-rays and γ-rays.

Methods of production
The waves in some of the regions are produced differently to other regions. For example whereas microwaves are produced by oscillating electrons in wires/aerials visible light is produced when electrons in atoms drop down energy levels. Here is a table with rough wavelength ranges for the regions with their methods of production.

The method of production is an important distinction where there is an overlap of wavelength. For example the x-ray and γ-ray regions overlap in wavelength; some x-rays have the same wavelength as some γ-rays. But we still call x-rays x-rays and not γ-rays because they were produced by a linear accelerator and not by nuclear decay.

Another thing to be aware of:
Frequency is related to wave energy, so high frequency γ-rays have a higher energy than all other electromagnetic waves. The frequency of waves in the UV part of the spectrum and above means they have enough energy to remove electrons from stable atoms. That makes UV, x-rays and γ-rays ionising radiation.

## According to the following reaction, how much energy is evolved during the reaction of 0.111 mol B2H6 and 0.252 mol Cl2 (Both gases are initially at STP)? The molar mass of B2H6 is 27.67 g/mol. B2H6(g) + 6 Cl2(g) → 2 BCl3(g) + 6 HCl(g) ΔH°rxn = -1396 kJ

Truong-Son N.
Featured 2 days ago

I'm going to guess it has something to do with which reagent you used as the limiting reagent, and the value you used for your enthalpy of reaction.

The limiting reagent, the one that is used up first, dictates how much product you will make. Here is a fairly quick way to figure out which reagent it is.

THE LIMITING REAGENT

You see how the equation you gave asks for one equivalent of ${\text{B"_2"H}}_{6} \left(g\right)$ (diborane)? Now compare to how many equivalents you need of ${\text{Cl}}_{2} \left(g\right)$ relative to the amount of ${\text{B"_2"H}}_{6} \left(g\right)$.

The coefficient in front of ${\text{B"_2"H}}_{6}$ implicitly says $1$, but the one in front of ${\text{Cl}}_{2} \left(g\right)$ says $6$... you need six equivalents.

Now, do you ACTUALLY have six times the number of $\text{mol}$s ${\text{Cl}}_{2} \left(g\right)$ as you have ${\text{B"_2"H}}_{6} \left(g\right)$? No, you have $\frac{0.252}{0.111} = \setminus m a t h b f \left(2.27\right)$ times as much as you have ${\text{B"_2"H}}_{6}$.

So, chlorine must be the limiting reagent. You don't have enough of it to react with that much diborane.

MOLAR ENTHALPY OF REACTION $\setminus m a t h b f \left(\Delta {H}_{\text{rxn}}^{\circ}\right)$ VS HEAT FLOW $\setminus m a t h b f \left(q\right)$

Ultimately, we need to determine the heat $\setminus m a t h b f \left(q\right)$ that is released from this reaction.

We have the molar enthalpy of reaction defined for ${25}^{\circ} \text{C}$ and $\text{1 bar}$ (not ${0}^{\circ} \text{C}$!!), $\textcolor{g r e e n}{\Delta {H}_{\text{rxn"^@ = -"1396 kJ/mol B"_2"H}} _ 6 \left(g\right)}$.

It is defined this way because the reaction you are referencing is based on $\setminus m a t h b f \left(\text{1 mol}\right)$ of $\setminus m a t h b f \left({\text{B"_2"H}}_{6}\right)$ and conventionally, the proper units are $\text{kJ/mol}$.

The units are NOT just $\text{kJ}$. That will matter!

The molar enthalpy of reaction is equal to the heat energy released from the reaction in constant pressure conditions divided by the $m o l$s of limiting reagent.

And there we go, that's where the limiting reagent comes in. :)

#\mathbf(DeltaH_"rxn"^@ = (q_"rxn")/("mols Limiting Reagent"))#

The idea is that you can:

1. Conveniently use a referenced $\Delta {H}_{\text{rxn}}^{\circ}$ from thermodynamic tables, which is on a per-mol basis (hence, $\text{kJ/mol}$), defined for specific atmospheric conditions (${25}^{\circ} \text{C}$ and $\text{1 bar}$), and based on the particular reagent which has a stoichiometric coefficient of $\setminus m a t h b f \left(1\right)$. In this case it is ${\text{B"_2"H}}_{6} \left(g\right)$.

2. Recognize that the enthalpy as-written ($- \text{1396 kJ/mol}$) is still in reference to $\setminus m a t h b f \left({\text{B"_2"H}}_{6}\right)$ being the limiting reagent, but we know that it is NOT the limiting reagent. So, we convert it to accommodate for that.

3. Scale it down to the size of your real reaction at the same atmospheric conditions to accommodate for the fact that you are not using exactly $\text{1 mol}$ of ${\text{Cl}}_{2} \left(g\right)$, but $\text{0.252 mol}$.

SCALING THE STANDARD REACTION DOWN TO YOUR REACTION

Since we currently are not talking about ${\text{B"_2"H}}_{6}$ as the limiting reagent, but about ${\text{Cl}}_{2} \left(g\right)$, we need to convert to get the accurate value for the molar enthalpy of reaction that is relative to ${\text{Cl}}_{2} \left(g\right)$.

$\textcolor{g r e e n}{\Delta {H}_{\text{rxn"^@) = (-"1396 kJ")/(cancel("1 mol B"_2"H"_6(g))) xx (cancel("1 mol B"_2"H"_6(g)))/("6 mols Cl}} _ 2 \left(g\right)}$

$= \textcolor{g r e e n}{- \text{232.67 kJ/mol}}$

(of $\textcolor{g r e e n}{{\text{Cl}}_{2} \left(g\right)}$)

That was the first step to the scaling: making sure you are referring to $\setminus m a t h b f \left(\text{1 mol}\right)$ of limiting reagent in your standard reaction.

Next, we can scale the enthalpy of reaction down to the molar heat released.

#(q_"rxn")/"mols Limiting Reagent" = DeltaH_"rxn"^@#

${q}_{\text{rxn" = DeltaH_"rxn"^@ xx "mols Limiting Reagent}}$

$\textcolor{b l u e}{{q}_{\text{rxn") = -"232.67 kJ/"cancel("mol Cl"_2(g)) xx "0.252" cancel("mols Cl}} _ 2 \left(g\right)}$

For this second scaling, you should see that we basically multiplied the enthalpy in $\text{kJ}$ by $\frac{0.252}{1}$, and you had $\text{0.252 mol}$s of ${\text{Cl}}_{2} \left(g\right)$.

So, what we've done is scaled the reaction down twice:

1. So that the amount of chlorine, the limiting reagent, referred to by the enthalpy of reaction is $\text{1 mol}$ instead of $\text{6 mol}$s.
2. So that the amount of chlorine used is $\text{0.252 mol}$ instead of $\text{1 mol}$.

The heat of reaction is thus:

$\textcolor{b l u e}{{q}_{\text{rxn}}} =$ $\textcolor{b l u e}{- \text{58.63 kJ}}$

Or, you could say that $\setminus m a t h b f \left(58.63 k J\right)$ of heat was released.

## What are the branches of chemistry and their definition?

Ernest Z.
Featured 1 week ago

The five major branches of chemistry are organic, inorganic, analytical, physical, and biochemistry. These divide into many sub-branches.

#### Explanation:

ORGANIC CHEMISTRY

Organic chemistry involves the study of the structure, properties, and preparation of chemical compounds that consist primarily of carbon and hydrogen.

Organic chemistry overlaps with many areas including

• Medicinal chemistry —the design, development, and synthesis of medicinal drugs. It overlaps with pharmacology (the study of drug action).
• Organometallic chemistry — the study of chemical compounds containing bonds between carbon and a metal.
• Polymer chemistry — the study of the chemistry of polymers.
• Physical organic chemistry — the study of the interrelationships between structure and reactivity in organic molecules.
• Stereochemistry — the study of the spatial arrangements of atoms in molecules and their effects on the chemical and physical properties of substances.

INORGANIC CHEMISTRY

Inorganic chemistry is the study of the properties and behaviour of inorganic compounds.

It covers all chemical compounds except organic compounds.

Inorganic chemists study things such as crystal structures, minerals, metals, catalysts, and most elements in the Periodic Table.

Branches of inorganic chemistry include:

• Bioinorganic chemistry — the study of the interaction of metal ions with living tissue, mainly through their direct effect on enzyme activity.

• Geochemistry — the study of the chemical composition and changes in rocks, minerals, and atmosphere of the earth or a celestial body.

• Nuclear chemistry — the study of radioactive substances.

• Organometallic chemistry — the study of chemical compounds containing bonds between carbon and a metal.

• Solid-state chemistry — the study of the synthesis, structure, and properties of solid materials.

ANALYTICAL CHEMISTRY

Analytical chemistry involves the qualitative and quantitative determination of the chemical components of substances.

Examples of areas using analytical chemistry include:

• Forensic chemistry — the application of chemical principles, techniques, and methods to the investigation of crime.

• Environmental chemistry —the study of the chemical and biochemical phenomena that occur in the environment.It relies heavily on analytical chemistry and includes atmospheric, aquatic, and soil chemistry.

• Bioanalytical Chemistry — the examination of biological materials such as blood, urine, hair, saliva, and sweat to detect the presence of specific drugs.

PHYSICAL CHEMISTRY

Physical Chemistry —the study of the effect of chemical structure on the physical properties of a substance.

Physical chemists typically study the rate of a chemical reaction, the interaction of molecules with radiation, and the calculation of structures and properties.

Sub-branches of physical chemistry include:

• Photochemistry — the study of the chemical changes caused by light.

• Surface chemistry — the study of chemical reactions at surfaces of substances. It includes topics like adsorption, heterogeneous catalysis, formation of colloids, corrosion, electrode processes, and chromatography.

• Chemical kinetics — the study of the rates of chemical reactions, the factors affecting those rates, and the mechanism by which the reactions proceed.

• Quantum chemistry — the mathematical description of the motion and interaction of subatomic particles. It incorporates quantization of energy, wave-particle duality, the uncertainty principle, and their relationship to chemical processes.

• Spectroscopy — the use of the absorption, emission, or scattering of electromagnetic radiation by matter to study the matter or the chemical processes it undergoes.

BIOCHEMISTRY

Biochemistry is the study of chemical reactions that take place in living things. It tries to explain them in chemical terms.

Biochemical research includes cancer and stem cell biology, infectious disease, and cell membrane and structural biology.

It spans molecular biology, genetics, biochemical pharmacology, clinical biochemistry, and agricultural biochemistry.

• Molecular biology — the study of the interactions between the various systems of a cell, such as the different types of DNA, RNA, and protein biosynthesis.

• Genetics — the study of genes, heredity, and variation in living organisms.

• Pharmacology — the study of mechanisms of drug action and the influence of drugs on an organism.
o Toxicology —a sub-branch of pharmacology that studies the effects of poisons on living organisms.

• Clinical biochemistry — the study of the changes that disease causes in the chemical composition and biochemical processes of the body.

• Agricultural biochemistry — the study of the chemistry that occurs in plants, animals, and microorganisms.

Thus, although there are FIVE main branches of chemistry, there are many sub-branches.

There is a huge overlap between Chemistry and Biology, Medicine, Physics, Geology, and many other disciplines.

Chemistry really is THE CENTRAL SCIENCE.

## How would you compare the strengths between ionic bonds, covalent bonds, hydrogen bonds, and van der Waals forces?

James C.
Featured 1 week ago

In relation to each other, covalent bonds are the strongest, followed by ionic, hydrogen bond, Dipole-Dipole Interactions and Van der Waals forces (Dispersion Forces).

#### Explanation:

Covalent Bonds :

These bonds are the strongest out of the list. These are referred to as intramolecular bonds, whilst the rest are referred to as intermolecular forces.

• Covalent bonds are the bonds between atoms created when the atoms share electrons. These bonds create such stability in the atoms, that they tend to be difficult to break, subsequently making them very strong.

• For example, diamonds are very strong, hard and have a extremely high melting point due to the the structure of the covalent network and the distribution of the shared electrons between the carbon atoms (covalent bond).

Ionic Bonds:

These bonds are the second strongest out of the list. These occur due to:

• Positive and negative ions attracting one another and binding together forming a new substance. This is called ionic bonding.

• For example, sodium chloride consists of Na+ ions and Cl- ions bound together.

• These are strong and kept together by the electro-static attraction between the negative and positively charged ions.

Hydrogen Bonds:

Hydrogen bonds are not accually bonds, and are accually a specific type of dipole-dipole interaction. They are stronger than common dipole-dipole They occur due to:

• Molecules having a permanent net dipole (difference in electronegativity) from hydrogen being covalently bonded to either fluorine, oxygen or nitrogen.

• For example, hydrogen bonds operate between water, ammonia and hydrogen fluoride.

Dipole-Dipole Interactions:

I added this to the explanation even thought it wasn't in the question as i thought it is a very important force in chemistry. Dipole-Dipole Interactions are stronger than Dispersion forces. They occur due to:

• Dipole-Dipole forces form when there is a large difference in electronegativity between two atoms joined by a covalent bond. The atoms share the electrons unequally.

• The more electronegative atom pulls the shared electrons toward itself. That means that one atom has a partial negative charge, and the other atom has a partial negative charge.

• For example, H-Cl has a dipole due to the difference in electronegativity between the hydrogen atom and chlorine atom. The electrons spend more of their time near the Cl atom.

• A dipole-dipole interaction is the attraction between two polar molecules. When they approach each other, the negative end one molecule attracts the positive end of the other.

Dispersion forces (Van der Waals forces):

Dispersion forces are very weak forces of attraction. They occur due to:

• Momentary dipoles occurring due to uneven electron distributions in neighbouring molecules as they approach one another.

• The weak residual attraction of the nuclei in one molecule for the electrons in a neighbouring molecule.

• The more electrons that are present in the molecule, the stronger the dispersion forces will be.

• Dispersion forces are the only forces in non-polar molecules.

## How many molecular orbitals will be formed by combination of the 3s and 3p atomic orbitals in 1.0 mol of Mg atoms? At 0 K, what fraction of these orbitals will be occupied by electron pairs?

Truong-Son N.
Featured 2 weeks ago

a) This is asking you to apply the Linear Combination of Atomic Orbitals (LCAO).

LINEAR COMBINATION OF ATOMIC ORBITALS (LCAO)

The idea is that $\setminus m a t h b f \left(n\right)$ number of atomic orbitals (AOs) yields $\setminus m a t h b f \left(n\right)$ number of molecular orbitals (MOs).

We can see this in context when we form the theoretical dimetal ${\text{Mg}}_{2}$ (just pretend it's in the gas phase if it bothers you to think about this; ${\text{Na}}_{2} \left(g\right)$ exists at very high temperatures):

1. Two $3 s$ AOs (one from each $\text{Mg}$) combine to form one ${\sigma}_{3 s}$ bonding MO and one ${\sigma}_{3 s}^{\text{*}}$ antibonding MO. That's a 2:2 conversion.
2. Six $3 p$ AOs (three from each $\text{Mg}$) combine to form one ${\pi}_{3 p x}$ bonding MO and one ${\pi}_{3 p x}^{\text{*}}$ antibonding MO, one ${\pi}_{3 p y}$ bonding MO and one ${\pi}_{3 p y}^{\text{*}}$ antibonding MO, and one ${\sigma}_{3 p z}$ bonding MO and one ${\sigma}_{3 p z}^{\text{*}}$ antibonding MO. Sum that up and you get a 6:6 conversion.
3. Since we used two $3 s$ and six $3 p$ AOs, that is $2 + 6 =$ eight AOs. We got out eight MOs---one each of the following: ${\sigma}_{3 s}$, ${\sigma}_{3 s}^{\text{*}}$, ${\pi}_{3 p x}$, ${\pi}_{3 p x}^{\text{*}}$, ${\pi}_{3 p y}$, ${\pi}_{3 p y}^{\text{*}}$, ${\sigma}_{3 p z}$, and ${\sigma}_{3 p z}^{\text{*}}$.

$\setminus m a t h b f \left(M {g}_{2}\right)$ MO DIAGRAM

Although the $3 p$ AOs are empty, as seen in the electron configuration for magnesium atom:

$\left[\text{Ne}\right] 3 {s}^{2}$

... pictorially this nevertheless looks like this:

With the molecular electronic configuration like so:

#(sigma_(1s))^2(sigma_(1s)^"*")^2(sigma_(2s))^2(sigma_(2s)^"*")^2color(blue)((sigma_(3s))^2(sigma_(3s)^"*")^2)#

where blue indicates the valence orbitals.

Naturally, this is for one ${\text{Mg}}_{2}$ diatomic molecule. We could have extended this to $n$ $\text{Mg}$ atoms, but I wanted to keep it as simple as possible.

For one $\text{Mg}$ atom, on the other hand (instead of two), divide the number of MOs by two to get four. Then, since we are talking about $\text{1 mol}$ of $\text{Mg}$ atoms, multiply the number of MOs/atom by $6.0221413 \times {10}^{23}$ to get:

$\setminus m a t h b f \left(\text{4 mol}\right)$s of $\text{MOs/Mg atom}$

b) Okay, so some terminology from Band Theory. It's not too bad.

THE HOMO-LUMO GAP

The Fermi level is where the highest-occupied molecular orbital (HOMO) currently lies at $\text{0 K}$.

That means the ${\sigma}_{3 s}^{\text{*}}$ antibonding orbital, the HOMO, is at the "Fermi level", the energy level that is closest to the band gap. The most likely electronic transition occurs from the HOMO to the lowest-unoccupied molecular orbital (LUMO), across this band gap.

This event constitutes conduction, and this energy gap is also called the HOMO-LUMO gap. The HOMO-LUMO gap is small for very conductive metals, as seen in the above diagram.

(When electron promotion occurs, it is said that each electron that moves into the empty orbitals above leaves a "hole" in the filled orbitals below the Fermi level, which is what is depicted on the right portion of the above diagram.)

FRACTION OF MOS FILLED

When we are at $\text{0 K}$, no electrons have been promoted yet (due to any thermal energy imparted due to an increase in temperature, for instance), so all the electrons in the MO diagram above are where they should be at $\text{0 K}$.

Having $2$ valence electrons per $\text{Mg}$ atom, we can fill $1$ MO. With $4$ MOs formed from the LCAO event, we would have needed $8$ valence electrons to fill $4$ MOs.

Hence, with $\text{1 mol}$ of $\text{Mg}$ atoms, we would fill $\text{1 mol}$ of MOs, out of the $\text{4 mol}$s we actually have.

Therefore, 2/8 = 25% of the MOs are occupied by electron pairs, i.e. filled.

## Calculate the pH of the solution after 5.0ml of 0.125 M KOH has been added to the 25.0mL solution of 0.250 M benzoic acid? Calculate the pH after 50.0ml of 0.125 M KCL has been added to the solution?Calculate the pH at the equivalence point?

Michael
Featured 2 weeks ago

(a)

$p H = 2.46$

(b)

$p H = 2.67$

(c)

$p H = 8.45$

#### Explanation:

a)

The reaction is:

${C}_{6} {H}_{5} C O O {H}_{\left(a q\right)} + K O {H}_{\left(a q\right)} \rightarrow {C}_{6} {H}_{5} C O O {K}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

The no. moles of benzoic acid is given by:

${n}_{b a} = c \times v = 0.25 \times \frac{25}{1000} = 0.00625$

The no. moles $K O H$ is given by:

${n}_{K O H} = 0.125 \times \frac{5}{1000} = 0.000625$

So the number of moles of benzoic acid remaining after some has been neutralised is given by:

${n}_{b a} = 0.00625 - 0.000625 = 0.005625$

The new total volume = $25 + 5 = 30 \text{ml"=0.03"L}$

So the new concentration of benzoic acid is given by:

$\left[{C}_{6} {H}_{5} C O O H\right] = {n}_{b a} / v = \frac{0.005625}{0.03} = 0.1875 \text{mol/l}$

Benzoic acid is a weak acid which dissociates according to the general equation:

$H X r i g h t \le f t h a r p \infty n s {H}^{+} + {X}^{-}$

Given its $p {K}_{a}$ value a handy expression to find its $p H$ is :

#pH=1/2(pK_a-log["Acid"])" "color(red)((1))#

$\therefore p H = \frac{1}{2} \left[4.2 - \log \left(0.1875\right)\right]$

$p H = \frac{1}{2} \left[4.2 - \left(- 0.7267\right)\right]$

$p H = 2.463$

(b)

The question is not clear if the KCl solution is added before or after the KOH. I'll assume its added after the KOH solution.

I will assume that the addition of $K C {l}_{\left(a q\right)}$ just dilutes the solution.

The new total volume =$5 + 25 + 50 = 80 \text{ml"=0.08"L}$

$\therefore \left[{C}_{6} {H}_{5} C O O H\right] = \frac{0.005625}{0.08} = 0.0703 \text{mol/l}$

Using $\textcolor{red}{\left(1\right)}$ gives:

$p H = \frac{1}{2} \left[4.2 - \log \left(0.0703\right)\right] = \frac{1}{2} \left[4.2 - \left(- 1.152967\right)\right]$

$p H = \frac{1}{2} \left(5.352967\right)$

$p H = 2.676$

So the dilution has raised the $p H$ slightly.

(c)

This neutralisation reaction is between a weak acid and a strong base.

This means that, even if they combine in the ratios specified by the stoichiometry of the equation, the resulting solution may not be neutral.

This is due to "salt hydrolysis". The benzoate ions formed formed are basic due to:

${C}_{6} {H}_{5} C O {O}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s {C}_{6} {H}_{5} C O O H + O {H}^{-}$

So the resulting solution at the equivalence point is slightly alkaline.

(This is why old fashioned soap - which was the salt of a weak acid and strong base - could sting your eyes)

To find the $p H$ of the resulting solution we need to find ${K}_{b}$ for the benzoate ion. We get this using:

$p {K}_{w} = p {K}_{a} + p {K}_{b}$

${K}_{w}$ is the ionic product of water and $= {10}^{- 14} {\text{mol"^2"/""l}}^{- 2}$ @298K.

$\therefore p {K}_{b} = 14 - 4.2 = 9.8$

Now we use the equivalent expression for bases as in $\textcolor{red}{\left(1\right)} :$

#pOH=1/2(pK_b-log["base"])" "color(red)((2))#

To find $\left[b a s e\right]$ which is $\left[\text{benzoate}\right]$ we know that we started with $0.00625$ moles of benzoic acid which, when neutralised gives $0.00625$ moles of benzoate ions.

This would therefore require $0.00625$ moles of $K O H$.

Since $c = \frac{n}{v}$ then $v = \frac{n}{c} = \frac{0.00625}{0.125} = 0.05 \text{L"=50"ml}$ of $K O H$

So the total volume at the equivalence point = $25 + 50 + 50 = 125 \text{ml"=0.125"L}$ (I added the 50ml of KCl)

#:.["benzoate"]=0.00625/0.125=0.05"mol/l"#

So from $\textcolor{red}{\left(2\right)} \Rightarrow$

$p O H = \frac{1}{2} \left[9.8 - \log \left(0.05\right)\right]$

$p O H = \frac{1}{2} \left(11.1\right)$

$p O H = 5.5$

$\therefore p H = 14 - 5.5 = 8.45$

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