Search found 66 matches
- Sun Jan 24, 2021 5:47 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium concentrations for acids
- Replies: 3
- Views: 11
Re: Equilibrium concentrations for acids
Yes, you would have to use Ka. If you were given Kb when you are solving for an acid, make sure you use that value to find the Ka using the formula (Ka)(Kb) = 10^-14
- Sun Jan 24, 2021 1:29 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Friday Lecture Question
- Replies: 1
- Views: 10
Re: Friday Lecture Question
It has to be converted from a liquid to gas because the table of bond enthalpies refers to breaking bonds in gases; therefore, you would need to add the enthalpy of phase change when something is in the liquid or solid phase.
- Sun Jan 24, 2021 1:19 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: When enthalpy is zero
- Replies: 4
- Views: 18
Re: When enthalpy is zero
Since it is the most stable state, the formation would not require any energy as compounds like O2 form naturally.
- Sat Jan 23, 2021 5:40 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Do concentrations change at equilibrium?
- Replies: 9
- Views: 38
Re: Do concentrations change at equilibrium?
Since the rates of the reactions are equal, the concentrations of the products and reactants at equilibrium should not be changing and are constant at equilibrium.
- Sat Jan 23, 2021 5:36 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Strong vs. weak acids and bases
- Replies: 3
- Views: 13
Re: Strong vs. weak acids and bases
I think it will be easier just to memorize the strong acids and bases since there are not a lot of them. All of the other acids and bases that are not strong can be considered weak.
- Sun Jan 17, 2021 1:15 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Q > K, Q< K
- Replies: 25
- Views: 75
Re: Q > K, Q< K
K is calculated when the reaction is at equilibrium that is why it is a constant and will not change. Q can be calculated at any time during the reaction. The value of Q is compared with K to show which direction the reaction will proceed or if the reaction is at equilibrium. When Q>K then there are...
- Sat Jan 16, 2021 11:19 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: ka vs kb
- Replies: 20
- Views: 60
Re: ka vs kb
Ka is the acidity constant and is found when calculating the equilibrium constant for weak acids. Kb (basicity constant) is the same thing but just for weak bases.
- Sat Jan 16, 2021 11:14 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Ka x Kb
- Replies: 11
- Views: 33
Re: Ka x Kb
In the autoprotolysis of water, the concentrations of H3O+ and OH- are both 1.0 X 10^-7. When these values are multiplied, you get 10^-14.
- Sat Jan 16, 2021 10:21 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE Box
- Replies: 5
- Views: 27
Re: ICE Box
To use the ICE box you would need the balanced chemical equation, and ICE corresponds to initial molar concentration, change in molar concentration, and equilibrium molar concentration. For the initial values, you should be given some information about the concentrations. If you were given the initi...
- Sat Jan 16, 2021 9:18 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Autoprotolysis
- Replies: 6
- Views: 36
Re: Autoprotolysis
Autoprotolysis is a reaction that occurs when a proton is transferred between two identical molecules. This reaction can occur in amphiprotic (proton acceptor and donor) solvents like water.
- Sun Jan 10, 2021 8:10 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: K values
- Replies: 4
- Views: 25
Re: K values
When the value of the equilibrium constant is between these values, it means that neither the reactants nor the products are strongly favored.
- Sun Jan 10, 2021 3:14 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Reaction Quotient Definition
- Replies: 7
- Views: 23
Re: Reaction Quotient Definition
The difference between Q and K is that K is the equilibrium constant, so it is calculated when the system is at equilibrium. Q, the reaction quotient is found during any stage of the reaction. They are calculated by the same equation, and this is why when you compare the two values, you can determin...
- Sun Jan 10, 2021 9:37 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Response of Equilibria to Change
- Replies: 8
- Views: 23
Re: Response of Equilibria to Change
The actual value of the equilibrium constant changes only due to a response in a change in temperature. Other factors will only potentially cause a shift in the position of equilibrium.
- Sun Jan 10, 2021 9:33 am
- Forum: Ideal Gases
- Topic: PV=nRT and concentration
- Replies: 27
- Views: 117
Re: PV=nRT and concentration
Concentration refers to the amount of a substance in a solvent. This is why it is calculated by the number of moles of the substance divided by the volume. You would use this when you want to convert between Kp and Kc. If you’re ever confused it’s good to know the units that are involved in the calc...
- Tue Jan 05, 2021 8:39 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium is shifted to the right/left
- Replies: 9
- Views: 37
Re: Equilibrium is shifted to the right/left
From the questions in the module, equilibrium is shifted to the right when K>1 whereas equilibrium is shifted to the left when K<1. This is because the equilibrium constant, K, refers to the relationship between product and reactant. Since products are in the numerator, when there are more products ...
- Sat Dec 12, 2020 6:59 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Protonation
- Replies: 4
- Views: 37
Re: Protonation
By the Bronsted Lowry definition, an acid donates a proton. This action is the acid deprotonating as it loses the proton. The Bronsted Lowry base will accept a proton from the acid, so it protonates.
- Sat Dec 12, 2020 5:13 pm
- Forum: Bronsted Acids & Bases
- Topic: metal/nonmetal oxides
- Replies: 3
- Views: 20
Re: metal/nonmetal oxides
Know that metal oxides form basic oxides whereas nonmetal oxides form acidic oxides. You should also remember which form amphoteric oxides.
- Sat Dec 12, 2020 12:48 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Calculation OH- sapling
- Replies: 4
- Views: 22
Re: Calculation OH- sapling
You can use the concentration of H+ given to calculate the pH by taking the negative log of the value. Then to find the pOH you take 14 minus the pH value found. To find the OH- concentration, you would just take 10 to the power of negative -pOH.
- Fri Dec 11, 2020 4:41 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Does en contriubiute to the coordination number twice?
- Replies: 2
- Views: 16
Re: Does en contriubiute to the coordination number twice?
en represents the neutral ligand ethylenediamine, and this ligand is a bidentate ligand. This means that it will contribute 2 to the coordination number, and it has two sites where it will bond to the transition metal.
- Fri Dec 11, 2020 4:37 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Chelating Ligands vs Polydentate ligands
- Replies: 5
- Views: 35
Re: Chelating Ligands vs Polydentate ligands
Ligands can be classified by how many times they bond with the transition metal. If it only forms one bond then it is monodentate. If it forms more than one bond, it is polydentate. Since it bonds more than once, it will form a chelate/ring of atoms.
- Sun Dec 06, 2020 8:48 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Sapling Week 9 HW #5
- Replies: 5
- Views: 25
Re: Sapling Week 9 HW #5
The coordination number should be six since en is bidentate so it bonds twice and there are two so that is four. You then have the two CO, so altogether there are six bonded to the TM.
- Sun Dec 06, 2020 12:44 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: pH and pOH
- Replies: 6
- Views: 45
Re: pH and pOH
pH is used to define the concentration of H3O+ whereas pOH is used to define the concentration of OH-. For both, you take the -log of the concentration of the ions.
- Sun Dec 06, 2020 12:36 pm
- Forum: Dipole Moments
- Topic: Dipole canceling
- Replies: 13
- Views: 81
Re: Dipole canceling
Yes, a nonpolar molecule can have polar bonds along with nonpolar bonds, but the polar bonds have dipoles that cancel out so that overall, the molecule has zero dipole moment.
- Sun Dec 06, 2020 12:28 pm
- Forum: Lewis Acids & Bases
- Topic: Acids question
- Replies: 4
- Views: 31
Re: Acids question
This formula is used to calculate concentrations of weak acids by setting up an equilibrium. This is only used for weak acids because they are not completely ionized in the solution, unlike strong acids.
- Sun Dec 06, 2020 12:20 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Ligand names
- Replies: 6
- Views: 46
Re: Ligand names
For the anions, they follow a pattern where they will end in -o (or -ido) most of the time. This ending is used to show that the ligand is directly bounded to the transition metal. A prefix is used to indicate the number of ligands, and they should be written in alphabetical order. For many of the l...
- Sun Dec 06, 2020 12:11 pm
- Forum: Naming
- Topic: Naming H2O
- Replies: 11
- Views: 90
Re: Naming H2O
I do not think it matters since both still represent a water molecule. The reason why they are written in different orders is that it represents which atom is bonded to the central atom. In the case of H20, since hydrogen can only form one bond, and it is already bonded to Oxygen, only the oxygen ca...
- Sun Nov 29, 2020 12:09 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Seesaw and T-shapes
- Replies: 4
- Views: 47
Re: Seesaw and T-shapes
Both seesaw and t-shaped molecules have five regions of electron density. The difference between them is that a seesaw shaped molecule has one lone pair and four bonded pairs. The VSEPR formula of a seesaw shaped molecule is AX4E. For a t-shaped molecule, the formula is AX3E2; there are two lone pai...
- Sun Nov 29, 2020 12:05 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Methane vs ammonia
- Replies: 5
- Views: 31
Re: Methane vs ammonia
Even though the two molecules both have 4 regions of electron density they are still different shapes due to one of the regions on ammonia being a lone pair of electrons. This lone pair causes repulsion because it is not bonded pushing the other atoms closer together resulting in the trigonal pyrami...
- Sat Nov 28, 2020 6:46 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Sapling Q.20
- Replies: 7
- Views: 53
Re: Sapling Q.20
You can tell whether a molecule is polar or not by looking at the lewis structure. Since you know that the molecular shape is tetrahedral and that all the atoms surrounding the central atom (As) are all the same, the molecule overall is nonpolar because the dipoles in the polar As and O bonds will c...
- Sat Nov 28, 2020 2:02 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Polarity and Shape
- Replies: 13
- Views: 93
Re: Polarity and Shape
You would have to draw it out to see if the dipoles cancel out. Both polar and nonpolar molecules can have polar bonds, but the difference is that in nonpolar molecules, you can see that the dipoles will cancel in the polar bonds pulling in opposite directions when you draw it.
- Sat Nov 28, 2020 1:56 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Sapling Wk8 Q. 18, Determining Molecular Shape
- Replies: 10
- Views: 70
Re: Sapling Wk8 Q. 18, Determining Molecular Shape
Yes, most of the time, the way the molecular formula is written indicates which atoms are bonded together. Especially when the same atoms are written out instead of just using a subscript to show how many of the atom there are.
- Sun Nov 22, 2020 7:22 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: BOND ANGLES FOR CERTAIN MOLECULAR SHAPES
- Replies: 4
- Views: 37
Re: BOND ANGLES FOR CERTAIN MOLECULAR SHAPES
You can't really calculate most bond angles as they have to be experimentally determined. All you can do is compare some bond angles such as looking at how many bonding pairs there are and how many lone pairs there are. For example, in a tetrahedral and in a trigonal pyramidal, there are 4 regions o...
- Sun Nov 22, 2020 6:03 pm
- Forum: Lewis Structures
- Topic: # of electrons in expanded valence
- Replies: 6
- Views: 53
Re: # of electrons in expanded valence
Since it is asking you about an expanded valence, you automatically know that the number of electrons it is asking for is greater than 8, the size of a normal octet. Therefore, you should also count the shared electrons.
- Sun Nov 22, 2020 5:59 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Trigonal Planar vs Angular
- Replies: 1
- Views: 19
Re: Trigonal Planar vs Angular
I think the electron arrangement being trigonal planar is because there are 3 regions of electron density (the two bonds and the unbonded electron). The shape would actually be angular because only two of them are bonded, and the extra electron bends the shape due to electron repulsion.
- Sun Nov 22, 2020 2:39 pm
- Forum: Trends in The Periodic Table
- Topic: Ionic Character
- Replies: 11
- Views: 93
Re: Ionic Character
Ionic character is determined by the partial charges that form between atoms that are bonded together; the more electronegative element will pull the electron towards itself making it slightly negative. This is ionic character because one atom is trying to take the electrons more and not wanting to ...
- Sun Nov 22, 2020 2:34 pm
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: London Dispersion
- Replies: 33
- Views: 165
Re: London Dispersion
Yes LDF occurs in all molecules because the electron clouds of atoms are always changing when atoms interact.
- Sun Nov 15, 2020 11:48 am
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 1E.21 (omit c)
- Replies: 3
- Views: 34
Re: 1E.21 (omit c)
In Bi, there 3 unpaired electrons because on the periodic table it is in the third column of the p block which has 3 orbitals (according to Hund's rule each orbital must have one electron before you can add a second). In Si, there are 2 unpaired electrons because it is in the second column of the p ...
- Sun Nov 15, 2020 10:43 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: electrons changing subshells
- Replies: 2
- Views: 13
Re: electrons changing subshells
Metals like copper and chromium take an electron in the d orbital that is expected to fill the s orbital because when d is half full or completely full it is more stable. So in the case of copper, instead of the expected configuration ending with 4s^2 3d^9, it becomes 3d^10 4s^1.
- Sun Nov 15, 2020 8:43 am
- Forum: Ionic & Covalent Bonds
- Topic: Delta Negative and Positive
- Replies: 3
- Views: 19
Re: Delta Negative and Positive
Whenever there are two atoms that covalently bonded, and the atoms are not the same, the electron pair in the bond is not shared equally. The electrons will be pulled towards the more electronegative atom because electronegativity is the electron pulling power of an atom. These are the elements in t...
- Sun Nov 15, 2020 8:35 am
- Forum: Octet Exceptions
- Topic: Identifying Radicals
- Replies: 24
- Views: 176
Re: Identifying Radicals
Usually, when you count up the valence electrons and there is an odd number you will have a radical. You can see this when you draw out the lewis structure and there is a single unpaired electron represented by a dot.
- Sun Nov 15, 2020 8:33 am
- Forum: Trends in The Periodic Table
- Topic: Question about Relationship Between Zeff, Atomic Radius, and Ionization Energy
- Replies: 2
- Views: 26
Re: Question about Relationship Between Zeff, Atomic Radius, and Ionization Energy
Zeff is the effective nuclear charge. As you go down the periodic table, there are more electrons in different shells, that is why the radius will increase. Since the electrons are now further away from the nucleus the effective nuclear charge will decrease. When going across the periodic table, eve...
- Sun Nov 08, 2020 8:12 am
- Forum: Lewis Structures
- Topic: Radicals
- Replies: 9
- Views: 91
Re: Radicals
Radicals are compounds that have electrons with unpaired spins, and it is an exception to the octet guideline. Because of this unpaired electron, the compound becomes highly reactive and unstable. When you draw the lewis structure, and there is an odd number of electrons, one of the electrons will ...
- Sun Nov 08, 2020 8:00 am
- Forum: Ionic & Covalent Bonds
- Topic: atomic radius
- Replies: 7
- Views: 49
Re: atomic radius
Atomic radius is just the measurement of half the distance of the nuclei between two of the same atoms that are touching. It is measured using two atoms because it is hard to measure the radius of one atom as the electron clouds do not have distinct boundaries.
- Sun Nov 08, 2020 7:54 am
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Polarizable
- Replies: 3
- Views: 16
Re: Polarizable
When a highly charged cation has the ability to distort the electron cloud of an anion, this process is known as polarization. Highly distorted electrons are described as highly polarizable. When atoms and ions have electron clouds that undergo large distortion, they have highly polarizing power.
- Thu Nov 05, 2020 11:26 am
- Forum: Lewis Structures
- Topic: Radicals
- Replies: 9
- Views: 91
Re: Radicals
Radicals are compounds that have electrons with unpaired spins, and it is an exception to the octet guideline. Because of this unpaired electron, the compound becomes highly reactive and unstable. When you draw the lewis structure, and there is an odd number of electrons, one of the electrons will r...
- Tue Nov 03, 2020 10:07 am
- Forum: Trends in The Periodic Table
- Topic: Ionization Energy
- Replies: 5
- Views: 28
Re: Ionization Energy
The farther away from the nucleus an electron is, the easier it is to be removed. Ionization energy refers to the energy required to remove an electron from an atom, so once, the first electron is removed from the neutral atom, the second electron is held on more tightly by the nucleus. The energy r...
- Sun Nov 01, 2020 1:25 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Sapling #24
- Replies: 16
- Views: 73
Re: Sapling #24
Just make sure the ending and the beginning of the wave would connect when you wrap it around in a circular shape. This is because an electron is described as having a circular standing wave around the nucleus.
- Sun Nov 01, 2020 12:08 pm
- Forum: Trends in The Periodic Table
- Topic: Sapling Weeks 2,3,4 Homework #18
- Replies: 4
- Views: 35
Re: Sapling Weeks 2,3,4 Homework #18
Basically, the cations are smaller than their parent atoms and anions are larger than their parent atoms. To order them by ionic radius, the ones that gain more electrons from their neutral states will have a larger radius because of additional shells added for the electrons.
- Sun Nov 01, 2020 11:55 am
- Forum: Trends in The Periodic Table
- Topic: Atomic Radius
- Replies: 30
- Views: 421
Re: Atomic Radius
The atomic radius will increase as you go down a group because there are additional shells that are farther from the nucleus. The reason why it will decrease across a period is that even though there might be more electrons, those electrons are in the same shell. An increase in the nuclear charge wi...
- Sat Oct 31, 2020 1:37 pm
- Forum: Properties of Light
- Topic: Relationship between wavelength and frequency
- Replies: 10
- Views: 91
Re: Relationship between wavelength and frequency
The relationship between wavelength and frequency is that they are inversely proportional so it is not necessarily the same amount. You can observe this from the equation speed of light = wavelength times frequency. If one increases, the other one will decrease.
- Fri Oct 30, 2020 6:58 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Nodal Plane Importance
- Replies: 9
- Views: 50
Re: Nodal Plane Importance
Nodal planes allow us to understand the different types of orbitals and why they have certain shapes. The number of nodal planes varies for different orbitals (s has none, p has one, d has two, f has three). In the case of the s orbital, you know there are no nodal planes because the probability of ...
- Sun Oct 25, 2020 8:02 am
- Forum: *Shrodinger Equation
- Topic: Schrodinger on MT
- Replies: 7
- Views: 89
Re: Schrodinger on MT
I think if we do have questions related to Shrodinger's it would mostly just be conceptual.
- Sat Oct 24, 2020 9:47 am
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Homework Problem 1A.15
- Replies: 2
- Views: 24
Re: Homework Problem 1A.15
From the problem, you know that the value of the final energy level is 1 because the ultraviolet spectrum correlates with the lyman series. It also states that it is the emission of energy so it is traveling down. After that, you can calculate the energy of the final with E= -hR/n^2. This value subt...
- Thu Oct 22, 2020 10:58 am
- Forum: Properties of Light
- Topic: Specifics on Light
- Replies: 5
- Views: 59
Re: Specifics on Light
yes, we should def know the order, and it is most important to know that visible light ranges from 700nm to 400nm. For the other ones, it's good to know the approximate ranges too just in case it asks you where it falls on the spectrum.
- Wed Oct 21, 2020 10:34 am
- Forum: Properties of Light
- Topic: Homework Problem 3
- Replies: 4
- Views: 42
Re: Homework Problem 3
This problem is conceptual. The wavelength and the frequency are both properties of each individual proton, so the amount does not matter as these properties will stay the same. The energy, however, is proportional to the number of photons, so increasing the value will affect it.
- Wed Oct 21, 2020 10:06 am
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Final vs Initial State
- Replies: 4
- Views: 40
Re: Final vs Initial State
It would depend on the context of the problem as you would have to determine which energy level it is traveling from and where it is traveling to. It would be the final energy level minus the initial energy level. When calculating this, I like to use the E=-hR/n^2 equation instead so I'm not confuse...
- Fri Oct 16, 2020 1:10 pm
- Forum: Properties of Light
- Topic: first worked example
- Replies: 3
- Views: 43
Re: first worked example
The 16 and 4 come from the values of n being squared as it was n=2 and n=4 in the problem. They are negative and in the denominator because of the equation that was being used: E= -(hr)/(n^2)
- Fri Oct 16, 2020 10:05 am
- Forum: Properties of Electrons
- Topic: Lyman vs. Balmer series
- Replies: 20
- Views: 164
Re: Lyman vs. Balmer series
The Lyman series is the transition of an electron to the ground state (n=1). This corresponds to the UV region because of the large delta E value between n=2 and n=1. The larger ΔE is, the higher the frequency; the UV region has a higher frequency. The Balmer series is the transition of an electron ...
- Wed Oct 14, 2020 11:29 am
- Forum: Properties of Light
- Topic: 1st worked example in lecture
- Replies: 4
- Views: 26
Re: 1st worked example in lecture
h is planck's constant and it is 6.626 x 10^-34 Js. c is the speed of light which is 3.00 x 10^8 m/s. I think both of these values will be given.
- Tue Oct 13, 2020 9:59 am
- Forum: Properties of Electrons
- Topic: keV
- Replies: 4
- Views: 26
Re: keV
keV is a kiloelectron volt. Usually, you would have to convert this to Joules when solving a problem.
- Mon Oct 12, 2020 11:03 am
- Forum: Properties of Light
- Topic: Light Intensity
- Replies: 7
- Views: 54
Re: Light Intensity
Yes, that is correct. Since the interaction between photons and electrons is one on one, increasing the intensity will not affect whether or not the electrons would be ejected as only the number of photons is increased. In order for electrons to be ejected, it depends on the energy of the individual...
- Wed Oct 07, 2020 11:44 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Fundamentals G5
- Replies: 4
- Views: 77
Re: Fundamentals G5
First I would convert all the units to L, mol, and g just to make it easier to solve for the different values. You are given the mass of 2.111g and you need to find the number of moles of the sodium carbonate because concentration is moles over volume. To find the number of moles you have to calcula...
- Wed Oct 07, 2020 11:08 am
- Forum: Molarity, Solutions, Dilutions
- Topic: G. 23
- Replies: 5
- Views: 58
Re: G. 23
Since you already have the molarity for the chloride ions in each solution, all you have to do is add them together in order to find the total value. The glucose portion is not important because it does not contain chloride ions, which is what you're focusing on.
- Wed Oct 07, 2020 9:49 am
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Review of Chemical and Physical Principles Fundamentals E.15
- Replies: 7
- Views: 44
Re: Review of Chemical and Physical Principles Fundamentals E.15
For this question, it is asking you what would the sulfide of this metal be if it had one. To solve it you would first need to figure out what element the metal (M) is. Then use that information to calculate the molar mass of the metal sulfide (MS).
- Mon Oct 05, 2020 11:04 am
- Forum: Balancing Chemical Reactions
- Topic: Law of Conservation?
- Replies: 16
- Views: 150
Re: Law of Conservation?
Law of Conservation is the idea that the total mass is the same before and after. This leads to the amount of atoms having to be the same values in order to add up to equal masses. The moles are only used in order to balance/adjust the number of atoms so they do not have to be equal as long as the m...
- Mon Oct 05, 2020 10:54 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Question about Mass%
- Replies: 14
- Views: 145
Re: Question about Mass%
Since the mass percentage compositions of the elements where calculated, you know how much of each element is in the compound as percentages. Using 100g instead of the actual 8.00g value makes it easier so you don't have to figure out the actual amounts of each element. The answer is not affected by...