[MnCl6] 3- is high spin and has five unpaired electrons while [Mn(CN)6] 3- has only two unpaired electrons.
A complex may be low spin or high spin depending on the kind of ligand attached to the central metal atom/ion. If the ligand is a weak field ligand, the complex may be high spin (maximum number of unpaired electrons). If the complex is low spin, there are few unpaired electrons (minimum number of unpaired electrons). In that case, the ligand is a strong field ligand.
In the octahedral geometry, [MnCl6] 3- is high spin and has five unpaired electrons since the chloride ion is a weak field ligand. On the other hand [Mn(CN)6] 3- has only two unpaired electrons because the cyanide ion is a strong field ligand.
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An example of kinetic energy being converted into heat energy
Answer:
if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less.
Explanation:
As you say, kinetic energy of large objects can be converted into this thermal energy. For example, if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less than.
A gas has a volume of 3.7 liters with a pressure of 1.75 atm. What is the pressure of the gas if its volume is raised to 4.5 L?
Answer: 251 K
Explanation: hope this helps :)
How many of sodium (Na) are needed to make 4.5 liters of a 1.5mol/L of Na solution?
Answer:
Explanation:
First you will find the mole from the molarity and then the desired mass from the mole.
How do you convert a temperature from Celsius to kelvin?
Answer:
Kelvin = Celsius + 273.15
Explanation:
Instead of 273.15 you can use just 273 as well
đặc điểm cơ bản nào để phân biệt vật thể tự nhiên và vật thể nhân tạo
Answer:
natural materials like cotton
the amount of solute dissolved in a solvent
A. Concentration.
B.alloy
C.mixture
D.solution
Answer:
solution
Explanation:
solvent +solute =solution
Learning Task 2: Read each statement or question below carefully and fill in the blank(s) with the best answer by choosing the words inside the box. Write your answers in a separate sheet of paper. cation 1 -ide -ine nonmetals O ion ionic compound anion metals root name 1. Any atom or molecule with a net charge, either positive or negative, is known as en 2. An atom that gains one extra electron forms an with a 1- charge. 3. A positive ion, called a is produced when one or more electrons are lost from a neutral atom. 4. Unlike a cation, which is named for the parent atom, an anion is named by taking the of the atom and changing the ending. 5. The name of each anions is obtained by adding the suffix to the root of the atom name. 6. The always form positive ions. 7. on the other hand, form negative ions by gaining electrons. 8. It is very important to remember that a chemical compound must have a net charge of
g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The height of the cylinder is 30 cm. The outside pressure is 105 Pa. The temperature of the gas is kept at 250 K throughout the experiment. The volume filled by the gas is 2.0 l. Now assume that solid cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. Cylinder and piston have the same diameter. Assume that the kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Calculate the change of entropy of the gas and of the environment. Please read this text very carefully
A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.
A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.
[tex]v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s[/tex]
The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:
[tex]K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J[/tex]
The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.
Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.
[tex]\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K[/tex]
The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, [tex]\Delta S_{env} = -1.18 J/K[/tex]
A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.
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answer this please........
Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if an additional 0.182mol of methane is added to the container under constant temperature and pressure? Give your answer in three significant figures.
The final volume of the methane gas in the container is 6.67 L.
The given parameters;
initial volume of gas in the container, V₁ = 2.65 Linitial number of moles of gas, n₁ = 0.12 moladditional concentration, n = 0.182 molThe total number of moles of gas in the container is calculated as follows;
[tex]n_t = 0.12 + 0.182 = 0.302 \ mol[/tex]
The final volume of gas in the container is calculated as follows;
[tex]PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L[/tex]
Thus, the final volume of the methane gas in the container is 6.67 L.
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) A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that the mineral grains just float. When a sample of the mixture in which the mineral calcite just floats is put in a special density bottle, the weight is 15.4448 g. When empty, the bottle weighs 12.4631 g, and when filled with water, it weighs 13.5441 g. What is the density of the calcite sample? (All measurements were carried out at 25 °C, and the density of water at 25 °C is 0.9970 g>mL)
At the left, grains of the mineral calcite float on the surface of the liquid bromoform (d = 2.890 g/mL) At the right, the grains sink to the bottom of liquid Chloroform (d = 1.444 g/mL). By mixing bromoform and chloroform in just the proportions required so that the grains barely float, the density of the calcite can be determined
Hey there!
It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.
In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:
[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]
Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:
[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]
Thus, the density of the calcite sample will be:
[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]
This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL
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9) Observe the halogens. Fluorine and chlorine are gases, bromine is a liquid, and
iodine is a solid all at room temperature.
Distinguish between these differences in phase.
A) From Fluorine to lodine, the number of electrons increases.
This allows the intermolecular forces to strengthen from
temporary dipole (dispersion) forces to permanent dipoles.
Therefore, iodine is a solid because it has the most electrons.
B) From Fluorine to lodine, the nucleus gets larger. More
neutrons, protons and electrons allow the atoms of these
elements to become more attracted to each other. This is also
the reason that the halogens are diatomic naturally.
) From Fluorine to lodine, the molecular mass increases. With
the increase in mass, there is an increase in protons. This causes
the strength of the intermolecular forces between particles to
increase as well.
D) From Fluorine to lodine, the molecular mass increases. The
electron polarizability increases with the mass increase. In turn,
this strengthens the temporary dipole (dispersion) forces
between particles
Iodine is much more easily polarizable than fluorine therefore temporary dipoles in the molecule are strengthened.
The halogens are members of group 17 in the periodic table. They are highly electronegative and seldom occur free in nature owing to their high level of reactivity.
We know that larger molecules are more easily polarized than smaller ones. Therefore, from Fluorine to lodine, the molecular mass increases. The electron polarizability increases with the mass increase. In turn, this strengthens the temporary dipole (dispersion) forces between particles. Hence the properties of halogens change smoothly down the group.
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If 25 g of Al was added to 90 g of HCl, what mass of H2 will be produced?
Answer:
im pretty sure its 2.7 gms
Explanation:
im not for sure tho
After being placed on the table where should the ocular lenses face?
What are the bond angles found in germanium disulfide ?
Answer:
Every germanium atom is tetrahedrally linked to four sulphur atoms, with an interatomic distance of 2.19A. The angle between the two sulphur bonds is 103°.
hope this helps
An interatomic distance of 2.19A exists between each germanium atom and the four sulphur atoms that are tetrahedrally connected to it. The two sulphur bonds form a 103° angle.
What is bond angle ?A complex molecule's or ion's bond angle is the angle between the two bonds, or the angle between two orbitals that contain bonding electron pairs surrounding the central atom. It is determined using a spectroscopic approach and measured in degrees.
Any angle between two bonds that share an atom is known as a bond angle, and it is often measured in degrees. The distance along the straight line between the nuclei of two bound atoms is known as a bond distance.
Bond angles also have a role on a molecule's structure. The angles between neighboring lines that form bonds are known as bond angles. The difference between linear, trigonal planar, tetrahedral, trigonal-bipyramidal, and octahedral crystals may be determined by the bond angle.
Thus, The two sulphur bonds form a 103° angle.
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What question would a student need to ask to form a compound with Group 16 nonmetals
A.
Will group 16 elements lose electrons to bond with group 2 in an XY format?
B.
Will group 16 elements gain electrons to bond with group 1 in an XY2 format?
C.
Will group 16 elements gain electrons to bond with group 2 in an XY format?
D.
Will group 16 elements lose electrons to bond with group 1 in an X2Y format?
Since nonmetals gain electrons, the correct question to ask about group 16 elements is; "Will group 16 elements gain electrons to bond with group 2 in an XY format?"
Group 16 elements are divalent and they form divalent negative ions. The periodic table is arranged in groups and periods. The elements in the same group have the same number of valence electrons. All elements in group 2 have six valence electrons.
If a wants to form a compound with the non metals of group 16, the correct question to ask is;"Will group 16 elements gain electrons to bond with group 2 in an XY format?"
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Who wants to simp for me??
Answer:
qrtyuioplkjhgfdssssssazxcvbn
Identify the strongest intermolecular force that is likely to affect each of the samples described below.
A mixture of chlorine gas (Cl) and fluorine gas (F): V London dispersion forces
COMPLETE
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A mixture of hydrogen sulfide (H2S) and hydrogen chloride (HCI): V dipole-dipole interactions
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COMPLETE
Guess
this
hydrogen bonding
A mixture of water (H2O) and ammonia (NH3):
Answer:
A mixture of chlorine gas (Cl2) and fluorine gas (F2):
✔ London dispersion forces
Explanation:
Compound A is neutral and Compound B is acidic. Both are water-insoluble solids. A and B are dissolved in dichloromethane (DCM) and extracted with aqueous base. The layers are then separated. What must be done to obtain the compound in the aqueous layer
In order to extract the compound in the aqueous layer, a strong acid must be added to the system.
Liquid - Liquid extraction is a common method for obtaining substances that can partition between two layers. In this case, compound A is neutral and compound B is acidic.
When the both compounds are dissolved in dichloromethane and extracted using an aqueous base, the acid substance will form a salt in the aqueous layer. In order to extract the compound in the aqueous layer, a strong acid must be added to the system.
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Define electrolysis
Electrolysis is the production of a chemical reaction by passing an electric current to liquid containing ions.
At 298 K, the reaction 2 HF (g) ⇌ H2 (g) + F2 (g) has an equilibrium constant Kc of 8.70x10-3. If the equlibrium concentrations of H2 and F2 gas are both 1.33x10-3 M, determine the initial concentration of HF gas assuming you only started with HF gas and no products initially.
This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.
Then, we can write the following equilibrium expression for hydrofluoric acid once the change, [tex]x[/tex], has taken place:
[tex][HF]=[HF]_0-2x[/tex]
Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):
[tex]8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }[/tex]
[tex][HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M[/tex]
Finally, the initial concentration of HF is calculated as follows:
[tex][HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M[/tex]
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https://brainly.com/question/13043707https://brainly.com/question/16645766What is a solute?
substance in which another substance dissolves and mixes evenly A
a solution unable to be separated by any means B
a mixture that can easily be separated with simple tools C
the substance that dissolves into another substance D
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
The Correct choice is ~ D
Solute is the substance that dissolves into another substance
What’s the answer to this? Pls help
Answer:
First choice
Explanation:
Calculate the percent by mass of each element in LiClO2.
% Li:
% Cl:
% O:
ok done. Thank to me :>
Answer: Mass percent of Li: total mass Li = 6.94 g/mol. % Li = mass Li mass LiClO 2 × 100 % Li = 6 . ...
Mass percent of Cl: total mass Cl = 35.45 g/mol. % Cl = mass Cl mass LiClO 2 × 100 % Cl = 35 . ...
Mass percent of O: total mass O = 32.00 g/mol.
Explanation:
Describing Chemical Reactions
Project: Modeling the Conservation of Mass
Answer:
Step 1: Prepare for the project.
a) Read the entire Student Guide before you begin this project.
b) If anything is unclear, be sure to ask your teacher for assistance before you begin.
c) Gather the materials you will need to complete this project.
Step 2: Review the chemical reaction.
a) The chemical reaction that you will model is shown below. Calcium oxide (CaO) is a white
solid with a crystalline structure. It is made by heating limestone, coral, sea shells, or chalk,
which are composed mainly of calcium carbonate (CaCO3). During the heating process,
carbon dioxide (CO2) is released and calcium oxide (CaO) is produced. Commercially,
calcium oxide is called lime. One of the oldest uses of lime is to make mortar, a substance
used in construction to secure bricks, stones, and blocks together.
CaCO3 CaO + CO2
Step 3: Build a model of the reactant.
a) Use gumdrops and toothpicks to build a model of CaCO3.
b) Select one color of gumdrop for calcium, a second color for carbon, and a third color for
oxygen. Be sure to include a key of your chosen color scheme with your model.
c) Use the periodic table to help you determine the number of bonds each atom will form.
Step 4: Build models of the products.
a) Use gumdrops and toothpicks to build a model of CaO and a model of CO2.
b) Be sure to use the same colors of gumdrops for calcium, carbon, and oxygen as you did in
Step 3. Include a key of your chosen color scheme with your models.
c) Use the periodic table to help you determine the number of bonds each atom will form.
Step 5: Type one to two paragraphs that describe your models and explain the conservation of
mass in the chemical reaction.
a) Create a new blank document. Type your name at the top.
b) Type one to two paragraphs that describe your models and relate them to the law of
conservation of mass. Your document should:
i. identify the names of the reactants and products in the reaction.
ii. identify the number of molecules that make up the reactants and products.
iii. identify the type and number of atoms in each molecule of the reactants and products.
iv. explain what happens during the chemical reaction.
v. explain how mass is conserved during the chemical reaction.
Step 6: Evaluate your project using this checklist.
If you can check each criterion below, you are ready to submit your project.
Did you create an accurate model of calcium carbonate (CaCO3)? Your model should include
the correct number of gumdrops for each element in calcium carbonate, consistent use of
colors for elements in calcium carbonate, and the correct number and placement of toothpicks
(bonds).
Did you create an accurate model of calcium oxide (CaO)? Your model should include the
correct number of gumdrops for each element in calcium oxide, consistent use of colors for
elements in calcium oxide, and the correct number and placement of toothpicks (bonds).
Did you create an accurate model of carbon dioxide (CO2)? Your model should include the
correct number of gumdrops for each element in carbon dioxide, consistent use of colors for
elements in carbon dioxide, and the correct number and placement of toothpicks (bonds).
Did you type one to two paragraphs that describe your models and relate them to the law of
conservation of mass? Your document should include the names of the reactants and
products in the reaction, the number of molecules that make up the reactants and products,
and the type and number of atoms in each molecule of the reactants and products. It should
also explain what happens during the chemical reaction and how mass is conserved during the reaction
Step 7: Revise and submit your project.
a) If you were unable to check off all of the requirements on the checklist, go back and make
b) When you have completed your project, submit your models to your teacher for grading.
c) Submit your document through the virtual classroom. Be sure that your name is on it.
Step 8: Clean up your workspace.
a) Clean up your workspace. Return any extra materials to your teacher and throw away any trash.
Explanation:
what is the name of something that starts with N and ends with G
Nutmeg
Nubbing
Notifying
Neezing
Nudging
Needing
Niffing
Narcotizing
Nanjing
Neighbouring
Nothing
Nimming
Neighboring
Netting
Nesting
Narcing
Nitpicking
Non-Living
Neutralizing
Nightwalking
Nonspeaking
An element with 5 valence electrons would be a
0 - 3 anion
O +5 anion
0-5 cation
O + 3 cation
Answer:
ask you subject teacher
Explanation:
Dissolution of KOH, ΔHsoln:
KOH(s) → KOH(aq) (10.1)
Neutralization of solid KOH, ΔHneut:
KOH(s) + HCl(aq) → H2O(l) + KCl(aq) (10.2)
1) Using Hess's law, show how to combine Reaction 10.1 and Reaction 10.2 to give
KOH(aq) + HCl(aq) → H2O(l) + KCl(aq) (10.3)
2)How should ΔHsoln and ΔHneut be combined (mathematically) to give the change in enthalpy for Reaction 10.3, ΔH?
Using Hess's law we found:
1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)
2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).
The reactions of dissolution (10.1) and neutralization (10.2) are:
KOH(s) → KOH(aq) ΔHsoln (10.1)
KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)
1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.
Hence, to get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
We need to add reaction 10.2 to the reverse of reaction 10.1
KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)
Canceling the KOH(s) from both sides, we get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:
[tex] \Delta H = \Delta H_{soln} + \Delta H_{neut} [/tex]
The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:
[tex] \Delta H = \Delta H_{neut} - \Delta H_{soln} [/tex]
Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.
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If an atom should form from its constituent particles,
(a) matter is lost and energy is taken in.
(b) matter is lost and energy is released. .
(c) matter is gained and energy is taken in.
(d) matter is gained and energy is released.
Answer:
(b) matter is lost and energy is released
Explanation:
When atoms are being formed from its constituent components it weighs less this is called mass defect so the answer would be (b) matter is lost and energy is released.
Why is the first one (A) correct?
Answer: yes it is correct
Explanation: the higher it is the cooler.
