Determine the kinds of intermolecular forces that are present in each element or compound. o2 nbr3 ch4 hf

Answers

Answer 1
O₂ (contain only dispersion force exist as it can't have hydrogen bond or dipole-dipole forces)

NB₃ (polar molecule so it contain dipole-dipole interaction in addition to dispersion force)

CH₄ (Non polar compound and has no hydrogen bond so only dispersion is present)

HF (It can form hydrogen bond because it is polar and contain F which has high electronegativity in addition to dipole-dipole and dispersion)
Answer 2
Final answer:

O2 experiences London dispersion forces, NBr3 experiences dipole-dipole interaction, CH4 experiences London dispersion forces, and HF experiences hydrogen bonding due to the presence of a hydrogen atom bonded to a highly electronegative atom.

Explanation:

The given molecules show different types of intermolecular forces. Oxygen gas (O2) is a non-polar molecule and it interacts through London dispersion forces, the weakest form of forces. Nitrogen tribromide (NBr3) has polar bonds, so this compound's main intermolecular force would be dipole-dipole interaction. Methane (CH4) is non-polar, so the intermolecular forces are due to London dispersion forces. Hydrogen Fluoride (HF) experiences hydrogen bonding because it contains a hydrogen atom bonded to a more electronegative atom, fluorine.

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Related Questions

Consider the following reversible reaction. mc030-1.jpg What is the equilibrium constant expression for the given system?

Answers

Since you have not included the given reaction, I am going to explain you how to solve these kind of problems.

1) The chemical equilibrium is a dynamic process. It means that in an equilibrim reaction there are two rectaions, the forward reaction and the reverse reaction whose velocities are the same.

2) The general equation of a a chemical reaction in equlibrium is:

aA + bB ⇄ cC + dD
Where A and B are the reactants, C and D are the products, and a, b, c, d, are the coefficientes in the balanced equation.

3) So, the equilibrium law is:

[tex]Keq= \frac{C]^c[D]^d}{[A]^a[B]^b} [/tex]

Where Keq is the constant of equilibrium

4) To complete the explanation, I am going to deal with an example:

i) Consider the equlibrium reaction between hydrogen and iodine:

H₂ (g) + I₂(g)   ⇄  2HI(g)

ii) The forward reaction is H₂ (g) + I₂(g)   →  2HI(g)

iii) The reverse reaction is 2HI (g) → H₂ (g) + I₂(g)

iv) The law of equilibrium is:

[tex]Keq= \frac{[HI]^2}{[H_2][I_2]} [/tex]


He rate constant for the forward reaction, k1, is 297 l·mol–1·min–1 and the rate constant for the reverse reaction, k–1, is 393 l·mol–1·min–1 at a given temperature. the activation energy for the forward reaction is 42.1 kj·mol–1, while the activation energy for the reverse reaction is 22.1 kj·mol–1. determine the equilibrium constant, k, of this reaction.

Answers

Equilibrium constant of reaction,
 = [tex] \frac{rate.constant.forward.reaction}{rate.constant.of.backward.reaction} [/tex]
= [tex] \frac{297}{393} [/tex]
=0.7557

The equilibrium constant is a number that shows the extent to which reactants are converted into products. The equilibrium constant of the reaction is 0.76.

What is equilibrium constant?

The equilibrium constant is a number that shows the extent to which reactants are converted into products.

We know that to obtain the equilibrium constant; K = k1/k-1 where;

k-1 = rate constant of reverse reaction

k1 = rate constant of forward reaction.

Hence;

K = 297 l·mol–1·min–1/393 l·mol–1·min–1

K = 0.76

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James is looking at a parallel circuit plan for lighting. There is a battery providing the power. There are switches labeled A,B,C,D that can be turned on to close the circuit. Which switch does not have to be on for light 3 to function?

Answers

the answer is D because for light 3 to function switch d does not  have to turn on

For light 3 to function, switch D does not have to be on. This diagram shows a parallel circuit that provides more than one way for the current to return to the power source.

Choose all the answers that apply. Muscles in the small intestine are _____. striated voluntary involuntary smooth cardiac

Answers

The muscles in the small intestine are smooth muscles and they are involuntary.

The muscles can be seen in the muscularis externa layer of the small intestine. Those muscles are arranged in two ways as outer is longitudinal and inner is circular way.  Those muscles help to do movements of small intestine.

Answer:

The answer is smooth and Involuntary.

Explanation:

Smooth muscles are involuntary which means they automatically do a function without the brain telling it to.

help again pleaseee!!

Answers

 I'm pretty sure it is d. a reaction cant just occur without access energy to aid in the reaction. think of it like glycolysis. atp goes through different processes to gain energy for the bodies processes. I'm sure it is d.
The correct answer is:
from the energy need to fuse the atoms together


Fusion is the process which powers the sun and the stars. It is energy that makes all life on earth possible. It is called 'fusion' because the energy is produced by fusing together light atoms, such as hydrogen, at the extremely high pressures and temperatures which exist at the centre of the sun (15 million ºC). At the high temperatures experienced in the sun any gas becomes plasma, the fourth state of matter (solid, liquid and gas being the other three).
The fusion reaction that is easiest to accomplish is the reaction between two hydrogen isotopes: deuterium, extracted from water and tritium, produced during the fusion reaction through contact with lithium. When deuterium and tritium nuclei fuse, they form a helium nucleus, a neutron and a lot of energy.

Acetyl-coa and succinyl-coa are both high-energy thioesters, but their chemical energy is put to different uses. explain how the chemical energy is used in the two cases.

Answers

Why are Acetyl-CoA and succinyl-CoA, both high energy thioesters, chemical energies put to different uses?
The energy released by hydrolysis of acetyl-CoA is needed for the condensation reaction that links acetyl moiety to oxaloacetate, yielding citrate. The energy released by hydrolysis of succinyl-CoA drives the phosphorylation of GDP, yielding GTP

Some reactions of the citric acid cycle are endergonic. Explain how the overall cycle is exergonic.
The sum of the energies of the individual reactions is -44.3kJ for each mole of acetyl-CoA that enters the cycle

Which enzymes of the citric acid cycle are missing from the glyoxylate cycle?
isocitrate dehydrogenase, alpha-ketoglutarate dehydrogenase, and succinyl-CoA synthetase

When the concentration of a is doubled,the rate for the reaction:2a+b → 2cquadruples.whentheconcentrationofbisdoubledtherateremainsthesame.whichmechanismbelowisconsistentwiththeexperimentalobservations?

Answers

A is second order, B is zero order. So overall the reaction is second order 

Calculate the freezing point of a solution that contains 8.0 g of sucrose (c12h22o11) in 100 g of h2o. kf for h2o = 1.86c/m

Answers

Answer is: the freezing point of the solution of sucrose is -0.435°C.
m(H₂O) = 100 g ÷ 1000 g/kg = 0.1 kg.
m(C₁₂H₂₂O₁₁) = 8.0 g. 
n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 8.0 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0233 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0233 mol ÷ 0.1 kg.
b(solution) = 0.233 m.
ΔT = b(solution) · Kf(H₂O).
ΔT = 0.233 m · 1.86°C/m.
ΔT = 0.435°C.
Tb = 0°C - 0.435°C = -0.435°C.

The freezing point is the temperature at which the fluid freezes to a solid form. The freezing point of the solution is -0.435 degrees celsius.

What is the freezing point?

The freezing point is the product of the molality, van 't Hoff factor, and the cryoscopic constant. It is given as,

[tex]\rm \Delta T_{F} = K_{F} \times b\times i[/tex]

Given,

Mass of water = 0.1 kg

Mass of sucrose = 8.0 gm

Moles of sucrose are calculated as:

[tex]\begin{aligned}\rm moles &= \dfrac{8.0}{342.3}\\\\&= 0.0233 \;\rm mol \end{aligned}[/tex]

The molality of sucrose is calculated as:

[tex]\begin{aligned}\rm b &= \dfrac{\text{moles of sucrose}}{\text{mass of water}}\\\\&= \dfrac{0.0233 \;\rm mol}{0.1}\\\\&= 0.233 \;\rm m\end{aligned}[/tex]

The freezing point depression is calculated as:

[tex]\begin{aligned}\rm \Delta T &= 0.233 \;\rm m \times 1.86\; ^{\circ} \;\rm C/m\\\\&= 0.435 ^{\circ}\;\rm C\end{aligned}[/tex]

Therefore, the freezing point of a solution is -0.435 degrees celsius.

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If 152 grams of ethane (c2h6) are reacted with 231 grams of oxygen gas, what is the mass of the excess reactant leftover after the reaction has reached completion? 2c2h6(g) + 7o2(g) → 4co2(g) + 6h2o(g)

Answers

Answer is: the mass of the excess reactant (ethane) leftover is 90.135 grams.
Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g).
m(C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.

The answer is:

the mass of the excess reactant  leftover after the reaction has reached completion is 90.135 grams

The explanation:

According to the reaction equation:

 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)

when m is the mass of C₂H₆ m(C₂H₆) = 152 g

So we need to get the number of moles of C₂H₆

n(C₂H₆) = mass C₂H₆ / molar mass of C₂H₆(M)

and when the molar mass of C₂H₆ = 30 g/mol

so, by substitution:

n(C₂H₆) = m(C₂H₆) / M(C₂H₆).

n(C₂H₆) = 152 g / 30 g/mol.

n(C₂H₆) = 5.067 mol.

Then

when the mass of O₂ m(O₂) = 231 g

so we need to get the number of moles of O₂

when nO₂ = mass O₂/ molar mass of O₂

when molar mass of O₂ = 32 g /mol

So, by substitution:

n(O₂) = 231 g / 32 g/mol.

n(O₂) = 7.218 mol

So O₂ is the limiting reactant

according to the chemical reaction we can get the molar ratio between the O₂and C₂H₆:

n O₂ : n C₂H₆    →  7.218 : n C₂H₆

      7 : 2                   7      :   2

∴ n(C₂H₆) = 2 * 7.218 mol / 7

∴ n(C₂H₆) = 2.0625 mol

The number of moles remaining n(C₂H₆) = 5.067 mol - 2.0625 mol

∴ n (C₂H₆) = 3.0045 mol

So the mass remains = moles remains * molar mass of C₂H₆

∴ m (C₂H₆) = 3.0045 mol * 30 g/mol = 90.135 g


Which element has the electron configuration shown above

Answers

The answer is Neon (Ne)

The 2p shell is the elements on period 2 from Boron to Neon on the periodic table. The 6 means it has 6 electrons and is sixth in line so it must be Neon.
One element that has the electron configuration (1s22s22p6) is Neon.

Click the "draw structure" button to launch the drawing utility. draw any one of the skeletal structures of a 2° alkyl bromide having the molecular formula of c6h13br and two stereogenic centers. indicate chirality by using wedge and hashed wedge notation. lone pairs do not need to be shown.

Answers

I have drawn a structure below having secondary (2°)  Alkyl Bromide, Two Stereogenic Centers and a molecular formula C₆H₁₃Br.

Below in structure it can be seen that the Br atom and CH₃- groups are pointing towards the viewer and shown by wedge bond and the protons at corresponding Chiral carbons are shown with dashed bonds pointing away from the viewer.

So, what is important in this question to identify ...

Secondary Alkyl Halide, an organic compound in which carbon to which halogen is bonded is further bonded to 2 carbon atoms.

And, Chiral carbon (Stereogenic Center) to which all the atoms or group of atoms attached are all different from each other.

The half-life of cobalt-60 is 5.20 yr. how many milligrams of a 2.000-mg sample remain after 10.50 years?

Answers

final amount =initial anount *(1/2)^(t/half life)

final amount =2.000 mg *(1/2)^(10.5/5.20) = 0.493 mg remain after 10.5 years

The (C4H1 0)3N represents a___ amine.
A. Primary
B. Secondary
C. Tertiary

Answers

C. tertiary :) hope this helps

Answer :The correct answer for (C₄H₁₀ )₃ N is C) TERTIARY AMINE .

The substitution level of nitrogen in amine or in simple words , description of carbon atoms attached to a given nitrogen in any amine is of 4 types :

1 ) Primary amine :

When the given nitrogen is attached to only one carbon then substitution level is known as PRIMARY amine.

Example : CH₃NH₂ : The nitrogen is attached to only one carbon (CH₃) . Hence it is primary amine

2) Secondary amine :

When the given nitrogen is attached to two carbons .

Example : (CH₃)₂NH : The nitrogen is attached to two carbon ( CH₃)₂ , hence it is secondary amine .

3) Tertiary amine :

When the given nitrogen is attached to three carbon atom .

Example : (CH₃)₃NH : The nitrogen is attached to 3 carbons atoms (CH₃)₃ . Hence it is a tertiary amine

4) Quaternary amine : when the nitrogen attached to 4 carbon atoms and amine posses a positive charge on it , its is known as quaternary amine .

Example : (CH₃)₄N⁺ : The nitrogen is attached to 4 carbons and nitrogen . hence it is quaternary amine .

(Image attached )

The given compound is (C₄H₁₀ )₃ N has 3 groups of C₄H₁₀ attached to N , means 3 carbons attached to N , ( image attached ) . Hence it can be said that (C₄H₁₀ )₃ N is TERTIARY AMINE .

Hence correct option is C) Tertiary amine .

The three beakers shown below contain solutions of [cof6]3–, [co(nh3)6]3+, and [co(cn)6]3–. based on the colors of the three solutions, which compound is present in each of the beakers?

Answers

[Co(CN)₆]³⁻ → Yellow
[Co(NH₃)₆]³⁺ → Orange
[CoF₆]³⁻ → Blue
Explanation:
- All the given compounds have octahedral geometry but the ligand in each are different with the same metal ion.

- Ligands strength order:     CN⁻ > NH₃ > F⁻ 

- The ligand CN will act as a strong field ligand so that the splitting is maximum when compared to NH₃ and F⁻

- If the splitting is more, the energy required for transition is more, and the wavelength is inversely proportional to energy.

- So CN complex will absorb at lower wavelength (yellow color)
Final answer:

The colors of the compounds in the beakers could potentially be: [CoF6]3– (green), [Co(NH3)6]3+ (yellow-orange), and [Co(CN)6]3– (red), although colors can vary based on different conditions.

Explanation:

The question asks us to identify the compounds present in three beakers based on the color of the solution. While this is generally impossible to answer with absolute certainty without more information or additional tests, we can make a fair guess based on some known color-characteristics of these compounds.

[CoF6]3– is generally green due to the color of the Cobalt(III) ion.

[Co(NH3)6]3+ is usually yellow to orange in color.

FInally, [Co(CN)6]3– commonly has a strong red color.

Note that in reality, actual colors can vary depending on concentration, temperature, and other factors.

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What is the molarity of 500 ml of a solution containing 85.0g AgNO3?

Answers

1.0 M, if you're just looking for an answer.

Formula mass of AgNO3 is 107.9 + 14 + (3 X 16) = 169.9 g.

85.0 g                           0.5 mole
-------                      =    ------------    = 1.0 M
169.9 (g/mole)              0.5 liter

Answer: 1 M

Explanation:

Molarity : It is defined as the number of moles of solute present per liter of the solution.

Formula used :

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n= moles of solute

[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{85.0g}{170g/mol}=0.5moles[/tex]  

[tex]V_s[/tex] = Volume of solution in ml

Now put all the given values in the formula of molarity, we get

[tex]Molarity=\frac{0.5moles\times 1000}{500ml}=1mole/L[/tex]

Therefore, the molarity of solution will be 1 M.

If I cut the height of an object in half, what will happen to its GPE?
A. Not enough info
B. Double
C. Disappear
D. Halve

Answers

the answer is D im 80% sure.
Final answer:

If you cut the height of an object in half, its gravitational potential energy (GPE) will also halve.

Explanation:

If you cut the height of an object in half, its gravitational potential energy (GPE) will also halve.

GPE is determined by the height of the object and its mass. When you cut the height in half, the GPE is reduced by half because the potential energy is directly proportional to the height.

For example, if a book has a certain GPE at a certain height, cutting the height in half will result in the book having half the GPE it had before.

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Glacier and erosion in a sentence

Answers

a glasier can cause an eroision miles away from eachother

Answer how many moles of sodium will react with 2.6l of cl2 gas at 1.15 atm

Answers

The chemical equation for the reaction is ,

                                      2 Na  +  Cl₂     →     2 NaCl

Before applying stoichiometry, first you will have to find the moles of Cl₂ at given conditions using Ideal gas equation.

                                        n  =  P V / R T

As, temperature is not given, so put standard Temperature i.e. 273 K, or if you are provided with it simply put its value So,

               n  =  (1.15 atm × 2.6 L) ÷ (0.0821 atm.L.mol⁻¹.K⁻¹ × 273 K)

               n  =  0.133 mol of Cl
Now,
According to Balance equation,

                          1 mole Cl₂ required  =  2 moles of Na
So,
                 0.133 mol Cl₂ will require  =  X moles of Na

Solving for X,
                             X  =  (0.133 mol × 2 mol) ÷ 1 mol

                             X  =  0.266 moles of Na

Name the salt k4[pt(co3)2f2] given that the carbonate ion acts as a monodentate ligand in the complex. the oxidation number of platinum is +2. enter the name.

Answers

The IUPAC name of compound is Potassium dicarbonatodifluoroplatinate (II)

*Note, since carbonate ion is acting as monodentate ligand, prefix of 'di' is used. In case, if it had been a bidentate ligand prefix 'bis' would have been used.

Answer:

Potassium dicarbonatedifluoroplatinate (II)

Explanation:

Hello,

Based on the IUPAC rules, the given compound is a complex called potassium dicarbonatodifluoroplatinate (II) as long as the carbonate ion is present twice as a monodentate ligand therefore it is preceded by a "di" prefix. In addition, two fluorines are also present with the "di" prefix as well as a platinum which complete the anionic section including +2 as the platinum oxidation state.

Best regards.

Complete the following table.
Acid Molarity Moles of H⁺ released per liter

HCl 1 ____

H2SO4 1 ____

H3PO4 1 ____

H2SO4 0.5 ____

H3PO4 3 ____

HNO3 2 ____

Answers

Answer: see the last column in the final table of the explanation below.

Explanation:

1) Arrange the data to understand the question:

Acid                 Molarity         Moles of H⁺ released per liter

HCl                      1                  ____

H2SO4                1                   ____

H3PO4                1                   ____

H2SO4                 0.5               ____

H3PO4                 3                   ____

HNO3                   2                  ____

2) All the given acids are strong acids, so you can consider they dissociate completely and release all the hydrogens present in the chemical formula.

3) The molarity formula is M = n/V, where n is the number of moles of solute and V is the volume of the solution in liters

From that, n = M×V

Go one by one with that formula.


Acid          Molarity    Moles of solute    number of H in            Moles H⁺                                  per liter                 the chemical formula  per liter

HCl            1             1M×1L = 1                  1                                  1×1 = 1

H₂SO4₄      1             1M×1L = 1                 2                                  2×1 = 2                 
H₃PO₄        1              1M×1L = 1                3                                 3×1 = 3

H₂SO₄         0.5           0.5M×1L = 0.5       2                                2×0.5 = 1

H₃PO₄        3               3M×1L = 3            3                                 3×3 = 9       
 
HNO₃         2               2M×1L = 2            2                                 2×2 = 4


Complete the following table.

Acid Molarity Moles of H⁺ released per liter

HCl 1

H2SO4 1

H3PO4 1

H2SO4 0.5

H3PO4 3

HNO3 2

[Answer]

1

2

3

1

9

2

Determine the overall charge on each complex.a) tetrachlorocuprate(i)b) pentaamminechlorocobalt(iii)c) diaquadichloroethylenediaminecobalt(iii)

Answers

Complex a: tetrachlorocuprate(i)
In present complex chlorine is negatively charged ligand (-1) and oxidation state of copper is +1. Therefore, total charge on complex = 4(-1) + 1 = -3

Complex b: pentaamminechlorocobalt(iii)
In present complex, ammine is a neutral ligand (charge = 0), chlorine is negatively charged ligand (charge = -1) and oxidation state of Co is +3. Therefore, total charge on complex is 5(0) + (-1) + (+3) = +2

Complex c: diaquadichloroethylenediaminecobalt(iii)
In present complex, aqua is a neutral ligand (charge = 0), chlorine is a negatively charged ligand (charge = -1), ethylenediamine is a neutral ligand (charge = 0) and oxidation state of cobalt is +3. Therefore, total charge on complex is 2(0) + 2(-1) + 2(0) + 3 = +1. 

Ksp [caso4] = 2.4 × 10-5. determine the molar solubility of caso4 in a solution containing 0.100 m na2so4.

Answers

Final answer:

The molar solubility of CaSO4 in a solution containing 0.100 M Na2SO4 is 2.4 x 10−4 M, calculated using the Ksp value and the concentration of sulfate ions due to the common ion effect.

Explanation:

The molar solubility of CaSO4 in a solution containing 0.100 M Na2SO4 can be calculated using the solubility product constant (Ksp) and the concept of common ion effect. The solubility product expression for CaSO4 can be written as Ksp = [Ca²+ ][SO4²−]. Since the sodium sulfate solution already contains sulfate ions, we will consider the concentration of [SO4²−] as 0.100 M given in the question. The equation simplifies to Ksp = [Ca²+]*0.100. Therefore, the molar solubility is [Ca²+]= Ksp/0.100 which is 2.4 x 10−5/0.100 = 2.4 x 10−4 M. This suggests that in a 0.100 M Na2SO4 solution, additional CaSO4 would start precipitating when the [Ca²+] reaches 2.4 x 10−4 M.

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The molar solubility of CaSO₄ in a solution containing 0.100 M Na₂SO₄, with a Ksp of 2.4 × 10⁻⁵ is 2.4 × 10⁻⁴ M, which is lower than in pure water due to the common ion effect.

Calculating the molar solubility of calcium sulfate (CaSO₄) in a solution containing 0.100 M sodium sulfate Na₂SO₄ given the solubility product constant (Ksp) for CaSO₄ is 2.4 × 10⁻⁵.

Because Na₂SO₄ dissociates into 2 Na⁺ and 1 SO₄²⁻ in solution, the concentration of SO₄²⁻ due to Na₂SO₄ is 0.100 M. Since the solution is already saturated with sulfate ions, the molar solubility of CaSO₄ will be different from its solubility in pure water. The common ion effect will reduce the solubility of CaSO₄ in the solution.

To find the molar solubility of CaSO₄, let's set up the equation based on the Ksp expression:

Ksp = [Ca²⁺] [SO₄²⁻]

Since the concentration of SO₄²⁻ from Na₂SO₄ is already 0.100 M, and the Ksp for CaSO₄ is 2.4 × 10⁻⁵, the molar solubility of CaSO₄ in this solution is calculated as:

Ksp = (s) × (0.100 M)

2.4 × 10⁻⁵ = (s) × (0.100 M)

s = 2.4 × 10⁻⁴ M

Therefore, the molar solubility of CaSO₄ in a 0.100 M Na₂SO₄ solution is 2.4 × 10⁻⁴M.

Consider the electrolysis of molten barium chloride (bacl2). (a) write the half-reactions. include the states of each species..

Answers

Molten barium chloride is separetes:
BaCl₂(l) → Ba(l) + Cl₂(g), 
but first ionic bonds in this salt are separeted because of heat: 
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).

Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

The anode is positive and the cathode is negative.

Final answer:

The electrolysis of molten barium chloride involves reduction of barium ions and oxidation of chloride ions, requiring two electrons to move through the circuit for one unit of the reaction.

Explanation:

During the electrolysis of molten barium chloride (BaCl2), barium ions (Ba2+) are reduced at the cathode, and chloride ions (Cl-) are oxidized at the anode. The half-reactions, showing the movement of electrons through the circuit, are as follows:

Cathode (reduction): Ba2+(l) + 2e- → Ba(l)Anode (oxidation): 2Cl-(l) → Cl2(g) + 2e-

The overall cell reaction is obtained by combining the half-reactions and balancing the electrons:

Ba2+(l) + 2Cl-(l) → Ba(l) + Cl2(g)

For each mole of barium ions reduced, two electrons are involved in the transfer. Similarly, for each mole of chlorine gas produced, two electrons are given up by chloride ions. As such, there would be two electrons that moved through the circuit for one unit of the reaction to occur.

Please help kind of confused here.
In the early 1900s, Henry Gwyn-Jeffreys Moseley added to the development of the modern periodic table. Which answer best explains how Moseley started his investigations?
1. by reviewing the work of those who contributed to the previous versions of the table
2. by making an entirely new organizational scheme for the previous versions of the table
3. by analyzing the work of a select group of contributors to the previous versions of the table
4. by rejecting the work of all those who had previously developed the earlier versions of the table

Answers

2. by making an entirely new organizational scheme for the previous versions of the table 

Moseley took Mendeleevs periodic table which was organized by weight and organized the elements instead by at atomic number

Answer:

The correct answer is 1. Moseley started his investigations by reviewing the work of those who contributed to the previous versions of the table.

Explanation:

Henry Gwyn Jeffreys Moseley was a British physicist. His most important scientific contribution was the proof of the correctness of the concept of atomic numbers in chemistry.

In 1913, using X-ray spectroscopy, he found a systematic relationship between the wavelength and the atomic number. Previously, it was assumed that the atomic number was an arbitrary number based on the order of the atomic masses, but had to be changed to bring an element into the right place in the periodic table. Moseley's discovery showed that atomic numbers had an experimentally measurable basis. In addition, he showed that there were gaps in numbers 43, 61 and 75 (known today as the radioactive elements technetium and promethium, as well as the stable but rare rhenium). His work was further proof of the then controversial atomic theory.

. how many lone pairs of electrons are present in the lewis structure of calcium sulfide?

Answers

Answer:
            Four

Explanation:
                     Calcium Sulfide is an ionic compound made up of Ca²⁺ and S²⁻.
Ca²⁺ is formed as,

                                   Ca    →    Ca²⁺  +  2 e⁻

These two electrons are accepte by Sulfur as,

                                     S  +  2 e⁻    →    S²⁻

So, before accepting 2 electrons S was having six valence electrons, after accepting two electrons from Ca it has 8 electrons which are present in four pairs as shown below,

Identify the nuclide produced when uranium-238 decays by alpha emission: 238 92u→42he + ?

Answers

Final answer:

The nuclide produced when uranium-238 decays by alpha emission is thorium-234. This process involves the emission of an alpha particle, which causes uranium-238 to lose 2 protons and 2 neutrons, resulting in thorium-234.

Explanation:

The nuclide produced when uranium-238 decays by alpha emission is thorium-234. This decay process involves the release of an alpha particle from uranium-238 nucleus. An alpha particle is equivalent to a helium nucleus - it contains 2 protons and 2 neutrons. Therefore, when uranium-238 (which has 92 protons and 146 neutrons) emits an alpha particle, it loses 2 protons and 2 neutrons, transforming into a new element with 90 protons and 144 neutrons, which is thorium-234 (234 90Th).

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What is the molar concentration of a phosphoric acid (h3po4) solution if 55.5 ml of the solution requires 35.5 ml of .150 m naoh to completely react with the phosphoric acid?

Answers

The  molar  concentration  of H3PO4  solution  is calculated  as below

find the  moles of NaOH  reacted =molarity x volume
= 35.5 x0.150 = 5.325  moles

write a balanced  reacting  equation
=NaOH +3 H3PO4 = Na3PO4 + 3H2O

by  use of reacting ratio between NaOH  to H3PO4  which  is 1:3  the moles of  H3Po4 is therefore
= 5.325moles/3= 1.775 moles

molar  concentration is therefore of H3PO4=  moles /volume
=1.775/55.3 = 0.0320 M of H3PO4

Write the net cell equation for this electrochemical cell. phases are optional. do not include the concentrations. sn(s)||sn2+(aq, 0.0155 m)∥∥ag+(aq, 3.50 m)||ag(s) net cell equation: sn +2ag^{+}->sn^{2+} +2ag sn+2ag+⟶sn2++2ag special δσω λμπ reset ( ) [ ] xyxyyyx⟶↽−−⇀ • (s) (l) (aq) (g) calculate e∘cell, δg∘rxn, δgrxn, and ecell at 25.0 ∘c, using standard potentials as needed

Answers

Answer:

E°cell = 0.94 V

Ecell = 1.00 V

ΔG = -1.9 × 10⁵ J

ΔG° = -1.8 × 10⁵ J

Explanation:

Let's consider this electrochemical cell:

Sn(s)|Sn²⁺(aq,0.0155M)||Ag⁺(aq, 3.50M)|Ag(s)

The corresponding half-reactions are:

Oxidation (anode): Sn(s) → Sn²⁺(aq) + 2 e⁻              E°red = -0.14 V

Reduction (cathode): 2 Ag⁺(aq) + 2 e⁻ → 2 Ag(s)    E°red = 0.80 V

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an = 0.80 V - (-0.14 V) = 0.94 V

We can find the cell potential using the Nernst equation.

Ecell = E°cell - (0.05916/n) . log Q

Ecell = 0.94 V - (0.05916/2) . log ([Sn²⁺]/[Ag⁺]²)

Ecell = 1.00 V

We can find ΔG and ΔG° using the following expressions.

ΔG = -n.F.Ecell = (-2mol).(96468J/mol.V).(1.00V) = -1.9 × 10⁵ J

ΔG° = -n.F.E°cell = (-2mol).(96468J/mol.V).(0.94V) = -1.8 × 10⁵ J

Final answer:

The net cell equation for the electrochemical cell is Sn(s) + 2Ag+ -> Sn₂+ + 2Ag. To find the standard cell potential, standard free energy change, free energy change, and cell potential at 25.0 °C, we use standard reduction potentials and the Nernst equation. Standard free energy change is calculated with ΔG°rxn = -nFE°cell, while the cell potential under non-standard conditions is found using the Nernst equation.

Explanation:

The net cell equation for the given electrochemical cell is Sn(s) + 2Ag+ (aq) → Sn₂+ (aq) + 2Ag(s). To calculate the standard cell potential (E°cell), standard free energy change (ΔG°rxn), free energy change (ΔGrxn), and the cell potential (Ecell) at 25.0 °C, we can use the standard reduction potentials and the Nernst equation. The standard cell potential is calculated by subtracting the standard reduction potential of the anode from that of the cathode.

The standard free energy change can be calculated from the standard cell potential using the formula ΔG°rxn = -nFE°cell, where n is the number of moles of electrons transferred in the reaction, and F is the Faraday constant. The cell potential under non-standard conditions (Ecell) can be determined using the Nernst equation, which incorporates the concentration of the ionic species involved in the half-reactions.

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What is the mass of 2.30 moles of magnesium chloride, mgcl2?

Answers

Mole Road Map:                            ÷            ×
Mass (grams) ⇄ Moles ⇄ Particles (atoms, molecules, etc.)
                        ×            ÷
                        ↓            ↓
             Molar Mass     6.02 × 10^23 (Avogadro's number)

Molar Mass of Magnesium Chloride: 1 Mg: 24.3
                                                        + 2 Cl: 2(35.5)
                                                          95.3 g/mol MgCl2
2.30 mol × 95.3 g/mol MgCl2 ≈ 219. grams MgCl2 (according to significant figures)

Which change in oxidation number represents reduction?

A) –1 to +1
B) –1 to –2
C) –1 to +2
D) –1 to 0

Answers

THE ANSWER IS B. The reducing of oxidation state (-1 to -2, a change of -1, a REDUCTION) is what makes it a reduction.
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