Bonding with intermolecular forces:

1.) is IBr a hydrogen bonding? yes or no?

2.) is IBr dipole-dipole interactions? Yes or no?

3.) is IBr a london dispersion forces? Yes or no?

Answer:

≅ 0.240 grams (3 sig. figs.)

Explanation:

Rxn:   CaCO₃  +          2HCl              => CaCl₂ + H₂O + CO₂

Given:    ?g          45.67ml(0.105M)

                            = 0.04567L x 0.105 mole/L

                            =  0.0048 mole HCl

Rxn ratio for CaCO₃ to HCl is 1:2

∴ moles CaCO₃ consumed = 1/2 of moles HCl used

=> 1/2(0.0048)mole CaCO₃ used = 0.0024 mole CaCO₃

mass CaCO₃ = 0.0024 mole CaCO₃  x  100.09 grams CaCO₃/mole CaCO₃

= 0.23998 grams CaCO₃ (calculator answer)

≅ 0.240 grams (3 sig. figs.)

Answer:

The radius of the atom Ta is 142pm

Explanation:

To solve this question we need to know that for a body-centered cubic lattice we have the equations:

X = 4/√3 R

Where X is edge length and R is radius of the atom in the lattice.

If adge length of unit cell is 328 pm:

328pm = 4/√3 R

328pm * √3 / 4 = R

142pm = R

The radius of the atom Ta is 142pm

In an exothermic reaction, the reactants have more energy than the products.

This experiment shows that the elements of pure solid substance in a system closed off from the surroundings has increased temperature very fast.

Pure substances are substances that are built up of only one kind of particle and have a fixed or abiding structure. Pure substances are further confidential as elements and compounds. An element is a substance that comprises only one type or kind of atom. a pure substance has a constant chemical composition.

No affair where you sample a substance, it is the same. For chemistry, the safest examples of pure substances are elements and compounds. So, examples cover gold, silver, helium, sodium chloride, and pure water.

So we can conclude that In chemistry, a pure substance is an element or compound made up of one type of particle.

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Answer:

V «T, 1/P, n (Product of

Temperature, Half of pressure, and Moles)

Explanation: If V o n, V « 1/P, and V T

then, V« T, 1/P, n

Answer:

Chlorine Ion

Explanation:

The chlorine ion is adding 1 valence electron in order to complete its outer most shell. It has a charge of 1- meaning it is adding an electron.

Answer:

3 m/s^2

Explanation:

m = 3 kg

F = 9 N

Use Newton's Second Law: F = ma

9 = 3a

a = 3 m/s^2

Considering the formal charges, the lewis structure in which the electronegative atom (O) contains a negative charge and the electropositive (N) contains a positive charge represents bonding in the N₂O. Therefore, option (B) is correct.

A formal charge in chemical bonding can be described as the charge assigned to an atom in a molecule by considering that electrons in all chemical bonds are equally shared between atoms, regardless of relative electronegativity.

The formal charge can be described as the difference between the number of valence electrons of an atom in a neutral state and the number assigned in a Lewis structure to that atom.

When determining the best Lewis structure for a molecule, the structure is selected such that the formal charge is as close to zero on each atom.

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Answer:

The atomic size decreases with an increase in atomic number when we move from left to right.

Explanation: Hope it helps you:))))))

                      Have a great day.

Answer:

36.66%

Explanation:

Step 1: Given data

Step 2: Calculate the mass of salt

The mass of the sample is equal to the sum of the masses of the components.

m(sample) = m(iron) + m(sand) + m(salt)

m(salt) = m(sample) - m(iron) - m(sand)

m(salt) = 2.875 g - 0.660 g - 1.161 g

m(salt) = 1.054 g

Step 3: Calculate the percent of salt in the sample

We will use the following expression.

%(salt) = m(salt) / m(sample) × 100%

%(salt) = 1.054 g / 2.875 g × 100% = 36.66%

The percent of salt in the original sample containing only iron, sand, and salt is 36.66%.

Salt is a mineral made up of primarily sodium chloride.

Seawater contains a vast amount of salt.

Given,

The total mass of the sample is 2.875 g

Mass of iron is  0.660 g

Mass of sand is 1.161 g

Step 1: To find the mass of salt

The total mass of the sample = the sum of the masses of the compounds

m(sample) = m(iron) + m(sand) + m(salt)

2.875 g =  0.660 g + 1.161 g + m(salt)

m(salt) = 2.875 g - 0.660 g - 1.161 g

m(salt) = 1.054 g

Step 2: Calculate the percent of salt in the sample

\(\%(salt) = \dfrac{m(salt) }{m(sample)} \times 100\%\\\\\\\%(salt) = \dfrac{1.054 g }{2.875 g} \times 100\% = 36.66\%\)

Thus, the percentage of salt in the sample is 36.66%.

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You know that one mole of tennantite contains 12 moles of copper, as given by the chemical formula of the ore C12As4S13 so the first thing to do here is to figure out how many moles of tennantite you have in that sample.

If heat is going into the system, then energy must have come out from the surroundings.

If heat is entering a system, it means that the system is gaining thermal energy, which can lead to an increase in temperature, changes in state, or other effects depending on the nature of the system and the heat transfer mechanism.

The first law of thermodynamics, also known as the law of conservation of energy, states that the total amount of energy in a closed system remains constant, and energy can neither be created nor destroyed, only transferred or converted from one form to another.

When heat enters a system, it is either used to increase the internal energy of the system or to perform work, such as moving a piston or driving an electrical generator.

The amount of heat transferred to a system can be quantified using the equation Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.

Therefore, if the system is gaining energy in the form of heat, then the surroundings (the rest of the universe) must be losing that same amount of energy. This is because energy is conserved.

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Answer: Diamond is the name ^-^

A food chain only follows just one path as animals find food. eg: A hawk eats a snake, which has eaten a frog, which has eaten a grasshopper, which has eaten grass. A food web shows the many different paths plants and animals are connected. eg: A hawk might also eat a mouse, a squirrel, a frog or some other animal.

Hope this helps ^_^

Answer:

Explanation:

Alkali metals   ------ outermost orbit containing one electron

ns²np¹

Alkaline metals -------- outermost orbit containing two electron

ns²np²

halogens --------------- outermost orbit containing seven electron

ns²np⁵

noble gas --------------- outermost orbit containing eight electron

ns²np⁶.

The speed of the molecules in the lake will decrease due to decrease in the kinetic energy.

Kinetic energy is the energy due to the motion of particles.

Kinetic energy increases with increase in temperature.

Since the temperature of the lake decreases ad it freezes, the kinetic energy decrease and the speed of the molecules in the lake will decrease.

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RbF is expected to have the largest difference between the experimental and ideal Van 't Hoff factors. Because RbF has a very high lattice energy due to the small size of the F- ion, making it more difficult for the ions to dissociate in water. Option D is correct.

The Van 't Hoff factor (i) is a measure of the number of particles into which a solute dissociates in a solution. The ideal Van 't Hoff factor (i) is the value expected for a completely dissociated solute, while the experimental Van 't Hoff factor (i) is the value determined by experiments.

For ionic compounds that dissociate in water, the experimental Van 't Hoff factor is often less than the ideal value due to ion pairing or other interactions that reduce the effective number of ions in solution.

The size of the difference between the experimental and ideal Van 't Hoff factors depends on the degree of ion pairing or other interactions in solution. The degree of ion pairing or other interactions in solution is influenced by factors such as the size and charge of the ions, as well as the solvent.

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Answer: The answer is incus

Explanations: The first law of thermodynamics states that energy cannot be created or destroyed, hoever, it can only change its state.

A) Incorrect - this is not related to thermodynamics

B) Incorrect - has no correlations to energy being conserved.

C) Correct - energy state of the heat is transfered from the system to surrounding, this is an example of thermodynamic

D) Incorrect - no energy is conserved here.

Only Ag2CO3 will precipitate from the solution.

When Na2CO3 is added to the solution, it will react with the Ag^+ and Pb^2+ ions to form precipitates of Ag2CO3 and PbCO3. The Mn^2+ ion concentration is not high enough to form a precipitate with Na2CO3.

First, let's calculate the initial concentration of Ag^+ and Pb^2+ ions in the solution:

Ag^+: 0.00050 M

Pb^2+: 0.00050 M

Next, we need to calculate the concentration of Na2CO3 after it is added to the solution. Since we added 10.00 ml of 1.0*10^-6 M Na2CO3 to a total volume of 500 ml, the final concentration of Na2CO3 is:

[Na2CO3] = (10.00 ml / 500 ml) * 1.010^-6 M

[Na2CO3] = 2.010^-8 M

Now we can use the Ksp values to determine which precipitates will form.

For Ag2CO3:

Ksp = [Ag^+]^2[CO3^2-]

8.110^-12 = (2x)^2 (2x)

8.110^-12 = 4x^3

x = 2.0*10^-4 M

Since the concentration of CO3^2- is higher than the solubility product, Ag2CO3 will precipitate.

For PbCO3:

Ksp = [Pb^2+][CO3^2-]

7.410^-14 = (0.00050 M)(2x)

x = 9.210^-11 M

Since the concentration of CO3^2- is lower than the solubility product, PbCO3 will not precipitate.

Therefore, the only precipitate that will form is Ag2CO3.

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An average brightness was calculated for each length of graphite tested to get a better understanding of the relationship between the length of the graphite and the brightness of the line it produced.

Graphite has many uses in various industries. Some of its uses include: Pencils, Lubricants, Refractories, Batteries, Electrodes, Nuclear reactors, Aerospace industry.

By calculating the average, it is possible to see if there is a trend in the data and if longer or shorter lengths of graphite produce brighter or duller lines. This information can be useful in determining the best length of graphite to use for a particular task or project.

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Here's link to the answer:

tinyurl.com/wtjfavyw

Including the title, abstract, introduction, techniques & materials, results, commentary, conclusion, & references, a lab report is divided into eight components.

This lab is equipped with all the tools required for sample preparation as well as the standards needed for a variety of fluid and solid sample analyses. Glassware, exhaust fans, a muffled furnace, centrifuges, aggregate mills, and ovens are among the conventional equipment found in the lab.

It gives you an opportunity to put your knowledge to use while reinforcing the topics you have studied in class. You will gain knowledge of certain crucial experimental skills that are required if you want to work as a successful chemist.

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An equilibrium reaction occurs when the rates of forward and backward reactions are equal. Therefore, the correct equation representing an equilibrium system is:

2Mg(s) + O₂(g) ⇌ 2MgO(s)

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Approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M.

To determine the mass of glucose (C6H12O6) required to prepare a 0.215 M solution in 5000 mL, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, let's convert the volume of the solution from milliliters (mL) to liters (L):

5000 mL = 5000/1000 = 5 L

Now, we can rearrange the formula to solve for moles of solute:

moles of solute = Molarity (M) x volume of solution (L)

moles of solute = 0.215 M x 5 Lmoles of solute = 1.075 mol

Since glucose (C6H12O6) has a molar mass of approximately 180.16 g/mol, we can calculate the mass of glucose using the equation:

mass of solute = moles of solute x molar mass of solute

mass of glucose = 1.075 mol x 180.16 g/mol

mass of glucose = 194.0 g (rounded to three significant figures)

Therefore, approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M. It's important to note that the molar mass of glucose used in this calculation may vary slightly depending on the level of precision required.

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The predominant form of ethylenediamine at pH 6.184 is H2NCH2CH2NH+3.

At pH 6.184, which is slightly acidic, the amino groups in ethylenediamine (H2NCH2CH2NH2) can partially protonate, resulting in the formation of ammonium ions (H+3NCH2CH2NH+3). The equilibrium between the neutral form and the protonated form is pH-dependent.

Therefore, at pH 6.184, the predominant form of ethylenediamine is H2NCH2CH2NH+3.

Answer:

Babygirl these are double DD'S

Explanation:

OooOOO I'm hurt oh my neck my neck and my back

The wavelengths of the corresponding elements with given electron affinities are, i) 9977.4nm,ii)849.12 nm, iii)613.9nm, iv)897.96 nm, v)365.02 nm, vi)343 nm, vii)368.3 nm, and viii)405.84 nm

The wavelength of each photon could be calculated as follows:

ΔE = hc/λ

Therefore for finding the wavelength of C with electron affinity as 122 we get

i) ΔE = 122x1000/6.023 x 10²³

       = 1992.3626x10⁻²³

ΔE = 1.9923 x 10⁻²⁰J

λ = hc/ΔE

λ = 6.626 x 10⁻³⁴x3x10⁸/1.9923x10⁻²⁰

  = 19.878/1.9923x10⁶

  = 9.9774 x 10⁻⁶

λ = 9977.4nm

Therefore, the wavelength of C is 9977.4nm

ii) For O we get,

ΔE = 141x1000/6.023 x 10²³

       = 23410.26x10⁻²³

ΔE = 23.41 x 10⁻²⁰J

λ = hc/ΔE

λ = 6.626 x 10⁻³⁴x3x10⁸/23.41x10⁻²⁰

  = 19.878/23.41x10⁶

  = 0.84912 x 10⁻⁶

λ = 849.12 nm

Therefore, the wavelength of O is 849.12 nm

iii) For Se we get

ΔE = 195x1000/6.023 x 10²³

       = 32375.89x10⁻²³

ΔE = 32.375 x 10⁻²⁰J

λ = hc/ΔE

λ = 6.626 x 10⁻³⁴x3x10⁸/32.375x10⁻²⁰

  = 19.878/32.375x10⁶

  = 0.61399 x 10⁻⁶

λ = 613.9nm

Therefore, the wavelength of Se is 613.9 nm

iv) For Si we get,

ΔE = 134x1000/6.023 x 10²³

       = 22248.049 x10⁻²³

ΔE = 22.248 x 10⁻²⁰J

λ    = hc/ΔE

λ = 6.626 x 10⁻³⁴x3x10⁸/22.248x10⁻²⁰

  = 19.878/22.248x10⁶

  = 0.89796 x 10⁻⁶

λ = 897.96 nm

Therefore, the wavelength of Si is 897.96 nm

v) For F we get,

ΔE = 328x1000/6.023 x 10²³

     = 54457.9x10⁻²³

ΔE = 54.457 x 10⁻²⁰J

λ = hc/ΔE

λ = 6.626 x 10⁻³⁴x3x10⁸/54.457x10⁻²⁰

  = 19.878/54.457x10⁶

  = 0.36502 x 10⁻⁶

λ = 365.02 nm

Therefore, the wavelength of F is 365.02 nm

vi) For Cl we get,

ΔE   = 349x1000/6.023 x 10²³

       = 57944.54 x 10⁻²³

ΔE = 57.944 x 10⁻²⁰J

λ = hc/ΔE

λ = 6.626 x 10⁻³⁴x3x10⁸/57.944x10⁻²⁰

  = 19.878/57.944x10⁶

  = 0.3430 x 10⁻⁶

λ = 343 nm

Therefore, the wavelength of Cl is 343 nm

vii) For Br we get,

ΔE = 325x1000/6.023 x 10²³

       = 53959.8 x10⁻²³

ΔE = 53.959 x 10⁻²⁰J

λ = hc/ΔE

λ = 6.626 x 10⁻³⁴x3x10⁸/53.959x10⁻²⁰

  = 19.878/53.959x10⁶

  = 0.3683 x 10⁻⁶

λ = 368.3 nm

Therefore, the wavelength of Cl is 368.3 nm

viii) For I we get,

ΔE = 295x1000/6.023 x 10²³

       = 48978.9x10⁻²³

ΔE = 48.9789 x 10⁻²⁰J

λ = hc/ΔE

λ = 6.626 x 10⁻³⁴x3x10⁸/48.9789x10⁻²⁰

  = 19.878/48.9789x10⁶

  = 0.40584x 10⁻⁶

λ = 405.84 nm

Therefore, the wavelength of F is 405.84 nm

Therefore, the wavelengths of the elements with given electron affinities were found in nanometers.

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Fluorine (F-) is the element with the fewest valence electrons. Seven electrons make up the outermost shell of fluorine, and one of them is unpaired. As a result, fluorine possesses seven valence electrons altogether.

Eight valence electrons are present in magnesium (Mg2+), nine are present in gallium (Ga+), eight are present in argon (Ar+), four are present in carbon (C+), six are present in sulphur (S2-), and seven are present in fluorine (F-).

Fluorine has a lower number of valence electrons than the other elements because it has a greater effective nuclear charge. This indicates that the fluorine atom will take electrons away from its outermost shell since it is more attracted to electrons than the other elements.

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