Need help ASAP please! It is hard for me to understandd

Need Help ASAP Please! It Is Hard For Me To Understandd

Answers

Answer 1
It’s all of the above

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please help me I am begging you.. )))): PLEASE HELP ME ~~~~~~~~~~~~~~~~~~~~~~ A football player experiences acute pain in his knee. Which of the following methods can a doctor use to diagnose the reason for the pain? --_-_-____- A.) Use infrared radiation from warm objects to look inside the knees. B.) Use radio waves emitted by radioactive substances to look at bones. C.) Use radiations emitted by very hot objects to penetrate the skin and bones. D.) Use x‒ray radiation to see if there are any fractured bones.

Answers

Answer:

D. Use x-ray radiation to see if there are any fractured bones.

Explanation:

The football player may have fractured a bone while he was practicing or playing, so it is best for the doctor to check if the player broke his bone or fractured it.

Many free radicals combine to form molecules that do not contain any unpaired electrons. The driving force for the radical–radical combination reaction is the formation of a new electron‑pair bond. Consider the formation of hydrogen peroxide. 2OH(g)⟶H2O2(g) Write Lewis formulas for the reactant and product species in the chemical equation. Include nonbonding electrons.

Answers

Answer:

In the attached image the Lewis equation is shown where it is shown how two oxygens react with two hydrogens to meet the octet of the electrons.

Explanation:

Hydrogen peroxide is one of the most named chemicals since it is not only sold as "hydrogen peroxide" in pharmacies but it is also one of the great weapons of immune defense cells to defend ourselves against anaerobic bacteria.

The disadvantage of this compound is that when dividing it forms free oxygen radicals that are considered toxic or aging for our body.

In the attached image below, you will see the Lewis equation is shown there. There, you will see how two oxygens react with two hydrogens to come about the octet of the electrons.

When two or more atoms bond with each other, they often form a molecule. When two hydrogens and an oxygen share electrons through covalent bonds, a water molecule is formed.

The octet rule is known as when most atoms want to gain stability in their outer most energy level by filling themselves that is the S and P orbitals of the highest energy level with eight electron.

HOOH is the compound  that is form. It is called Hydrogen peroxide. This because it is has reactive oxygen species and the simplest peroxide.

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what volume(in dm³) of 0.30 moldm-³ Nacl solution can be prepared from 0.060mol of solute?

Answers

Answer:

0.2 dm³

Explanation:

Concentration = no. of moles of solute / volume

Concentration: 0.30 mol dm-³

no. of moles of solute: 0.060mol

0.3 = 0.06 / v

v = 0.2 dm³

The volume(in dm³) of 0.30 moldm-³ Nacl solution can be prepared from 0.060mol of solute is 0.2 dm³.

What is volume?

Volume is the space occupied by a three-dimensional object. A mixture of two or more components is referred to as a solution.

A solute in these combinations is a substance that dissolves in a solvent, which is a different substance.

Nacl is a common salt used in daily basis for cooking purpose.

Given, the 0.30 moldm-³ Nacl solution

Solute is 0.060mol

Concentration = no. of moles of solute divided by the volume of the Nacl solution.

Concentration is 0.30 mol dm-³

No. of moles of solute is 0.060mol

0.3 = 0.06 / v

The volume = 0.2 dm³

Therefore, the volume(in dm³) of 0.30 moldm-³ Nacl solution can be prepared from 0.060mol of solute is 0.2 dm³.

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Write the empirical formula

Answers

Answer:

[tex]Fe(C_{2}H_{3}O_{2})_{3}\\Fe (OH)_{3}\\\\NH_{4}(C_{2}H_{3}O_{2})\\\\NH_{4}OH[/tex]

Explanation:

[tex]Fe^{3+}(C_{2}H_{3}O_{2})^{-}_{3}--->Fe(C_{2}H_{3}O_{2})_{3}\\Fe^{3+} (OH^{-})_{3}--->Fe (OH)_{3}\\\\NH_{4}^{+}(C_{2}H_{3}O_{2})^{-}--->NH_{4}(C_{2}H_{3}O_{2})\\\\NH_{4}^{+}OH^{-} ---> NH_{4}OH[/tex]

Using the provided table, determine the enthalpy for the reaction
2 NH3 (g) + 3 N20 (g) 4 N2 (g) + 3 H20 (1)

Answers

Answer:

ΔH°r = -1009.8 kJ

Explanation:

Let's consider the following balanced reaction.

2 NH₃(g) + 3 N₂O(g) ⇒ 4 N₂(g) + 3 H₂O(l)

We can calculate the standard enthalpy of the reaction (ΔH°r) using the following expression.

ΔH°r = [4 mol × ΔH°f(N₂(g)) + 3 mol × ΔH°f(H₂O(l))] - [2 mol × ΔH°f(NH₃(g)) + 3 mol × ΔH°f(N₂O(g))]

ΔH°r = [4 mol × 0 kJ/mol + 3 mol × (-285.8 kJ/mol)] - [2 mol × (-46.2 kJ/mol) + 3 mol × 81.6 kJ/mol]

ΔH°r = -1009.8 kJ

Determine the radius of an Al atom (in pm) if the density of aluminum is 2.71 g/cm3 . Aluminum crystallizes in a face centered cubic structure with an edge leng

Answers

Answer:

143pm is the radius of an Al atom

Explanation:

In a face centered cubic structure, FCC, there are 4 atoms per unit cell.

First, you need to obtain the mass of an unit cell using molar mass of Aluminium  and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.

Mass of an unit cell

As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:

4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g

Edge length

As density of aluminium is 2.71g/cm³, the volume of an unit cell is:

1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³

And the length of an edge of the cell is:

∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m

Radius:

As in FCC structure, Edge = √8 R, radius of an atom of Al is:

4.044x10⁻¹⁰m = √8 R

1.430x10⁻¹⁰m = R.

In pm:

1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =

143pm is the radius of an Al atom

The radius of the atom of Al in the FCC structure has been 143 pm.

The FCC lattice has been contributed with atoms at the edge of the cubic structure.

The FCC has consisted of 4 atoms in a lattice.

The mass of the unit cell of Al can be calculated as:

[tex]\rm 6.023\;\times\;10^2^3[/tex] atoms = 1 mole

4 atoms = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles

The mass of 1 mole Al has been 26.98 g/mol.

The mass of [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles × 26.98 g

The mass of 1 unit cell of Al has been = 1.792 [tex]\rm \bold{\times\;10^-^2^2}[/tex] g.

The volume of the Al cell can be calculated as:

Density = [tex]\rm \dfrac{mass}{volume}[/tex]

Volume = Density × Mass

The volume of Al unit cell = 2.71 g/[tex]\rm cm^3[/tex] × 1.792 [tex]\rm \times\;10^-^2^2[/tex] g

The volume of Al cell = 6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex]

The volume of the cube has been given as:

Volume = [tex]\rm edge\;length^3[/tex]

6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex] = [tex]\rm edge\;length^3[/tex]

Edge length = [tex]\rm \sqrt[3]{6.613\;\times\;10^-^2^3}[/tex] cm

Edge length = 4.044 [tex]\rm \times\;10^-^8[/tex] cm

Edge length = 4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.

In an FCC lattice structure, the radius of the atom can be given by:

Edge length = [tex]\rm \sqrt{8\;\times\;radius}[/tex]

4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = [tex]\rm \sqrt{8\;\times\;radius}[/tex]

Radius = 1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.

1 m = [tex]\rm 10^1^2[/tex] pm

1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = 143 pm.

The radius of the atom of Al in the FCC structure has been 143 pm.

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The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35. (a) Find the energy of the third excited rotational state; that is, the J

Answers

Answer:

the energy of the third excited rotational state [tex]\mathbf{E_3 = 16.041 \ meV}[/tex]

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = [tex]\dfrac{m_1 \times m_2}{m_1 + m_2}[/tex]

μ = [tex]\dfrac{1 \times 35}{1 + 35}[/tex]

μ = [tex]\dfrac{35}{36}[/tex]

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = [tex]\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg[/tex]

μ  = 1.6139 × 10⁻²⁷ kg

[tex]r_o = 127 \ pm = 127*10^{-12} \ m[/tex]

The rotational level Energy can be expressed by the equation:

[tex]E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)[/tex]

where ;

J = 3 ( i.e third excited state)  &

[tex]I = \mu r^2_o[/tex]

[tex]E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)[/tex]

[tex]E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)[/tex]

[tex]E_3= 2.5665 \times 10^{-21} \ J[/tex]

We know that :

1 J = [tex]\dfrac{1}{1.6 \times 10^{-19}}eV[/tex]

[tex]E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV[/tex]

[tex]E_3 = 16.041 \times 10 ^{-3} \ eV[/tex]

[tex]\mathbf{E_3 = 16.041 \ meV}[/tex]

g The most common position for an double bond in an unsaturated fatty acid is delta _________(fill in the number).

Answers

Answer:

The most common position for an double bond in an unsaturated fatty acid is delta 9 (Δ⁹)

Explanation:

Unsaturated fatty acids are carboxylic acids which contains one or more double bonds. The chain length as well as the number of double bonds is written separated by a colon. The positions of the double bonds are specified starting from the carboxyl carbon, numbered as 1, by superscript numbers following a delta (Δ). For example, an 18-carbon fatty acid containing a  single double bond between carbon number 9 and 10 is written as 18:1(Δ⁹).

In most monounsaturated fatty acids, the double bond is between C-9 and C-10 (Δ⁹), and the other double bonds of polyunsaturated fatty acids are generally Δ¹² and Δ¹⁵. This positioning is due to the nature of the biosynthesis of fatty acids. In the mammalian hepatocytes, double bonds are introduced easily into fatty acids at the Δ⁹ position, but cannot introduce additional double bonds between C-10 and the methyl-terminal end. However, plants are able to introduce these additional double bonds at the  Δ¹² and Δ¹⁵ positions.

What is the number of Cl atoms in a 784 gram pure sample of NC13?

Answers

Answer:

idk

Explanation:

idk cool pee bee mee nee hee gee fee kee

In a 784 gram pure sample of NC13, there are approximately 1.33 x 10²⁵ chlorine atoms.

To determine the number of chlorine (Cl) atoms in a given sample, we need to utilize the Avogadro's number and the molar mass of chlorine.

The molar mass of chlorine (Cl) is approximately 35.45 grams/mol. To calculate the number of moles in the sample, we divide the given mass by the molar mass:

Number of moles of Cl = Mass / Molar mass

Number of moles of Cl = 784 g / 35.45 g/mol

Number of moles of Cl ≈ 22.08 mol

According to Avogadro's number, there are 6.022 x 10²³ entities (atoms, molecules, or formula units) in 1 mole of a substance. Therefore, to find the number of chlorine atoms, we multiply the number of moles by Avogadro's number:

Number of Cl atoms = Number of moles of Cl x Avogadro's number

Number of Cl atoms = 22.08 mol x 6.022 x 10²³ atoms/mol

Number of Cl atoms ≈ 1.33 x 10²⁵ atoms

Therefore, in a 784 gram pure sample of NC13, there are approximately 1.33 x 10²⁵ chlorine atoms.

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Concerning Boyle's Law, if you had a gas at a pressure of 101 kPa and decreased the volume of the container holding the gas to one quarter from where it started, what would be the new pressure of the gas

Answers

Answer:

[tex]P_2=404 kPa[/tex]

Explanation:

Hello,

In this case, the Boyle's is mathematically defined via:

[tex]P_1V_1=P_2V_2[/tex]

Which stands for an inversely proportional relationship between volume and pressure, it means the higher the volume the lower the pressure and vice versa. In such a way, since the volume is decreased to one quarter, we can write:

[tex]V_2=\frac{1}{4} V_1[/tex]

We can compute the new pressure:

[tex]P_2=\frac{P_1V_1}{V_2} =\frac{P_1V_1}{\frac{1}{4} V_1} =\frac{101kPa*V_1}{\frac{1}{4} V_1} \\\\P_2=4*101kPa\\\\\\P_2=404 kPa[/tex]

Which means the pressure is increased by a factor of four.

Regards.

A mineral that helps in clotting of blood________.
please tell me what is the answer of this question.

Answers

Answer:

Vitamin K

Explanation:

this is the answer

Which best describes the trends in electonegativity on the periodic table

Answers

Hey! :)

__________ ☆ ☆__________________________________

Answer:

The answer is Electronegativity increases up and to the right

Explanation:

When you move from left to right it increases ( in the periodic table )

But when you move down the table electronegativity decreases.

So “ Electronegativity increases up and to the right” describes the trends the best.

Hope this helps! :)

____________☆ ☆________________________________

By, BrainlyMember ^-^

Good luck!

What is the percent yield for a process in which 10.4g of CH3OH reacts and 10.1 g of CO2 is formed
according to the following equation:
2CH3OH(I) + 302(g) → 2C02(g) + 4H2O(1)
A. 70.7%
B. 103%
C. 97.1%
D. 52.1%
E. 37.9%

Answers

Answer:

A. 70.7%

Explanation:

In the first step lets compute the molar mass of CH₃OH and CO

Molar Mass of CH₃OH =  1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)

                                     = 32.042 g/mol

Molar Mass of CO₂      = 1(12.01 g/mol) + 2(16.00 g/mol)  

                                     = 44.01 g/mol

                                   

Mass of only one reactant i.e. CH₃OH is given so  it must be the limiting reactant. Next, the theoretical yield is calculated directly as follows:

Given mass of CH₃OH is 10.4 g. So we have:

                                     10.4g CH₃OH

Convert grams of CH₃OH to moles of CH₃OH utilizing molar mass of CH₃OH as:

                          1 mol CH₃OH / 32.042 g CH₃OH

Convert CH₃OH to moles of CO₂ using mole ratio as:

                             2 mol CO₂ / 2 mol CH₃OH

Convert moles of  CO₂ to grams of  CO₂ utilizing molar mass of  CO₂ as:

                           44.01 g/mol CO₂ / 1 mol CO₂

Now calculating theoretical yield using above steps:

[ 10.4 g CH₃OH ]  [1 mol CH₃OH / 32.042 g CH₃OH ]  [2 mol CO₂ / 2 mol CH₃OH]  [44.01 g/mol CO₂ / 1 mol CO₂]

Multiplication is performed here. We are left with 10.4 and 44.01 g CO₂ from numerator terms in the above equation and 32.042 from denominator terms after cancellation process of above terms. So this equation becomes:

= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042

= 457.704/32/042

=  14.28 g CO₂

Theoretical yield =  14.28 g CO₂  

Finally compute the percent yield for a process in which 10.4g of CH₃OH reacts and 10.1 g of CO₂ is formed:

percent yield = (actual yield / theoretical yield) x 100

As we have calculated theoretical yield which is 14.28 g CO₂ and actual yield is 10.1 g CO₂ So,

percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%

                       = 0.707 x 100%

                       = 70.7 %

Hence option A 70.7% yield is the correct answer.

The percent yield for a process is:

A. 70.7%

Calculation for percent yield:

In the first step, lets compute the molar mass of CH₃OH and CO

Molar Mass of CH₃OH =  1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)

= 32.042 g/mol

Molar Mass of CO₂ = 1(12.01 g/mol) + 2(16.00 g/mol)  

= 44.01 g/mol

Mass of only one reactant i.e. CH₃OH is given so it must be the limiting reactant.

Given mass of CH₃OH= 10.4 g.

Converting into number of moles:

1 mol CH₃OH / 32.042 g CH₃OH

Convert CH₃OH to moles of CO₂ using mole ratio as:

2 mol CO₂ / 2 mol CH₃OH

Convert moles of CO₂ to grams of  CO₂ utilizing molar mass of  CO₂ as:

44.01 g/mol CO₂ / 1 mol CO₂

Calculation for theoretical yield:

[ 10.4 g CH₃OH ]  [1 mol CH₃OH / 32.042 g CH₃OH ]  [2 mol CO₂ / 2 mol CH₃OH]  [44.01 g/mol CO₂ / 1 mol CO₂]

= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042

= 457.704/32/042

=  14.28 g CO₂

Theoretical yield =  14.28 g CO₂  

Adding values in percent yield formula:

Percent yield = (actual yield / theoretical yield) / 100

Percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%

= 0.707 x 100%

= 70.7 %

Hence, option A is the correct answer.

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. Explain why, in the sample calculations, 0.1 g of the unknown produced a GREATER freezing point depression than~e same mass of naphthalene.

Answers

Answer

Naphthalene is a non electrolyte

If the unknown compound is an electrolyte it gives 2 or more ions in solution

( NaCl >> Na+ + Cl- => 2 ions

Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)

the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )

For naphthalene

delta T = 1.86 x m

for a salt that gives 2 ions

delta T = 1.86 x m x 2

hence the lowering in freezion point of unkown is greater then napthalene

A 3.35 g sample of an unknown gas at 81 ∘C and 1.05 atm is stored in a 1.75 L flask.
What is the density of the gas?
density:
g/L
What is the molar mass of the gas?
molar mass:

Answers

Answer:

Molar mass = 52.96g/mol

density = 1.91g/L

Explanation:

using ideal gas equation

PV=nRT

Ideal gas law is valid only for ideal gas not for vanderwaal gas. Density and mass of unknown gas is 1.91g/L and 52.96g/mol respectively. The equation used to solve this is PM=dRT.

What is ideal gas equation?

Ideal gas equation is the mathematical expression that relates pressure volume and temperature.

Mathematically the relation between Pressure, Molar mass and temperature can be given as

PV=nRT

This equation is arranged as

PM=dRT

Where,

P=pressure of gas = 1.05 atm

M= molar mass=?

d= density=?

R = Gas constant = 0.0821 L.atm/K.mol

T=temperature=354K

density=mass÷ volume

density=3.35 g÷1.75 L

density = 1.91g/L

PM=dRT

P×M=d×R×T

1.05 atm ×M= 1.91g/L× 0.0821 ×354K

M=52.96g/mol

Therefore, density and mass of unknown gas is 1.91g/L and 52.96g/mol respectively.

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To condition the buret, add a small volume of ____________ to the buret and rotate the barrel ____________ so that the liquid makes contact with the full inner surface of the barrel. Complete this action _________ and discard each volume in a designated waste container.

Answers

Answer:

Acid, Horizontally, Drain off the acid by opening the stopcock

Explanation:

A buret is a calibrated glass appratus used to measure and deliver accurate volume of liquid, usually acids, in acid-base titrations.

Before usually the buret to carry out your titration reaction, it is advised to condition or prepare your buret for use.

To condition the buret, add a small volume of the acid to the buret and rotate the barrel horizontally to ensure that the liquid ( acid) makes contact with the inner surface of the barrel o the buret.

This procedure is done to wash off any previous acid or liquid that the buret had been used for in previous titrations.

Complete this action by opening the stopcock of the buret to drain off the acid and discarding each volume in a designated waste container.

Note that acids could be corrosive and dangerous to the skin and so should be handled with great care

Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page?

Answers

The question is incomplete; the complete question is: Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? Same (angles do not change) Different (angles change) Answer Bank | H2O | CO2, SO2, XeF2, BF3 CIF3, NH3, CH4, SF4, XeF4, BrF5, PCI5,SF6

Answer:

Compounds whose real bond angle are the same as ideal bond angle;

SF6, BF3, CH4, PCI5

Compounds whose real bond angles differ from ideal bond angles;

H2O, CO2, SO2, XeF2, CIF3, NH3, SF4, XeF4, BrF5

Explanation:

According to the valence shell electron pair repulsion theory (VSEPR), molecules adopt various shapes based on the number of electron pairs on the valence shell of the central atom of the molecule. The electron pairs usually orient themselves as far apart in space as possible leading to various observed bond angles.

The extent of repulsion of lone pairs is greater than that of bond pairs. Hence, the presence of lone pairs on the valence shell of the central atom in the molecule distorts the bond angles of molecules away from the ideal bond angles predicted on the basis of valence shell electron pair repulsion theory.

For instance, methane is a perfect tetrahedron having an ideal bond angle of 109°28'. Both methane and ammonia are based on a tetrahedron, however, the presence of a lone pair of electrons on nitrogen distorts the bond angle of ammonia to about 107°. The distortion of lone pairs in water is even more as the bond angles of water is about 104°.

Which neutral atom has the following electron configuration?
1s2 2s2 2p6 3s2 3p6 4s2 3d10 5p6 6s2

Answers

Answer: The answer is Strontium(Sr) i.e there are 38 electrons in this electronic configuration.

How To Find: You should count the number of electrons in each orbitals.

For example:

1s2 has two electrons in its orbital.Count similarly the number of electrons for every orbital.

When aqueous silver nitrate solution is mixed with barium chloride solution, a white precipitate forms along with barium nitrate solution. Write a balanced chemical equation for the reaction with appropriate states of matter.

Answers

Answer: [tex]2AgNO_3(aq)+BaCl_2(aq)\rightarrow 2AgCl(s)+Ba(NO_3)_2(aq)[/tex]

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The reaction for aqueous silver nitrate solution with barium chloride solution produces barium nitrate solution and a white precipitate of silver chloride.

Thus the balanced chemical reaction will be:

[tex]2AgNO_3(aq)+BaCl_2(aq)\rightarrow 2AgCl(s)+Ba(NO_3)_2(aq)[/tex]

Which of the following is NOT a type of crystal structure? A. None of these B. Metallic C. Ionic D. Macromolecular (giant covalent)

Answers

Answer:

A. None of these.  

Explanation:

A crystal structure is an arrangement of atoms or ions in a repeating three-dimensional array.

B. is wrong. Metal atoms, such as gold, arrange themselves into a crystal structure.

C. is wrong. Ionic solids, such as sodium chloride, arrange themselves into a crystal structure.

D. is wrong. Macromolecules (network solids), such as diamond, arrange themselves into a crystal structure.

The correct answer is None of these.  

What is a crystal structure?A crystal structure is a three-dimensional collection of atoms or ions that repeats itself.Metal atoms(gold), Ionic solids (sodium chloride), and Macromolecules(network solids) arrange themselves into a crystal structure.

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A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calculate the mole fraction and molality of benzene in this solution.

Answers

Answer:

[tex]x_B=0.0769[/tex]

[tex]m=0.990m[/tex]

Explanation:

Hello,

In this case, we can compute the mole fraction of benzene by using the following formula:

[tex]x_B=\frac{n_B}{n_B+n_C}[/tex]

Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:

[tex]n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol[/tex]

Thus, we compute the mole fraction:

[tex]x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769[/tex]

Next, for the molality, we define it as:

[tex]m=\frac{n_B}{m_C}[/tex]

Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:

[tex]m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg[/tex]

Or just 0.990 m in molal units (mol/kg).

Best regards.

Considering the definition of mole fraction and molality:

the mole fraction of benzene is 0.077.the molality of benzene is 0.9908 [tex]\frac{moles}{kg}[/tex].

You know that:

Mass of benzene = 6.24 gramsMass of cyclohexane= 80.74 gramsMolar mass of benzene= 78.11 g/moleMolar mass of cyclohexane= 84.16 g/mole

Mole fraction

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

Being the molar mass and the mass in the solution of each compound, the number of moles of each compound can be calculated as:

Benzene: [tex]\frac{mass of benzene}{molar mass of benzene} =\frac{6.24 grams}{78.11 \frac{grams}{mole} } = 0.08 moles[/tex]Cyclohexane:[tex]\frac{mass of cyclohexane}{molar mass of ciclohexane} =\frac{80.74 grams}{84.16\frac{grams}{mole} } = 0.96 moles[/tex]

So, the total moles of the solution can be calculated as:

Total moles = 0.08 moles + 0.96 moles = 1.04 moles

Finally, the mole fraction of benzene can be calculated as follow:

[tex]\frac{number moles of benzene}{total moles} =\frac{0.08 moles}{1.04 moles} = 0.077[/tex]

Finally, the mole fraction of benzene is 0.077.

Molality

Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.

The Molality of a solution is determined by the expression:

[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]

In this case, you know:

number of moles of solute= 0.08 moles Mass of solvent = 80.74 g = 0.08074 kg (being 1000 g=1 kg)  

Replacing:

[tex]Molality benzene=\frac{0.08 moles}{0.08074 kg}[/tex]

Molality benzene= 0.9908 [tex]\frac{moles}{kg}[/tex]

Finally, the molality of benzene is 0.9908 [tex]\frac{moles}{kg}[/tex].

Learn more about:

mole fraction brainly.com/question/14434096?referrer=searchResults brainly.com/question/10095502?referrer=searchResults molality brainly.com/question/20366625?referrer=searchResults brainly.com/question/4580605?referrer=searchResults

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is

Answers

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

[tex][A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M[/tex]

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

[tex]\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%[/tex]

Hello please help me on this question Describe how you would separate and recover iodine from an impure aqueous solution of iodine

Answers

Answer:

First the aqueous solution of iodine is heated mildly and then collection of the iodine crystals is done from its vapors.

Explanation:

Iodine is one of the elements that can get recovered easily from a given solution by going through the process of mild heating. For doing this, first, the aqueous solution is heated mildly over a low flame with a dish placed over the flame. As the process of mild heating continues, the fumes of the iodine start to originate that slowly get condense around the dish's cooler parts.  

With condensation, the formation of pure iodine crystals takes place. These iodine crystals can now be extracted easily by a physical method.  

Question 4 (2 points)
CuO(s) + H2(g)
Cu(s) +
H2O(1)
Balance the equation

Answers

Answer:

CuO(s) + H₂(g) --> Cu(s) + H₂O(l)

Explanation:

It is already balanced. You can see that the values of the elements of the reactants are equal to the values of the elements of the products.

At 25.0°C the Henry's Law constant for methane CH4 gas in water is ×1.410−3/Matm.
Calculate the mass in grams of CH4 gas that can be dissolved in 75.mL of water at 25.0°C and a CH4 partial pressure of 0.68atm. Round your answer to 2 significant digits.

Answers

Answer:

1.1 × 10⁻³ g

Explanation:

Step 1: Given data

Henry's Law constant for methane (k): 1.4 × 10⁻³ M/atm

Volume of water (=volume of solution): 75 mL

Partial pressure of methane (P): 0.68 atm

Step 2: Calculate the concentration of methane in water (C)

We will use Henry's law.

[tex]C = k \times P = 1.4 \times 10^{-3}M/atm \times 0.68atm = 9.5 \times 10^{-4}M[/tex]

Step 3: Calculate the moles of methane in 75 mL of water

[tex]\frac{9.5 \times 10^{-4}mol}{L} \times 0.075 L = 7.1 \times 10^{-5}mol[/tex]

Step 4: Calculate the mass corresponding to 7.1 × 10⁻⁵ mol of methane

The molar mass of methane is 16.04 g/mol.

[tex]7.1 \times 10^{-5}mol \times \frac{16.04g}{mol} = 1.1 \times 10^{-3} g[/tex]

50.0ml each of 1.0M Hcl and 1.0M Naoh at room temperature (20.0c) are mixed the temperature of the resulting Nacl solutions increase to 27.5c
the density if the resulting Nacl solutuion 1.02 g/ml
the specific heat of the resulting Nacl solutions is 4.06j/gc
calculate the heat of neutralisation of hcl and naoh in kj/mol nacl products​

Answers

Answer:

62.12kJ/mol

Explanation:

The neutralization reaction of HCl and NaOH is:

HCl + NaOH → NaCl + H₂O + HEAT

You can find the released heat of the reaction and heat of neutralization (Released heat per mole of reaction) using the formula:

Q = C×m×ΔT

Where Q is heat, C specific heat of the solution (4.06J/gºC), m its mass and ΔT change in temperature (27.5ºC-20.0ºC = 7.5ºC).

The mass of the solution can be finded with the volume of the solution (50.0mL of HCl solution + 50.0mL of NaOH solution = 100.0mL) and its density (1.02g/mL), thus:

100.0mL × (1.02g / mL) = 102g of solution.

Replacing, heat produced in the reaction was:

Q = C×m×ΔT

Q = 4.06J/gºC×102g×7.5ºC

Q = 3106J = 3.106kJ of heat are released.

There are 50.0mL ×1M = 50.0mmoles = 0.0500 moles of HCl and NaOH that are reacting releasing 3.106kJ of heat. That means heat of neutralization is:

3.106kJ / 0.0500mol of reaction =

62.12kJ/mol is heat of neutralization

What is the freezing point of an aqueous solution that boils at 105.9 ∘C? Express your answer using two significant figures.

Answers

Answer:

THE FREEZING POINT OF THE AQUEOUS SOLUTION IS - 7.3 °C

Explanation:

To solve this problem, we must know the following variables:

Normal boiling point of water (solvent) = 100 °C

The molar boiling point elevation constant of water  = 1.51 °C /m

Normla freezing point of water ( solvent) = 0 °C

The molar freezing point depression constant = 1.86 °C /m

The boiling point of the aqueous solution = 105.9 °C

Molarity = xM

Change in boiling point = boiling point of solution - boiling point of water

Change in boiling point = 105.9 - 100 °C

= 5.9 °C

From the formula:

Change in boiling point = i * Kb * M

Re- arranging the formula by making M the subject of the equation, we have:

M = change in boiling point / Kb

i = 1

M = 5.9 °C / 1.51 °C/m

M = 3.907 M

Then, we calculate the freezing point:

Change in freezing point = i * Kb * M

= 1 * 1.86 °C/m * 3.907 M

= 7.267 °C

Hence, the freezing point = freezing point of water - change in freezing point

Freezing point = 0 °C - 7.267 °C

Freezing point = - 7.267 °C

Freezing point = -7.3 °C

A compound containing only carbon and hydrogen and which has only single bonds between atoms is classified as an Group of answer choices

Answers

Answer:

Alkanes

Explanation:

Alkanes are hydrocarbons containing only C-H and C-C single bonds. They're saturated compounds and they form a homologous series with the general formula CₙH₂ₙ₊₂, where n is the number of carbon atoms. Example of members in this group are

Methane = CH₄

Ethane = C₂H₆

Propane = C₃H₈

All alakane compounds ends with with the suffix "-ane" and this differentiate them during naming from other compounds.

Identify the acid, base, conjugate acid and conjugate base in the following reactions:
a. NH_3(aq) + CH_3COOH(aq) NH_4^+ (aq) + CH_3COO^-(aq)
b. HClO_4(aq) + NH_4(aq) ClO_4^- (aq) + NH_4^+ (aq)

Answers

Answer:

a. NH₃ : base

CH₃COOH (acetic acid) : acid

NH₄⁺ : conjugate acid

CH₃COO⁻ : conjugate base

b. HClO₄ (perchloric acid) : acid

NH₃ : base

ClO₄⁻ : conjugate base

NH₄⁺ : conjugate acid

Hope this helps.

In the purification process, the organic layer is washed sequentially with 2 M NaOH, water, and saturated aqueous sodium chloride. What is the purpose of each of this wash and why is cold 2 M NaOH recommended

Answers

Answer:

Its washed with 2M Naoh to remove acidic impurities like sulphuric acid and HBr

Its washed in water to remove water soluble impurities

Its washed with Nacl to remove large quantity of water that may be in the organic layer

Explanation:

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