What happens to electrons in the photoelectric effect?
A. They are added to atoms.
B. They are knocked away from atoms.
C. They are turned into light.
D. They are made to orbit faster.

The high concentration of salt forces water out of the cells lining your stomach and intestine.

Answer:

a.

[tex]M_{Co^{2+}}=0.5M\\ \\M_{NO_3^{-}}=1.0M[/tex]

b.

[tex]M_{Fe^{3+}}=1.0M\\ \\M_{ClO_4^{-}}=3.0M[/tex]

Explanation:

Hello,

a. In this case, the ions are cobalt (II) and nitrate, for which, one mole of cobalt (II) nitrate contains one mole of cobalt (II) and two moles of nitrate (see subscripts), therefore, concentrations turn out:

[tex]M_{Co^{2+}}=0.5\frac{molCo(NO_3)_2}{L}* \frac{1molCo}{1molCo(NO_3)_2}=0.5M\\ \\M_{NO_3^{-}}=0.5\frac{molCo(NO_3)_2}{L}* \frac{2molNO_3^{-}}{1molCo(NO_3)_2}=1.0M[/tex]

b. In this case, the ions are iron (III) and chlorate, for which one mole of iron (III) is contained in one mole of iron (III) chlorate and three moles of chlorate are in one mole of iron (III) chlorate (see subscripts), therefore, the concentrations turn out:

[tex]M_{Fe^{3+}}=1.0\frac{molFe(ClO_4)_3}{L}* \frac{1molFe^{3+}}{1molFe(ClO_4)_3}=1.0M\\ \\M_{ClO_4^{-}}=0.5\frac{molFe(ClO_4)_3}{L}* \frac{3molClO_4^{-}}{1molFe(ClO_4)_3}=3.0M[/tex]

Regards.

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

Answer:

Vitamin K

Explanation:

this is the answer

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]

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!

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.

Answer:

Explanation:

We shall find out the molecular formula of the substance .

Ration of number of atoms of C , O and H

= [tex]\frac{37.48}{12} :\frac{49.93}{16} :\frac{12.58}{1}[/tex]

= 3.12 : 3.12 : 12.58

= 1 : 1 : 4

volume of gas at NTP

= 250 x 273 / 350 mL .

= 195 mL .

Molecular weight of the substance = .2804 x 22400 / 195 g

= 32. approx

Let the molecular formula be

(COH₄)n  

n x 32 = 32

n = 1

Molecular formula = COH₄

The compound appears to be CH₃OH

a )

CO + 2H₂ = CH₃OH

28g     4g          32g

59      8

For 8 kg hydrogen , CO required = 56 kg

CO is in excess .  hydrogen is the limiting reagent .

mass of product formed

= 32 x 8 / 4

= 64 kg

b )

percentage yield = product actually formed / product to be formed theoretically  x 100

= 59.6 x 100 / 64

= 93.12 %

c )

2CH₃OH + 3O₂ = 2CO₂ + 4H₂O .

64 g                     2 x 22.4 L

Gram of gas in 1 gallon of fuel

= .7914 x 3785

= 2995.5 g

CO₂ produced at NTP by 2995.5 g CH₃OH

= 2 x 22.4 x 2995.5 / 64 L

= 2096.85 L

At 27° C and 766 mm Hg , this volume is equal to

2096.85 x 300 x 760 / 273 x 766

= 2286.18  L .

d )

C₈H₁₈  =  8CO₂

114g           8 x 22.4 L

gram of fuel per unit gallon

= .6986 x 3785

= 2644.2g

gram of CO₂ produced by 1 gallon of fuel  at NTP

= 8 x 22.4 x 2644.2 / 114

= 4156.5 L

So it produces more CO₂ .

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 =

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.  

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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%

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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The molecule shown below is a Saturated Alkane, because are hydrocarbons that have only single bonds and open chains.

Alkanes are acyclic and saturated hydrocarbons, that is, they are compounds formed only by carbon and hydrogen atoms, open chain and with only single bonds between their carbons.

In this case, Saturated alkanes are hydrocarbons whose carbon atoms are linked only with single bonds.

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The molecule shown below is a Saturated Alkane, because are hydrocarbons that have only single bonds and open chains.

Alkanes are saturated hydrocarbons. This means that their carbon atoms are joined to each other by single bonds.

Alkanes are saturated hydrocarbons. This means that contains only carbon and hydrogen atoms bonded by single bonds only. The general formula for an alkane is [tex]C_nH_{2n+2}[/tex].

In this case, Saturated alkanes are hydrocarbons whose carbon atoms are linked only with single bonds.

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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

Answer:

a) 101%

b)59.7%

Explanation:

The equation for the thermal decomposition of baking soda is shown;

2NaHCO3 → Na2CO3 + H2O + CO2

Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles

From the reaction equation;

2 moles of baking soda yields 1 mole of sodium carbonate

0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.

Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.

%yield = actual yield/theoretical yield ×100

% yield = 0.991/0.9805 ×100

%yield = 101%

Since ;

2NaHCO3 → Na2CO3 + H2O + CO2

And H2O + CO2 ---> H2CO3

Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3

Molar mass of H2CO3= 62.03 gmol-1

Molar mass of baking soda= 84 gmol-1

Therefore, mass of baking soda=

0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3

% of NaHCO3= 0.88/1.473 × 100 = 59.7%

The decomposition reaction of baking soda is a reaction in which water and carbon dioxide ae given off as gaseous products.

Reasons:

Mass of baking soda = 1.555 g

Mass of Na₂CO₃ produced = 0.991 g

Required:

Calculation for the theoretical yield

Solution:

Theoretical yield (mass) of Na₂CO₃ produced is found as follows;

Molar mass of Na₂CO₃ = 105.9888 g/mol

Molar mass of NaHCO₃ = 84.007 g/mol

[tex]\displaystyle 1.555 \, g \, NaHCO_3 \times \frac{1 \, mol \, NaHCO_3}{84.007 \, g \, NaHCO_3} \times \frac{1 \, mol \, Na_2CO_3}{2 \, mol \, NaHCO_3} \times 105.9888 \ g \approx 0.9809 \, g \, Na_2CO_3[/tex]

The theoretical yield of Na₂CO₃ ≈ 0.9809 grams.

The percentage yield is given as follows;

[tex]\displaystyle Percentage \ yield = \mathbf{\frac{Actual \, Yield}{Theorectical \, Yield} \times 100 \%}[/tex]

The percentage yield of Na₂CO₃ is therefore;

[tex]\displaystyle Percentage \ yield \ of \ Na_2CO_3= \frac{0.991}{0.9809} \times 100 \% \approx \underline{ 101.02 \%}[/tex]

(Some baking soda may remain if the reaction is not completed)

6. Mass of the unknown mixture of baking soda = 1473 g

Mass loss from the mixture = 0.325 g

Required:

The percentage of NaHCO₃ in the mixture.

Solution:

The chemical in the mass loss from heating the NaHCO₃ = H₂CO₃

Molar mass of H₂CO₃ = 62.03 g/mol

[tex]\displaystyle \mathrm{Number \ of \ moles \ of \ H_2CO_3 \ produced} = \frac{0.325 \, g}{62.03 \, g/mol} \approx 5.2394 \times 10^{-3} \ moles[/tex]

The chemical reaction is presented as follows;

The number of moles of NaHCO₃ in the mixture is therefore;

Mass of NaHCO₃ in the mixture is therefore

[tex]\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \mathbf{ \frac{Mass \ of \ NaHCO_3}{Mass \ of \ mixture} \times 100}[/tex]

Which gives;

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Answer:

1: 2-bromo-3-chloropentane

Explanation:

find longest carbon chain =5

place the Br and Cl on the carbon chain

follow naming rules I guess

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]

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

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:

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:

You know that:

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:

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 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:

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].

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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.

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) =

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.

[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.

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]

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.

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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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.

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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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:

The question is incomplete as some part is missing:

concentration in octanol Partition Ratio = concentration in water

a) What are the intermolecular forces of attraction between octanol molecules? Explain.

b) Which of the intermolecular forces of attraction identified in (a) account for most of the interactions between octanol molecules? Explain. Use the immiscibility in water and the data included in figures 1 and 2 as evidence to support your answer.

c) Would a compound with a partition coefficient less than one be more hydrophobic or more hydrophilic than one with a partition coefficient greater than 10? Explain.

d) Would nonane (figure 2) be more soluble in water or octanol? Explain.

e) Draw another structure for a compound with the same chemical formula as nonane (CH20) that has a lower boiling point. Explain.

f) Are any of the C atoms in the structure you drew for CH20 sp?hybridized? Explain.

Octanol Boiling point = 195°C Figure 2 Nonane (CH20) Boiling point = 151°C

Answer:

1. The forces between octanol molecules would be attractive. These forces include Vanderwaal forces, H-bonds due to the presence of highly polar O-H group.

2. H-bonding ahould account for most of the attractive forces. The O-H bond should behave like and dipole, oxygen of one molecule attracts the hydrogen of the neighbouring molecule forming D-H...A links throughout (D stands for donor of H-Bond and A for acceptor for H-Bond).

3. Partition coefficient less than 1 will be more hydrophilic, generally drugs with low partition coefficients are regarded as hydrophilic. As parition coefficient of 10 mean more of the solute is dispersed in octanol as compared to water.

4. Nonane is non polar, so it would not dissolve in water. It follows the rule like dissolves like. Polar substances dissolve in polar solvents. 1-octanol is able to bind with water through hydrogen bonds thus its soluble in water but nonane doesn't. Nonane will forms a different layer from water.

5) no all carbons in 2-methyloctane are single bonded. Thus sp3 hybrid. A sp2 hybridised carbon would have a double bond C=C.

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.  

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]

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.

Answer:

k= 1.5

[A] = 1 M

[B] = 3 M

m = 2

n = 1

Explanation:

rate = k[A]”[B]"

The rate of the reaction is 4.5 mol L⁻¹s⁻¹.

Rate of a reaction is defined as the change in concentration of any one of the reactants or products of the reaction, in unit time.

Here,

The concentration of A, [A] = 1 M

The concentration of B, [B] = 3 M

The partial order with respect to A, m = 2

The partial order with respect to B, n = 1

The rate constant of the reaction, k = 1.5

The rate of the reaction,

r = k[A]^m [B}^n

r = 1.5 x 1² x 3

r = 4.5 mol L⁻¹s⁻¹

Hence,

The rate of the reaction is 4.5 mol L⁻¹s⁻¹.

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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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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°.