Suppose that you add 28.8 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f Kf of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.06 oC compared to pure benzene. What is the molar mass of the unknown compound

Answer:

The base is involved in the rate determining step of an E2 reaction mechanism

Explanation:

Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;

Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.

For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;

Rate = k[alkyl halide] [base]

The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.

Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.

Answer:

The new solution is 1.4% m/V

Explanation:

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:

[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]

We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

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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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Answer:

[tex]14\\8[/tex]O

Explanation:

The top number always represents the mass number.

The bottom number always represents the atomic number.

The element always goes after the numbers.

If charge is present, that comes after the element.

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:

7p

Explanation:

principal quantum number is 7

n=7( principle shell)

angular momentum quantum number gives sub shell

l = 1 means it is p orbital

so answer is 7p orbital

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

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:

Option B

Explanation:

Scientific question are answered through experimentation, through testing the theory about the physical world.

Answer: its A

through observing and measuring the physical world

Explanation:

Answer:

B. Valence Electron Pairs

Explanation:

Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.

Answer:

The element is strontium and the number of neutrons it have is 51.

Explanation:

Based on the given information, the ionic compound is,  

XCl₂ ⇔ X₂⁺ + 2Cl⁻

X2+ is the ion of the mentioned element

As mentioned in the given question, the number of electrons of the element X is 36 and as seen from the reaction the charge present on the ion is +2. Now the atomic number will be,  

No. of electrons = atomic number - charge

36 = atomic number - 2

Atomic number = 38

Based on the periodic table, the atomic number 38 is for strontium element, and the sign of strontium is Sr. Hence, the element X is Sr.  

Now based on the given information, the mass number of the element is 89. Now the no. of neutrons will be,  

No. of neutrons = mass number - atomic number

= 89 - 38

= 51 neutrons.  

Answer:

WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.

Explanation:

Equation for the reaction:

K2SO4 + H20 ------->2 K+ + SO4^2-

When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.

1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-

At STP, 1 mole of K2SO4 will be the molar mass of the substance

Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol

Molar mass = 174 g/mol

So therefore;

1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion

When 63.7 g is used; we have:

174 g = 2 moles of K+

63.7 g = ( 63.7 * 2 / 174) moles of K+

= 0.73 moles of K+

Forr sulfate ion, we have:

174 g = 1 mole ofSO4^2-

63.7 g = (63.7 * 1 / 174) moles of SO4^2-

= 0.366 moles of SO4^2-

In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.

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]

Answer:

[tex]3Sr^{+2}+6Cl^{-}+6Li^{+}+2PO_{4}^{3-}-->Sr_{3}(PO_{4})_{2}+6Li^{+}+6Cl^{-}\\\\3Sr^{+2}+2PO_{4}^{3-} --->Sr_{3}(PO_{4})_{2}[/tex]

First equation is the complete ionic equation.

Second equation is the net ionic equation.

Answer:

Equilibrium constant Kc = [x]² / [A - x] [B - x]

Explanation:

The equilibrium constant is defined as the ratio of the concentration of the products to that of the reactants at equilibrium

ie Kc = [products] / [reactants].

The balanced equation of the reaction is given as : A + B ⇄ AB

At the beginning of the reaction,

Initial concentration I = A = 1M

                                       B = 1M

                                      AB = 0M

After a period of time and assuming 'x' to be the concentration of product AB formed, the concentrations become

                                         C = reactant A = [A - x] M

                                                 rectant B =   [B - x] M

                                              Product AB =  [x] [x] M

At equilibrium, the concentrations are,

                                            E  = rectant A = [A - x] M

                                                   reactant B = [B - x] M

                                                   product AB = [x]² M

therefore , the equilibrium constant, Kc  = [products]/[reactants]

                                                                   = [x]² / [A - x] [B - x]

Answer:

pH = 3.39

Explanation:

The equilibrium in water of ascorbic acid (With its conjugate base) is:

H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)

Where the acidic dissociation constant is written as:

Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]

H₂O is not taken in the Ka expression because is a pure liquid.

As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:

[H₂C₆H₆O₆] = 2.5x10⁻³M - X

[HC₆H₆O₆⁻] = X

[H₃O⁺] = X

Replacing in the Ka expression:

7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]

1.975x10⁻⁷ - 7.9x10⁻⁵X = X²

0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷

Solving for X:

X = -0.00048566→  False solution, there is no negative concentrations

X = 0.00040666 → Right solution

As [H₃O⁺] = X, [H₃O⁺] = 0.00040666

pH is defined as -log [H₃O⁺];

pH = -log 0.00040666,

Answer:

Option B. 0.136 g

Explanation:

The balanced equation for the reaction is given below:

2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)

Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:

Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol

Mass of AgNO3 from the balanced equation = 2 x 170 = 340g

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of Ag2O = (108x2) + 16 = 232g/mol

Mass of Ag2O from the balanced equation = 1 x 232 = 232g

Summary:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH.

Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.

From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.

Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.

In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.

The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted to produce 232g of Ag2O.

Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.

Therefore, 0.136g of Ag2O was produced from the reaction.

Answer:

Polarity in chemistry referred to physical properties of compounds related to solubility, melting and boiling properties.

Polarity of black pepper can be seen when black pepper is sprinkled on water. The balck pepper float on water and get displaced if touched.

It means black pepper is non-polar and have no difference in electronegativity between bonded atoms. Black pepper is so light in weight and non-polar, the surface tension of water keep it floating in the water.

Answer:

A. [OH⁻] = 2.188x10⁻³M

B. pKb = 6.02

Explanation:

When hydrazine is in equilbrium with water, its reaction is:

N₂H₄(aq) + H₂O(l) ⇄ HN₂H₄⁺(aq) + OH⁻(aq)

Where Kb, is defined as the ratio between concentrations in equilibrium of the species, thus:

Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]

A. From pH, you can find [OH⁻], thus:

pH = -log [H⁺]

11.34 = -log [H⁺]

4.57x10⁻¹² = [H⁺]

As 1x10⁻¹⁴ = [OH⁻] [H⁺]

1x10⁻¹⁴ / 4.57x10⁻¹² = [OH⁻]

B. Concentrations in equilibrium of the species are:

[N₂H₄] = 5.0M - X

[HN₂H₄⁺] = X

[OH⁻] = X

Where X is reaction coordinate

As [OH⁻] = 2.188x10⁻³M

X = 2.188x10⁻³M

Replacing:

[N₂H₄] = 5.0M - 2.188x10⁻³M = 4.9978M

[HN₂H₄⁺] = 2.188x10⁻³M

[OH⁻] = 2.188x10⁻³M

Replacing in Kb expression:

Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]

Kb = [2.188x10⁻³M] [2.188x10⁻³M] / [4.9978M]

Kb = 9.577x10⁻⁷

pKb is defined as -log Kb

pKb = -log 9.577x10⁻⁷

Answer:

10.328 m

Explanation:

normal atmospheric pressure = 101325 Pa

density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3

pressure = pgh

where p = density

g = acceleration due to gravity = 9.81 m/s^2

h = height of column

imputing values, we have

101325 = 1000 x 9.81 x h

height of column h = 101325/9810 = 10.328 m

Answer: The amount of sample left after 8323 years is 4.32g

Explanation:

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]

[tex]k=\frac{0.693}{8694years}=7.97\times 10^{-5}years^{-1}[/tex]

b) amount left after 8323 years

[tex]t=\frac{2.303}{7.97\times 10^{-5}}\log\frac{8.30g}{a-x}[/tex]

[tex]8323=\frac{2.303}{7.97\times 10^{-5}}\log\frac{8.30g}{a-x}[/tex]

[tex]0.285=\log\frac{8.30}{a-x}[/tex]

[tex]\frac{8.30}{a-x}=1.92[/tex]

[tex](a-x)=4.32g[/tex]

The amount of sample left after 8323 years is 4.32g

Answer:

3,4,1 and 6,5,2

Explanation:

In the electromagnetic spectrum the arrangement of the waves in increasing frequencies and decreasing wavelengths are as follows;

Radio waves

Microwaves

Infrared waves

Visible light rays

Ultraviolet rays

X-rays

Gamma rays

(a simple mnemonic is RMIVUXG)

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:

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:

0.48 dm3  (or 480 cm3)

Explanation:

First find the original no. of moles existing in the sulphuric acid:

no. of moles = volume (in dm3) x concentration

                     = 120/1000 x 10

                     = 1.2 mol

Then let the total volume of the diluted acid be v dm3.

Since

Concentration = no. of moles / volume,

so by substituting the given information,

2 = 1.2 / v

v = 0.6 dm3

Hence, the volume of water required

= 0.6 - 120/1000

= 0.48 dm3  (or 480 cm3)

Considering the definition of dilution, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].

First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one and consists of reducing the amount of solute per unit volume of solution. This is accomplished simply by adding more solvent to the solution in the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

In this case, you know:

Replacing in the definition of dilution:

10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³= 2 [tex]\frac{mol}{dm^{3} }[/tex]× Vf

Solving:

Vf= (10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³) ÷2 [tex]\frac{mol}{dm^{3} }[/tex]

Vf= 600 cm³

In summary, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].

Learn more about dilution:

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:

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.

For more information about the FCC structure, refer to the link:

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Answer:

B. Reactant + Reactant -> Product + Product

Explanation:

Reactants are substances that- as the name suggests- reacts with other substances at the beginning of a reaction

Products are substances that are produced as a result of the reaction

Typically, when writing a chemical reaction, an arrow is used to show the direction the reaction is moving.  In this case, the arrows in options B and C suggest that the reaction only moves in one direction- forwards

And as mentioned above, reactants are the substances at the start of the reaction, they're what mixes together to form a new product.  

To keep things simple:

Products can't be at the beginning of a reaction since they weren't formed yet.

Similarly, reactants can't be part of the products since they already existed and didn't need to be made. In a lot cases, the reactants would be completely used up to make the products

As such, only one possible chemical reaction would follow that reasoning:

    Reactant + Reactant ->  Product + Product

Reactant + Reactant → Product + Product is the correctly written chemical reaction. Hence, option B is correct.

A chemical equation is a mathematical expression of the chemical reaction which represents the product formation from the reactants.

In an equation, the reactants are written on the left-hand side and the products are written on the right-hand side demonstrated by one-headed or two-headed arrows.

Hence, option B is correct.

Learn more about the chemical equation here:

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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]