Select the correct answer
What determines the average kinetic energy of the particles in a gas?
ОА
the number of collisions
OB.
the number of particles
OC. the size of the particles
OD. the temperature

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]

Answer:

conductivity increasing order CH₃-CH₂-OH < HF< NaCl

Explanation:

NaCl is the better conductor comparing with remaining two. it is strong electrolyte. dissociation percent always nearly eqaul to 100% but HF is weaker acid than NaCl and dissociation percent <100% . So, the no of ions furnished by HF less than that of NaCl

CH₃-CH₂-OH organic compound . in general it is not treated as an electrolyte and it cannot carry any charge. If it  carries, it is very very less compared to remaining two

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:

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:

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:

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:

Boron and Aluminium

Explanation:

Boron and Aluminium are present in Group 13 of the modern periodic table. Group 13 (IUPAC System) can also be referred to as Group III-A. Logically, Boron and Aluminum can't be placed alongwith elements such as Yttrium as they don't exhibit properties of a transition metal.

Answer:

1) HCl (Hydrochloric acid reacts with NH3 and forms dense fumes)

2) Methane (It is from the group of alkanes and is a green house gas)

3) Deliquescent substance (It is a solid which when kept in open forms a solution after sometime)

4) Brass (It is a solid-in-solid solution used to make electrical fittings)

5) Aluminium

Question 2:

1)    Al₂O₃ + 2NaOH    ⇒   2NaAlO₂  (Sodium Aluminate) + H₂O

2)   Zn +  H₂SO₄   (dilute)  =>  ZnSO₄  (Zinc Sulphate) + H₂

Answer:

equations

Aluminium oxide and Sodium hydroxide react to form water and sodium aluminate

Zinc reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas

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:

D . Sulphur

Explanation:

the element with a 3p4 valence configuration, look in period 3 and group XVI, and that is ...

S, sulfur.

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; 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°.

Answer: 10 moles/kg.

Explanation:

Given, Mass of solute = 24.2 g

Molar mass of solute = 24.2 g/mol

[tex]\text{Moles of solute =}\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}\\\\=\dfrac{24.2}{24.2}=1[/tex]

Mass of solvent = 100.0g = 0.1 kg  [1 g=0.001 kg]

[tex]\text{Molality}=\dfrac{\text{Moles of solute}}{\text{kilograms of Solvent}}\\\\=\dfrac{1}{0.1}\\\\=10\ moles/kg[/tex]

Hence, the molality of the resulting solution is 10 moles/kg.

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.

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:

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:

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:

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:

B. 1.82(10³) seconds (1820 seconds)

Explanation:

Speed of Light: 3.00(10⁸) m/s or 3.00(10⁵) km/s

We know the distance from the Moon to Earth as 5.46(10⁸) km

We simply divide the distance by the speed of light:

[tex]\frac{5.46(10^8)}{3.00(10^5)}[/tex]

We should get 1820 seconds or B. 1.82(10³) as our answer.

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:

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! :)

____________☆ ☆________________________________

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Good luck!

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:

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:

MgSo4.7H2O = Magnesium sulfate

Cs3PO4.H2O = Cesium Phosphate

Hope this helps!

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:

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

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.

Answer:

C. An electron at this electrode has a higher potential energy than it has at a standard hydrogen electrode.

Explanation:

The standard hydrogen electrode (SHE) is used to measure the electrode potential of substances. The standard hydrogen electrode is arbitrarily assigned an electrode potential of zero. Recall that electrode potentials are always measured as reduction potentials in electrochemical systems.

For an electrode that has a negative electrode potential, electrons at this electrode have a higher potential energy compared to electrons at the standard hydrogen electrode. Electrons flow from this electrode to the hydrogen electrode.

On the other hand, a positive electrode potential implies that an electron at this electrode has a lower potential energy than it has at a standard hydrogen electrode. Hence electrons will flow from the standard hydrogen electrode to this electrode.

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:

[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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