Determine the percent sulfuric acid by mass of a 1.61 m aqueous solution of H2SO4. %

Answer:

The value of the dissociation constant will be:[tex]3.94\times 10^{-4}[/tex].

Explanation:

                [tex]AsH\rightleftharpoons As^++H^+[/tex]

At ,t= 0      c              0     0

At eq'm   (c-x)            x     x

Concentration of aspirin = c

[tex]c=\frac{2.00 g}{180 g/mol\times 0.600 L}=0.01851 M [/tex]

Expression for dissociation constant will be given as:

[tex]K_a=\frac{[H^+][As^+]}{[AsH]}=\frac{x\times c}{(c-x)}=\frac{x^2}{(c-x)}[/tex]..(1)

The pH of the solution  = 2.60

The pH of the solution is due to free hydrogen ions whcih come into solution after partial dissociation of aspirin.

[tex]pH=2.60=\log[H^+]=-\log[x][/tex]

[tex]x=0.002511 M[/tex]

Putting value of x in (1).

[tex]K_a=\frac{x^2}{(c-x)}=\frac{(0.002511 M)^2}{(0.01851 M-0.002511 M)}[/tex]

[tex]K_a=3.94\times 10^{-4}[/tex]

The value of the dissociation constant will be:[tex]3.94\times 10^{-4}[/tex].

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

                                  = 2.75 x 10²³ molecules XeF₆.

To determine the number of molecules of XeF₆,

First, we will determine the number of moles of F₂ present in the 12.9L of F₂

From the ideal gas equation

PV = nRT

Where P is the pressure

V is the volume

n is the amount of substance ( number of moles)

R is ideal gas constant

T is the temperature

From the question,

P = 2.6 atm

V = 12.9 L

R = 0.082057 L atm mol⁻¹ K⁻¹

T = 298 K

Putting these values into the equation

PV = nRT

2.6 × 12.9 = n × 0.082057 × 298

33.54 = n × 24.452986

∴ n = 33.54 ÷ 24.452986

n = 1.3716116 moles

The number of moles of F₂ present is 1.3716116 moles

From the given equation of reaction

Xe(g) +3F₂(g)→XeF₆(g)

1 mole of Xe reacts with 3 moles of F₂ to produce 1 mole of XeF₆

∴ 1.3716116 moles of F₂ will give (1.3716116/3) moles of XeF₆

Hence, the number of moles of XeF₆ produced = 0.457204 moles

To determine the number of molecules formed,

From the formula

Number of molecules = number of moles × Avogadro's number

∴ Number of molecules of XeF₆ formed = 0.457204 × 6.02214 ×10²³

Number of molecules of XeF₆ formed = 2.7533 × 10²³ molecules

Number of molecules of XeF₆ formed ≅ 2.75 × 10²³ molecules

Hence, the number of molecules of XeF₆ that are formed from 12.9 L of F₂ (at 298 K and 2.6 atm) is 2.75 × 10²³ molecules.

The correct option is E) 2.75 × 1023 molecules XeF6

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Answer: Correct answer is a - bromo-3-isopropylcyclodecane

Answer:

The polymerization of amino acids produces polypeptides.

Explanation:

The polymerization of amino acids producing polypeptide is known as translation. mRNA produces amino acids.

The polymerization of amino acids produces polypeptides. Hence, option D is correct.

Polymerization is defined as "A chemical process that combines several monomers to form a polymer or polymeric compound."

Polymerization of amino acids is the formation of proteins. Amino acids are small molecules consisting of an amino group, a carboxyl group, a hydrogen atom, and an ‘R’ group.  

In polymerization, two amino acids make a bond with each other with a peptide bond, which is formed between the amino group of one amino acid and the carboxyl group of another amino acid.

Hence, the polymerization of amino acids produces polypeptides which are also known as translation. mRNA produces amino acids.

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Answer & Explanation:

1- The sulfur present in the crude oil can poison the catalysts that are used in the refining of crude oil into the many useful products obtained from crude oil.

2- The crude oil, containing sulfur compounds, introduced into a car or truck engine, it will again interfere with the catalytic converter installed in the exhaust system of the vehicle.

3-  if sulfur is present in fuel, when the fuel is burned it will be converted to SO₂ which is a known precursor to acid rain.

4- Sulfur compounds can lead to increased corrosion in distillation equipment at the high temperatures used.

The correct answer is that H2S, mercaptans, sulphur compounds, and salt are removed from crude oils to prevent corrosion, environmental pollution, and to meet product specifications.

H2S (hydrogen sulphide), mercaptans (thiols), and other sulphur compounds are removed from crude oil for several reasons:

1. Corrosion Prevention: H2S is a highly corrosive gas that can cause significant damage to pipelines, storage tanks, and refinery equipment. It can lead to the formation of sulphuric acid when it reacts with moisture, which accelerates corrosion. Removing H2S and other sulphur compounds helps to mitigate this risk.

2. Environmental Concerns: Sulphur compounds in fuels contribute to the formation of acid rain when burned, as they can react with moisture and oxygen in the atmosphere to form sulphuric acid. Additionally, H2S is toxic and can be harmful to human health and the environment if released into the atmosphere.

3. Product Specifications: Many refined products, such as gasoline, diesel, and jet fuel, have strict sulphur content specifications. Low sulphur fuels burn cleaner and are less harmful to the environment. The removal of sulphur compounds is necessary to meet these specifications.

4. Safety: H2S is also a toxic gas that can be lethal at high concentrations. Its removal is essential for the safety of workers and for the safe operation of refineries and transportation systems.

5. Catalyst Poisoning: Sulphur compounds can act as poisons for catalysts used in various refining processes, reducing their effectiveness and lifetime. Removing sulphur before processing increases the efficiency and longevity of these catalysts.

Salt is removed from crude oil primarily because it can cause corrosion and scaling in refinery equipment, which can lead to reduced efficiency and increased maintenance costs. Additionally, salt can contaminate the final products, affecting their quality and usability.

The process of removing these impurities typically involves a combination of physical and chemical treatments, such as distillation, adsorption, hydrodesulphurization, and caustic washing, depending on the nature and concentration of the contaminants.

Answer : The number of moles of [tex]H_2O[/tex] produced are, 0.66 mole.

Explanation : Given,

Given moles of [tex]NH_3[/tex] = 0.44 mole

The given balanced chemical reaction is,

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

From the given balanced chemical reaction, we conclude that

As, 4 moles of [tex]NH_3[/tex] react to give 6 moles of [tex]H_2O[/tex]

So, 0.44 moles of [tex]NH_3[/tex] react to give [tex]\frac{6}{4}\times 0.44=0.66[/tex] moles of [tex]H_2O[/tex]

Therefore, the number of moles of [tex]H_2O[/tex] produced are, 0.66 mole.

Answer:

The 1st functional group is secondary amine and the 2nd fun. group is tertiary amine.

Explanation:

Lidocaine does not contain amide functional group in its composition as nitrogen is associated only with carbon.

In lidocaine, Functional group 1 is a secondary amine because it has two hydrocarbon groups attached to the nitrogen atom, and Functional group 2 is an amide due to the presence of a carbonyl group bonded to the nitrogen atom.

To classify the nitrogen-containing functional groups in lidocaine, it is important to understand the structure of amines and amides. Amines are classified as primary, secondary, or tertiary by the number of hydrocarbon groups attached to the nitrogen atom, with a primary amine having one alkyl or aryl group, a secondary amine having two, and a tertiary amine having three. An amide is a functional group with a carbonyl group (C=O) bonded to a nitrogen atom. In the structure of lidocaine, Functional group 1 is a secondary amine because it has two carbon atoms directly bonded to the nitrogen atom. Functional group 2 is an amide, identifiable by its carbonyl group bonded to the nitrogen atom.

Answer : The work done on the surroundings is, 709.1 Joules.

Explanation :

The formula used for isothermally irreversible expansion is :

[tex]w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)[/tex]

where,

w = work done

[tex]p_{ext}[/tex] = external pressure = 1.00 atm

[tex]V_1[/tex] = initial volume of gas = 1.00 L

[tex]V_2[/tex] = final volume of gas = 8.00 L

Now put all the given values in the above formula, we get :

[tex]w=-p_{ext}(V_2-V_1)[/tex]

[tex]w=-(1.00atm)\times (8.00-1.00)L[/tex]

[tex]w=-7L.atm=-7\times 101.3J=-709.1J[/tex]

The work done by the system on the surroundings are, 709.1 Joules. In this, the negative sign indicates the work is done by the system on the surroundings.

Therefore, the work done on the surroundings is, 709.1 Joules.

The work done by the gas on the surroundings when it expands is -709.1 J.

The subject of this question is the work done by a gas during isothermal expansion. In physics, the work done by a gas is given by the formula W = -PΔV, where P is the pressure and ΔV is the change in volume. For this question, the external pressure P is 1.00 atm (which is equivalent to 101.3 J/L), the initial volume V₁ is 1.00 L, and the final volume V₂ is 8.00 L.

Thus, the change in volume ΔV is V₂ - V₁ = 8.00 L - 1.00 L = 7.00 L. Substituting into the formula for work, we find W = -(1.00 atm)(7.00 L) = -7.00 L⋅atm.

One last step needed is to convert from liters-atmospheres to joules. As given in the question, 1 L⋅atm is approximately 101.3 J, so W = -7.00 L⋅atm * 101.3 J/L⋅atm = -709.1 J. The negative sign indicates work was done by the gas on the surroundings.

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Final answer:

To find the arsenic concentration in ppb, we convert the volume of the sample to liters and then use the formula Concentration (ppb) = (Mass of As (ng) / Volume of water (L)) x 1 billion. For a 14.3 cm3 sample containing 159.5 ng As, the arsenic concentration is 11,153.85 ppb.

Explanation:

To find the concentration of arsenic (As) in the lake water sample in parts per billion (ppb), we need to use the given information about the mass of arsenic and the volume of the water sample. Since we are provided with a 14.3 cm3 sample containing 159.5 ng As and assuming that the density of the lake water is 1.00 g/cm3, we can calculate the concentration.

First, we need to convert the volume of the lake water from cm3 to liters since 1 cm3 is equal to 1 mL and there are 1,000 mL in a liter. For the 14.3 cm3 sample, this converts to:

14.3 cm3 × (1 mL/1 cm3) = 14.3 mL

14.3 mL × (1 L/1,000 mL) = 0.0143 L

Next, we can calculate the concentration of arsenic in the water sample in ppb using the following equation:

Concentration (ppb) = (Mass of As (ng) / Volume of water (L)) × 1 billion

Filling in the values we have:

Concentration (ppb) = (159.5 ng / 0.0143 L) × 1,000,000,000 ng/billion

Concentration (ppb) = 11,153.85 ppb

The arsenic concentration in the sample is therefore 11,153.85 ppb.

Explanation:

It is known that atmospheric pressure is equal to 760 torr.

And, at atmospheric pressure that is, 760 torr the boiling point of pure water is 100 degree celsius.

So, calculate the boiling point of pure water at 653.7 torr as follows.

                   [tex]\frac{100^{o}C}{760 torr} \times 653.7 torr[/tex]

                = [tex]0.131 \times 653.7 torr[/tex]

                = [tex]86.01^{o}C[/tex]

Therefore, we can conclude that the boiling point of pure water at 653.7 torr is [tex]86.01^{o}C[/tex].

Answer: The molecular formula for the given organic compound is [tex]C_2H_2O_4[/tex]

Explanation:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=21.33g[/tex]

Mass of [tex]H_2O=4.366g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.33 g of carbon dioxide, [tex]\frac{12}{44}\times 21.33=5.82g[/tex] of carbon will be contained.

In 18g of water, 2 g of hydrogen is contained.

So, in 4.366 g of water, [tex]\frac{2}{18}\times 4.366=0.485g[/tex] of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5.82g}{12g/mole}=0.485moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = [tex]\frac{0.485}{0.485}=1[/tex]

For Hydrogen  = [tex]\frac{0.485}{0.485}=1[/tex]

For Oxygen  = [tex]\frac{0.969}{0.485}=1.99\approx 2[/tex]

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is [tex]C_1H_{1}O_2=CHO_2[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

[tex]n=\frac{90.04g/mol}{45g/mol}=2[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4[/tex]

Thus, the molecular formula for the given organic compound is [tex]C_2H_2O_4[/tex]

Answer : The net ionic equation will be,

[tex]3Cu^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Cu_3(PO_4)_2(s)[/tex]

Explanation :

First we have to balance the chemical reaction.

The given balanced ionic equation will be,

[tex]3CuCl_2(aq)+2K_3PO_4(aq)\rightarrow Cu_3(PO_4)_2(s)+6KCl(aq)[/tex]

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The ionic equation in separated aqueous solution will be,

[tex]3Cu^{2+}(aq)+6Cl^-(aq)+6K^+(aq)+2PO_4^{3-}(aq)\rightarrow Cu_3(PO_4)_2(s)+6K^+(aq)+6Cl^-(aq)[/tex]

In this equation, [tex]K^+\text{ and }Cl^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]3Cu^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Cu_3(PO_4)_2(s)[/tex]

Answer:

True => ΔH°f for C₆H₆ = 49 Kj/mole

Explanation:

See Thermodynamic Properties Table in appendix of most college level general chemistry texts. The values shown are for the standard heat of formation of substances at 25°C. The Standard Heat of Formation of a substance - by definition - is the amount of heat energy gained or lost on formation of the substance from its basic elements in their standard state. C₆H₆(l) is formed from Carbon and Hydrogen in their basic standard states. All elements in their basic standard states have ΔH°f values equal to zero Kj/mole.

Answer:

Case I => %C-14 remaining = 77% => Age of artifact = 2200 yrs

Case II => %C-14 remaining = 6.2% => Age of artifact = 23,000 yrs

Explanation:

Given:

Half-Life C-14 = 5730 yrs

=> Rate Constant = k = 0.693/t(1/2) = (0.693/5730)yrs⁻¹ = 1.2 x 10⁻⁴ yrs⁻¹

NOTE => All radioactive decay is 1st order kinetics.

=> A = A₀eˉᵏᵗ  (classic 1st order decay equation)

- A = remaining activity

-A₀ = initial activity

- k = rate constant

- t = time of decay (or, age of object of interest; i.e., not everything is organic but the 1st order decay equation is good for non-organic objects (rocks) also. Analysts just use a different decay standard => K-40 → Ar-40 + β).

Solving the decay equation for time (t) ...

t = ln(A/A₀)/-k

Applying to problem cases...

Case I => %C-14 remaining = 77%

t = ln(A/A₀)/-k = ln(77/100)/-1.2x10⁻⁴ years = 2178 yrs ~ 2200 yrs

Case II => %C-14 remaining = 6.2%

t = ln(A/A₀)/-k = ln(6.2/100)/-1.2x10⁻⁴ years = 23,172 yrs ~ 23,000 yrs

Answer:

[tex]\boxed{\text{a) 2160 yr ago; b) 23 000 yr ago}}[/tex]

Explanation:

Two important equations in radioactive decay are

[tex](1) \qquad \ln \dfrac{N_{0} }{N_{t}} = kt\\\\(2) \qquad t_{\frac{1}{2}} = \dfrac{\ln2}{k }[/tex]

We use them for carbon dating.

a) The dye

Data:

[tex]t_{\frac{1}{2}} = \text{5730 yr}\\\\N_{t} = 0.77 N_{0}[/tex]

Calculations:

[tex]\text{From Equation (2)}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{\text{5730 yr}} = 1.21 \times 10^{-4} \text{ yr}^{-1}\\\\\text{From Equation (1)}\\\\\ln \dfrac{N_{0} }{0.77N_{0}} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\\ln\dfrac{1}{0.77} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\-\ln0.77 = 0.261 = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\t = \dfrac{0.261}{ 1.21 \times 10^{-4} \text{ yr}^{-1}} = \textbf{2160 yr}\\\\\text{The cloth was painted } \boxed{\textbf{2160 yr}}\text{ ago}[/tex]

b) The wood

Data:

[tex]N_{t} = 0.062 N_{0}[/tex]

Calculations:

[tex]\text{From Equation (1)}\\\\\ln \dfrac{N_{0} }{0.062N_{0}} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\\ln\dfrac{1}{0.0.062} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\-\ln0.062 = 2.78 = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\t = \dfrac{2.78}{ 1.21 \times 10^{-4} \text{ yr}^{-1}} = \textbf{23 000 yr}\\\\\text{The wood was cut} \boxed{\textbf{23 000 yr}}\text{ ago}[/tex]

Answer:

0.5

Explanation:

for more info see the attachment.

The specific heat capacity of the unknown metal is 0.5Jg-¹°C-¹

SPECIFIC HEAT CAPACITY:

Q = mc∆T

Where;

METAL:

LIQUID:

Therefore, the specific heat capacity of the unknown metal is 0.5Jg-¹°C-¹.

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Answer:

[tex]\boxed{\text{NaOH}}[/tex]

Explanation:

[tex]\rm \underbrace{\hbox{NaOH }}_{\hbox{base}} + \underbrace{\hbox{HCl}}_{\hbox{acid}} \longrightarrow \, NaCl + H$_{2}$O[/tex]

A hydroxide of a metal is a base.

[tex]\text{The base is }\boxed{\textbf{NaOH}}[/tex]

B is wrong. HCl is an acid.

C is wrong. NaCl is a salt.

D is wrong. Water is neutral.

Answer:

A. NaOH

A. Is The Correct Answer

B. HCl

C. NaCl

D. H2O

Answer:

The correct answer is C: hydrogen bonds

Explanation:

Water's surface tension and heat storage capacity are accounted for by its hydrogen bonds. Molecules of water are very strongly attracted to each other through hydrogen bonding.

Hydrogen bonds are constantly formed and broken in water molecules. Surface tension results from this hydrogen bonding which means that Water has a higher storage capacity for heat.

That is why at night, the Earth gets colder must faster than water.

Water takes time to slowly release heat to keep the atmospheric temperature moderate at night.

Final answer:

Hydrogen bonds are responsible for water's high surface tension and heat storage capacity. They result in strong cohesion between water molecules and a significant amount of energy is required to change water's state due to these bonds.

Explanation:

Water's surface tension and heat storage capacity are accounted for by its hydrogen bonds. The correct answer to the student's question is C) hydrogen bonds. These bonds occur because the hydrogen atoms in a water molecule have a slight positive charge, while the oxygen atom has a slight negative charge. Each water molecule can form four hydrogen bonds with surrounding molecules, leading to a high degree of cohesion. This cohesion is the main reason behind water's significant surface tension.

Additionally, because of the strength of these bonds, water has a high heat capacity and high heat of vaporization. It requires a large amount of heat energy to break the hydrogen bonds for water to change state, which explains why water is able to absorb and store a lot of heat without undergoing a significant increase in temperature.

Answer: The correct answer is Option B.

Explanation:

Precipitate is defined as the insoluble solid substance which is formed when two different aqueous solutions are mixed. It settles down at the bottom of the solution.

For the given chemical equation:

[tex]FeCl_2(aq.)+(NH_4)_2SO_4(aq.)\rightarrow FeSO_4(aq.)+2NH_4Cl(aq.)[/tex]

The products formed in the reaction are ferrous sulfate and ammonium chloride. Both the products are soluble in aqueous solutions. Thus, no precipitate will be formed in the reaction.

Hence, the correct answer is Option B.

Answer:

C₆H₄O₂

Explanation:

Given parameters:

Mass of the unknown compound = 0.315g

Atoms contained in the compound C, H and O

Mass of CO₂ produced = 0.771g

Mass of H₂O = 0.105g

Molar mass of compound = 108gmol⁻¹

The complete combustion of most hydrocarbon compounds like this would produce carbon dioxide and water only.

Solution

We first find the mass of the Carbon, hydrogen and Oxygen in the compounds given.

Mass of carbon in compound = [tex]\frac{Molar mass of Carbon}{Molar mass of CO_{2} }[/tex] x mass of CO₂

Molar mass of C = 12

Molar mass of CO₂ = 12 + (16x2) = 44

Mass of CO₂ produced = 0.771g

Mass of carbon in compound =  [tex]\frac{12}{44 }[/tex] x 0.771=0.2103g

Mass of H in compound =  [tex]\frac{Molar mass of H}{Molar mass of H_{2}O }[/tex] x mass of H₂O

Molar mass of H in H₂O = 2

Molar mass of H₂O = 2 + 16 = 18

Mass of H₂O = 0.105g

Mass of H in compound =  [tex]\frac{2}{18}[/tex] x 0.105 = 0.012g

Now, to find the mass of oxygen in the compound, we sum the mass H and C and subtract from the mass of the compound given:

Mass of oxygen = 0.315 - (0.2103 + 0.012)= 0.0927g

We then continue to derive the empirical formula of the compound.

                                    C                       H                         O

mass of

atoms                      0.2103                0.012                   0.0927

Moles                  0.2103/12             0.012/1                 0.0927/16

                               0.018                   0.012                    0.006

Dividing by

the smallest       0.018/0.006       0.012/0.006           0.006/0.006

                                   3                           2                                1

The empirical formula of the compound is C₃H₂O

From the empirical formula, we can find the compound from the given molar mass:

Molecular formula = (Empirical formula)n

Where n is the number of repeating times of the empirical formula present in one mole of the molecule.

Therefore n = [tex]\frac{molar mass of the compound}{molar mass of the empirical formula of the compound}[/tex]

 Molar mass of the empirical formula C₃H₂O = (12x3) + (2x1) + (16) = 54g/mol

               n = [tex]\frac{108}{54}[/tex] = 2

The molecular formula of the compound is = 2(C₃H₂O) = C₆H₄O₂

Final answer:

To identify the unknown organic compound from the combustion analysis data, one must calculate the moles of carbon and hydrogen from the CO2 and H2O produced, determine the empirical and eventually the molecular formula, and verify which compound options match the calculated formula and given molar mass.

Explanation:

When an unknown organic compound undergoes complete combustion, the products are CO2 and H2O. The amounts of these products can be used to determine the empirical formula of the compound.

First, from the 0.771 g of CO2 produced, we can calculate the moles of carbon, since each mole of CO2 contains one mole of carbon atoms. Similarly, from the 0.105 g of H2O produced, we can determine the moles of hydrogen, as each mole of H2O contains two moles of hydrogen atoms.

To find the identity of the unknown compound, we need to calculate the moles of each element and then the empirical formula. After finding the empirical formula, we can use the given approximate molar mass to find the molecular formula, which, in turn, will help us identify the compound. The compounds given as the options for the identity of the unknown must be checked against the calculated molecular formula.

Answer:

[tex]\boxed{1.19 \times 10^{-4} \text{ s}^{-1}}[/tex]

Explanation:

1. Determine the order of reaction

The question gives us a hint: the units of k are s⁻¹. This suggests a first order rate law.

To confirm, we plot ln(p) vs t.  We should get a straight line with slope = -k.

Here are your data with the pressures converted to natural logarithms.

[tex]\begin{array}{rcc}\textbf{t/s} & \textbf{p/mmHg} &\textbf{ln(p)}\\0 & 400 & 5.99\\2000 & 316 & 5.76\\4000 & 248 & 5.51\\6000 & 196 & 5.28\\8000 & 155 & 5.04\\10000 & 122 & 4.80\\\end{array}[/tex]

We get the graph shown below.

2. Determine the rate constant

The points fit so well that we can just use the end points to determine the slope.

[tex]\text{slope} = \dfrac{y_{2} - y_{1}}{ x_{2} - x_{1} } = \dfrac{4.80 - 5.99}{10 000 - 0} = -1.19 \times 10^{-4} \text{ s}^{-1}}\\\\k = \boxed{1.19 \times 10^{-4} \text{ s}^{-1}}[/tex]

The value of the rate constant for the reaction based on the given pressures  =  [tex]1.19 * 10^{-4} s^{-1}[/tex]

First step : determine the order of reaction

The reaction is a first-order reaction because the S.I. unit of K = s⁻¹  in the question

next step : Plot the value of  In( p ) vs Time

Where:  P = pressure ( mmHg )

            In ( p ) =  natural Logarithm of P values

Graph is attached below

Final step : calculate the value of the slope of the graph

slope = Δ y / Δ x

         = ( 4.8 - 5.99 ) / ( 10000 - 0 )

         = -1.19 / 10000  

∴ The value of  K ( rate constant )  = 1.19  *  10^{-4} s^{-1}

Hence we can conclude that the The value of the rate constant for the reaction based on the given pressures  =  [tex]1.19 * 10^{-4} s^{-1}[/tex]

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Answer:

It will decrease by 2 units.

Explanation:

The Henderson-Hasselbalch equation for a buffer is

pH = pKa + log(base/acid)

Let's assume your acid has pKa = 5.

(a) If the base: acid ratio is 1:1,

pH(1) = 5 + log(1/1) = 5  + log(1) = 5 + 0 = 5

(b) If the base: acid ratio is 1:100,

pH(2) = 5 + log(1/100) = 5  + log(0.01) = 5 - 2 = 3

(c) Difference

ΔpH = pH(2) - pH(1) = 5 - 3 = -2

If you increase the acid:base ratio to 100:1, the pH will decrease by two units.

Answer:

D : An electron may act with either particle-like or wave-like characteristics.  

Explanation:

This is the whole basis of the Schrödinger equation.

The other options are correct, but they do not state the dual nature of the electron.

Answer:

For part A: The number of moles of [tex]NO_2[/tex] is 2.0 moles.

For part B: The number of moles of [tex]NO_2[/tex] is [tex]1.1\times 10^1[/tex] moles.

Explanation:

For the given chemical equation:

[tex]2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)[/tex]

By Stoichiometry of the reaction:

2 moles of [tex]N_2O_5[/tex] produces 4 moles of [tex]NO_2[/tex]

So, 1.0 moles of [tex]N_2O_5[/tex] will produce = [tex]\frac{4}{2}\times 1.0=2.0moles[/tex] of [tex]NO_2[/tex]

Hence, the number of moles of [tex]NO_2[/tex] expressed in two significant figures are 2.0 moles.

By Stoichiometry of the reaction:

2 moles of [tex]N_2O_5[/tex] produces 4 moles of [tex]NO_2[/tex]

So, 5.4 moles of [tex]N_2O_5[/tex] will produce = [tex]\frac{4}{2}\times 5.4=10.8moles[/tex] of [tex]NO_2[/tex]

To express it in two significant figures, we round this value of 10.8 mol to 11 mole and then express it in scientific notation.

Scientific notation is defined as the way of representing a number which have very large value or very small value and is written in the decimal form.

Hence, the number of moles of [tex]NO_2[/tex] expressed in two significant figures are [tex]1.1\times 10^1[/tex] moles.

In the given reaction, 2 moles of NO2 are produced for every mole of N2O5. So, in part A, 1 mole of N2O5 will produce 2 moles of NO2, and in part B, 5.4 moles of N2O5 will produce around 11 moles of NO2.

The question is asking about the stoichiometry of a chemical reaction - specifically, how many moles of nitrogen dioxide (NO2) are formed from a given number of moles of dinitrogen pentoxide (N2O5). The balanced chemical equation is 2N2O5(g)→4NO2(g)+O2(g).

Part A: With 1.0 mol of N2O5, according to the stoichiometric ratio in the balanced equation (2:4), you would get 2 mol of NO2 for every 1 mol of N2O5, so you would have 2.0 moles of NO2.

Part B: With 5.4 mol of N2O5, again using the stoichiometric ratio, you would get 2 * 5.4 = 10.8 ≈ 11 moles of NO2 (rounded to two significant figures).

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Answer: The correct answer is Option D.

Explanation:

Atomic mass of an atom is defined as the sum of number of neutrons and number of protons that are present in an atom. It is represented as 'A'.

Atomic number = Number of protons + Number of neutrons

We are given:

Number of protons = 6

Number of neutrons = 6

Number of electrons = 6

Atomic mass = 6 + 6 = 12

Hence, the correct answer is Option D.

To neutralize the 0.50 M H2SO4, calculate the moles of NaHCO3 required using the mole ratio from the balanced equation and then convert the moles to grams.

To neutralize the 0.50 M H2SO4, we need to understand the stoichiometry of the reaction.

From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaHCO3.

Using the molarity and volume of the H2SO4 solution, we can calculate the moles of H2SO4, and then use the mole ratio to find the moles of NaHCO3 required to neutralize it. Finally, we can convert the moles of NaHCO3 to grams by using the molar mass of NaHCO3.

The correct answer is C) 19 g.

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Answer:

0.04 m³.

Explanation:

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

(P₁V₁) = (P₂V₂).

V₁ = 0.025 m³, P₁ = 1.5 x 10⁶ Pa,

V₂ = ??? m³, ​P₂ = 0.95 x 10⁶ Pa.

∴ V₂ = (P₁V₁)/(P₂) = (0.025 m³)(1.5 x 10⁶ Pa)/(0.95 x 10⁶ Pa) = 0.04 m³.

Answer: The final volume of the container is [tex]0.040m^3[/tex]

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.

[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.

We are given:

[tex]P_1=1.50\times 10^6Pa\\V_1=0.0250m^3\\P_2=0.950\times 10^6\\V_2=?m^3[/tex]

Putting values in above equation, we get:

[tex]1.50\times 10^6Pa\times 0.0250m^3=0.950\times 10^6Pa\times V_2\\\\V_2=\frac{1.50\times 10^6\times 0.0250}{0.950\times 10^6}=0.040m^3[/tex]

Hence, the final volume of container is [tex]0.040m^3[/tex]

Answer:

See Explanation

Explanation:

a. pH of 1M HOAc(aq)

                HOAc      ⇄      H⁺   +      OAcˉ

C(eq)         1.0M                 x               x

Ka = [H⁺][OAc⁻]/[HOAc] = x²/1.0M = 1.85x10⁻⁵

=> x = [H⁺] = SqrRt([HOAc]Ka) = SqrRt[(1M)(1.85x10ˉ⁵)] = 4.30x10ˉ³M

=> pH = -log[H⁺] = -log(4.30x10ˉ³) = 2.37

b. pH of 0.10M CH₃NH₃OH(aq)

                CH₃NH₃OH => CH₃NH₃⁺ + OHˉ; Kb = 4.4x10ˉ⁴

C(eq)             0.10M                x              x

=> Kb = [CH₃NH₃⁺][OH⁻]/[CH₃NH₃] = x²/0.10M

=> x = [OHˉ] = SqrRt([CH₃NH₃OH]Kb) = SqrRt[(0.10M)(4.4x10ˉ⁴)] = 6.63x10ˉ³M

=> pOH = -log[OHˉ] = -log(6.63x10⁻³) = 2.18

=> pH = 14 – pOH = 14 – 2.18 = 11.82

c. pH of 0.30M HOAc/0.10M OAcˉ(aq)

                HOAc      ⇄      H⁺   +      OAcˉ

C(eq)        0.30M               x           0.10M

=> Ka = [H⁺][OAcˉ]/[HOAc] => [H⁺] = Ka[HOAc]/[OAcˉ]

          = 1.85X10ˉ⁵(0.30M)/(0.10M) = 5.55X10ˉ⁵M

=> pH = -log[H⁺] = -log(5.55x10ˉ⁵) = 4.26

To calculate pH for acetic acid, protonated methylamine, and a formic acid/formate mixture, we utilize an ICE table for the weak acids and the Henderson-Hasselbalch equation for the buffer solution, considering their respective pKa values.

To calculate the pH values of the given solutions, we apply the pH formula and utilize ICE (Initial, Change, Equilibrium) tables for weak acids and bases. Using the provided pKa values, we can determine the pH for each solution:

These calculations showcase the principles of acid-base equilibrium and buffer systems in solution chemistry.

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Answer: The chemical reaction is given below.

Explanation:

When solid sodium metal reacts with water molecule to produce aqueous sodium hydroxide and hydrogen gas. The equation for this follows:

[tex]2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)[/tex]

By Stoichiometry of the reaction:

2 moles of solid sodium metal reacts with 2 moles of water molecule to produce 2 moles of sodium hydroxide and 1 mole of hydrogen gas.

Sodium metal is present in solid state, Water molecule is present in liquid state, Sodium hydroxide is present in aqueous state and hydrogen is present in gaseous state.

A chemical reaction between sodium (Na) and water (H₂O) yields hydrogen gas (H₂) and aqueous sodium hydroxide (NaOH). The reaction is highly exothermic as sodium hydroxide reacts vigorously with water, generating a lot of heat. The balanced chemical equation is '2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g)'.

To answer your question, we need to first understand the reaction occurring in this process. This is a classic example of a reaction between a metal and water, specifically it's an alkali metal (sodium) reacting with water. Upon addition of solid sodium to liquid water, the sodium displaces hydrogen from the water, leading to the production of hydrogen gas and aqueous sodium hydroxide.

The balanced chemical equation is thus written as: 2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g). In this equation, (s) represents solid, (l) represents liquid, (aq) represents aqueous or a material dissolved in water, and (g) represents gas.

Sodium hydroxide NaOH is a strong base and it reacts vigorously with water, generating a great deal of heat and forming very basic solutions. In fact, 40 grams of sodium hydroxide can dissolve in only 60 grams of water at 25 °C.

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Answer:

moles of Cu° needed =  ½ (5.8 moles) = 2.9 moles Cu° needed.

Explanation:

Cu + 2AgNO₃  =>  Cu(NO₃)₂ + 2Ag

? moles Cu + 5.8 moles AgNO₃ =>    Cu(NO₃)₂ + 2Ag

Note coefficient of Cu° vs coefficient of AgNO₃

=> 1 mole Cu° < 2 moles AgNO₃ ...

=> Since moles of Cu° are smaller than moles of AgNO₃ in the given equation, the moles of Cu° needed to react with 5.8 moles AgNO₃ will be smaller than the 5.8 by the ratio of coefficients that will make 5.8 smaller. That is ...

moles of Cu° needed =  ½ (5.8 moles) = 2.9 moles Cu° needed.

Note: Using 2/1(5.8) will make value greater than 5.8 and incorrect.

One can also set up a ratio relationship as follows...

1 mole Cu° <=> 2 moles AgNO₃ (fm equation)

? mole Cu° <=> 5.8 moles AgNO₃ (problem)

=> (1 mole Cu°)/X=(2 mole AgNO₃)/(5.8 moles AgNO₃)

=> X = (1 mole Cu°)(5.8 mole AgNO₃)/2 mole AgNO₃

        = ½(5.8) mole Cu° = 2.9 mole Cu°

Final answer:

2.9 moles of Cu are needed to react with 5.8 moles of AgNO3 according to the stoichiometry of the balanced chemical reaction.

Explanation:

To determine how many moles of Cu (copper) are required to react with 5.8 moles of AgNO3 (silver nitrate), we use the stoichiometry of the balanced chemical reaction:

Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag

This balanced equation tells us that one mole of Cu reacts with two moles of AgNO3. If we have 5.8 moles of AgNO3, then the amount of Cu required is calculated by dividing the moles of AgNO3 by 2:

(5.8 moles AgNO3) / (2 moles AgNO3/moles Cu) = 2.9 moles of Cu

Therefore, 2.9 moles of Cu are needed to react with 5.8 moles of AgNO3.