A reaction was performed in which 3.4 g of benzoic acid was reacted with excess methanol to make 1.2 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.

Answer : The volume of [tex]CO_2[/tex] produced are, 26.7 liters.

Explanation :

First we have to calculate the moles of [tex]C_{10}H_{22}[/tex]

[tex]\text{Moles of }C_{10}H_{22}=\frac{\text{Given mass }C_{10}H_{22}}{\text{Molar mass }C_{10}H_{22}}=\frac{16.9g}{142g/mol}=0.119mol[/tex]

Now we have to calculate the moles of [tex]CO_2[/tex].

The given combustion reaction is:

[tex]2C_{10}H_{22}+31O_2\rightarrow 22H_2O+20CO_2[/tex]

From the balanced chemical reaction we conclude that,

As, 2 moles of [tex]C_{10}H_{22}[/tex] react to give 20 moles of [tex]CO_2[/tex]

So, 0.119 moles of [tex]C_{10}H_{22}[/tex] react to give [tex]\frac{20}{2}\times 0.119=1.19[/tex] moles of [tex]CO_2[/tex]

Now we have to calculate the volume of [tex]CO_2[/tex] produced.

As we know that, 1 mole of gas occupies 22.4 L volume of gas.

As, 1 mole of [tex]CO_2[/tex] gas occupies 22.4 L volume of [tex]CO_2[/tex] gas.

So, 1.19 mole of [tex]CO_2[/tex] gas occupies 1.19 × 22.4 L = 26.7 L volume of [tex]CO_2[/tex] gas.

Therefore, the volume of [tex]CO_2[/tex] produced are, 26.7 liters.

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

[tex]deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol[/tex]

b) To calculate the constant we have the following expression:

[tex]lnK=-\frac{deltaG_{rxn} }{RT}[/tex]

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

[tex]lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066[/tex]

c) The equilibrium pressure of O₂ over M is:

[tex]K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm[/tex]

Answer:

.1m HCl  0.001HCl  .00001m HCl  Distilled water  .00001NaOH   .001NaOH   .01NaOH

Explanation:

Thanks to the person who posted the answers, I just wrote it out so its easier to see. :D

In the given compounds, the most acidic compound is 0.1m HCl and most basic compound is 0.01NaOH.

Acids are those compounds whose has a pH value in the range from 0 to 7 and bases are those compounds which has a pH range from 7 to 14.

pH of any solution will be calculated as:

pH = -log[H⁺], where

[H⁺] = concentration of H⁺ ion and this concentration is present in the form of molarity (molar concentration).

So, pH value is directly proportional to the concentration value of H⁺ ion as concentration of H⁺ ion decreases so acidity also decreases and sequence of given compounds from most acidic to most basic is represented as:

0.1m HCl > 0.001HCl > 0.00001m HCl > Distilled water > 0.00001NaOH > 0.001NaOH > 0.01NaOH.

Hence 0.1m HCl is most acidic and 0.01 NaOH is most basic.

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Answer: The mass of [tex]AlCl_3[/tex] theoretically be made can be, 37.4 grams.

Explanation : Given,

Mass of [tex]Al[/tex] = 20.0 g

Mass of [tex]Cl_2[/tex] = 30.0 g

Molar mass of [tex]Al[/tex] = 27 g/mol

Molar mass of [tex]Cl_2[/tex] = 71 g/mol

First we have to calculate the moles of [tex]Al[/tex] and [tex]Cl_2[/tex].

[tex]\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{20.0g}{27g/mol}=0.741mol[/tex]

and,

[tex]\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}=\frac{30.0g}{71g/mol}=0.422mol[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]

From the balanced reaction we conclude that

As, 3 moles of [tex]Cl_2[/tex] react with 2 moles of [tex]Al[/tex]

So, 0.422 moles of [tex]Cl_2[/tex] react with [tex]\frac{2}{3}\times 0.422=0.281[/tex] moles of [tex]Al[/tex]

From this we conclude that, [tex]Al[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Cl_2[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]AlCl_3[/tex]

From the reaction, we conclude that

As, 3 moles of [tex]Cl_2[/tex] react to give 2 moles of [tex]AlCl_3[/tex]

So, 0.422 moles of [tex]Cl_2[/tex] react to give [tex]\frac{2}{3}\times 0.422=0.281[/tex] mole of [tex]AlCl_3[/tex]

Now we have to calculate the mass of [tex]AlCl_3[/tex]

[tex]\text{ Mass of }AlCl_3=\text{ Moles of }AlCl_3\times \text{ Molar mass of }AlCl_3[/tex]

Molar mass of [tex]AlCl_3[/tex] = 133 g/mole

[tex]\text{ Mass of }AlCl_3=(0.281moles)\times (133g/mole)=37.4g[/tex]

Therefore, the mass of [tex]AlCl_3[/tex] theoretically be made can be, 37.4 grams.

Answer:

Volume stays.

Explanation:

If aerosol can throwing into the fire than the temperature of the aerosol can & the temperature of the contents of the gas in an aerosol will also increase, and it induces the pressure to rise.

Against the sides of the can, the gas molecules will smash rapidly with each other which are present inside an aerosol can, and the proportion of gas will stay the same as earlier. According to the law of conservation of matter, as a result of the heat, the gas particles can not be eradicated or expanded.

Answer: The amount of heat energy in joules required to raise the temperature is 526 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed = ?

m= mass of substance = 7.40 g

c = specific heat capacity = [tex]4.184J/g^0C[/tex]

Initial temperature of the water = [tex]T_i[/tex] = 29.0°C

Final temperature of the water = [tex]T_f[/tex]  = 46.0°C

Change in temperature ,[tex]\Delta T=T_f-T_i=(46.0-29.0)^0C=17.0^0C[/tex]

Putting in the values, we get:

[tex]Q=7.40g\times 4.184J/g^0C\times 17.0^0C[/tex]

[tex]Q=526J[/tex]

The amount of heat energy in joules required to raise the temperature is 526 Joules

The amount of heat energy will be "526 J".

Given:

Mass,

Initial temperature,

Final temperature,

Change in temperature,

              [tex]= 46.0-29.0[/tex]

              [tex]= 17.0^{\circ} C[/tex]

The amount of heat energy,

→ [tex]Q = mc \Delta T[/tex]

By substituting the values, we get

      [tex]= 7.40\times 4.184\times 17.0[/tex]

      [tex]= 526 \ J[/tex]

Thus the above answer is right.

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Answer: This mechanism has no catalyst

Explanation:

The proposed mechanism is :

Step 1 : [tex]H_2(g)+ICl(g)\rightarrow HI(g)+HCl(g)[/tex]    (slow)

Step 2: [tex]HI(g)+ICl(g)\rightarrow I_2(g)+HCl(g)[/tex]    (fast)

The combined chemical equation will be :

[tex]H_2(g)+2ICl(g)\rightarrow I_2(g)+2HCl(g)[/tex]

A catalyst is a substance which enhances the rate of chemical reaction without being consumed in the chemical reaction. Thus catalyst gets used up in first step and gets regenerated in second.

HI is formed as an intermediate as it gets formed in first step and gets used up in the second step.

Here there is no substance which is used as a catalyst.

Answer:

Higher oxidation state metals form stronger bong with ligands

Explanation:

Ligand strength are based on oxidation number, group and its properties

Statement e is not an accurate description because the ordering of the spectrochemical series depends on multiple factors, including ligand size, pi-donor or pi-acceptor abilities, and the metal's identity and oxidation state, in addition to electronegativity of the donor atom.

The correct answer to the question "Which of the following statements is not an accurate description of a factor that contributes to the ordering of the spectrochemical series for ligands and metals?" is e. The electronegativity of the donor atom is the most important factor for ligands.

The spectrochemical series is an ordering of ligands based on the magnitude of the ligand field splitting parameter (Δoct) they produce in coordination complexes. Contrary to option e, it is not just the electronegativity of the donor atom but also a series of other factors such as the size of the donor atom, the ligand's field strength (pi-donor or pi-acceptor abilities), and the identity of the metal ion and its oxidation state that contribute to the ordering of the spectrochemical series. For example, small neutral ligands with highly localized lone pairs, such as NH₃, produce larger Δoct values which illustrates more than just electronegativity playing a role.

Answer:

0.72g of gallium

Explanation:

Equation of the reduction reaction:

Ga^3+(aq) +3e --------> Ga(s)

If it takes 3F coulumbs of electricity to deposit 70g of gallium (relative atomic mass of gallium)

Then (0.710 × 70×60) coulombs of electricity will deposit (0.710 × 70×60) × 70/3F

But F = 96500C or 1Faraday

Therefore:

(0.710 × 70×60) × 70/3×96500

Mass of gallium deposited= 0.72g of gallium

Answer:

Lemon

HCI

Blood

Saliva

Bleach

NaOH

Explanation:

Blood 7.35-7.45

Bleach 12.6

Saliva 6.2-7.6

Lemon 2-3

HCI 3.01

NaOH 13

Answer:

The amount of heat is 84.4894kJ

Explanation:

Given data:

mass of octane=148g

The moles of octane is:

[tex]n_{octane} =148g*\frac{1mol}{114.23g} =1.2956moles[/tex]

The melting point is=-56.82°C

ΔHfus=20.73kJ/mol

The heat of fusion is:

[tex]Q_{fus} =n_{octane} *delta-H_{fus} =1.2956*20.73=26.8578kJ[/tex]

The heat to bring the octane to a temperature of 117.8°C is:

[tex]Q_{2} =mCp(117.8-(56.82))=148*2.23*(117.8+56.82)=57631.5848J=57.6316kJ[/tex]

The total heat in the process is:

Qtotal=Qfus+Q₂=26.8578+57.6316=84.4894kJ

Answer:

[tex]\large \boxed{\text{0.0503 g}}[/tex]

Explanation:

The limiting reactant is the reactant that gives the smaller amount of product.

Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:   39.10    80.41                2.016  

            2K  +  2HBr ⟶ 2KBr + H₂

m/g:     5.5      4.04

a) Limiting reactant

(i) Calculate the moles of each reactant  

[tex]\text{Moles of K} = \text{5.5 g} \times \dfrac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} \times \dfrac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}[/tex]

(ii) Calculate the moles of H₂ we can obtain from each reactant.

From K:  

The molar ratio of H₂:K is 1:2.

[tex]\text{Moles of H}_{2} = \text{0.141 mol K} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol K}} = \text{0.0703 mol H}_{2}[/tex]

From HBr:  

The molar ratio of H₂:HBr is 3:2.  

[tex]\text{Moles of H}_{2} = \text{0.049.93 mol HBr } \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol HBr}} = \text{0.024 97 mol H}_{2}[/tex]

(iii) Identify the limiting reactant

HBr is the limiting reactant because it gives the smaller amount of NH₃.

b) Excess reactant

The excess reactant is K.

c) Mass of H₂

[tex]\text{Mass of H}_{2} = \text{0.024 97 mol H}_{2} \times \dfrac{\text{2.016 g H}_{2}}{\text{1 mol H}_{2}} = \textbf{0.0503 g H}_{2}\\ \text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$ }[/tex]

Answer:

(4) increasing the temperature  

Explanation:

PCl₅(g) + energy ⇌ PCl₃(g) + Cl₂(g)

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

If you increase the temperature, the position of equilibrium will move to the right to get rid of the added heat.

(1) is wrong. Adding a catalyst does not change the position of equilibrium.

(2) is wrong. If you add more PCl₃, the position of equilibrium will move to the left to get rid of the added PCl₃.

(3) is wrong. The left-hand side has fewer moles of gas. If you increase the pressure, the position of equilibrium will move to the left to relieve the pressure.

Answer: Increasing the temperature

Explanation: If you increase the temperature, the position of equilibrium will shift to the right so it can eliminate the heat.

Answer:The Zinc Reacts With The Hydrochloric Acid Producing Zinc Chloride And Hydrogen Gas, And Leaving The Copper Behind. A. If 25.0 G Of Zinc ... Zn+ 2 HCI --> ZnCl2 + H2 (answer .771 G H2) B. If The Reaction Yields . ... If 25.0 g of zinc are in a sample of bronze, determine the theoretical yield of hydrogen gas. Zn+ 2 HCI

The equilibrium constant, Kc, for the given reaction can be calculated using the solubility product constant, Ksp, and the overall formation constant, Kf. Substituting the given values, we find that Kc is equal to 3.186 × 10⁻⁵.

The equilibrium constant, Kc, for the reaction AgCl (s) + Cl⁻ (aq) ⇌ AgCl⁻₂ (aq) can be calculated using the solubility product constant, Ksp, for AgCl and the overall formation constant, Kf, for AgCl⁻₂.

The equilibrium constant is given by the product of Ksp and Kf:

Kc = Ksp * Kf

Substituting the values given, Ksp = 1.77 × 10⁻¹⁰ and Kf = 1.8 × 10⁵, we can calculate the value of Kc:

Kc = (1.77 × 10⁻¹⁰) * (1.8 × 10⁵) = 3.186 × 10⁻⁵

Answer:

No

Explanation:

We first find the solubility of the copper sulphate salt

Number of moles= 4.6 moles

Volume =1.75 L

Molarity= number of moles/ volume = 4.6/1.75= 2.6 molL-1

Hence, only 2.6 moles of Copper II sulphate dissolves in 1.75L of water. Hence 4.6 moles of copper II Sulphate dies not dissolve in 1.75 L of water.

Answer:

Explanation:

find the solution below

The organic compounds carboxylic acid, amine, and alcohol contain different numbers of sigma and pi bonds resulting from the overlapping of atomic orbitals. In these provided examples, the carboxylic acid contains nine sigma and two pi bonds, the amine has twelve sigma and two pi bonds, and the alcohol has nine sigma and 0 pi bonds.

Sigma

and

pi bonds

occur in various types of organic molecules, including carboxylic acids, amines, and alcohols. A sigma (σ) bond is formed by the head-on overlapping of atomic orbitals, whereas a pi (π) bond is formed by the lateral overlap of two p orbitals. In the carboxylic acid H2CCHCH2COOH, there are nine sigma and two pi bonds. As for the amine HCCCH2CHCHCH2NH2, there are twelve sigma and two pi bonds. Lastly, an alcohol molecule, such as CH3OH, consists of 9 sigma bonds and 0 pi bonds. It's essential to note that a double bond contains one sigma and one pi bond, while a triple bond contains one sigma and two pi bonds.

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Answer:

Molarity= 1.7M

Explanation:

n= m/M = 25/58.5= 0.427mol

Applying

n=CV

0.427= C×0.25

C= 1.7M

Answer:

Rain Water: Acidic

Cola: Acidic

Tomato Juice: Acidic

Liquid Drain Cleaner: Basic

Rain Water: Acidic

Cola: Acidic

Tomato Juice: Acidic

Liquid Drain Cleaner: Basic

If the value of pH is 7, then the solution is neutral, if greater than 7 then basic, and if less than 7 then acidic.

The relation between pH and pOH is as follows:-

[tex]pH+pOH=14[/tex]

The value of pH for rain water is 14-8.5=5.5.

So, the rain water is acidic.

The value of pH for cola is 14-11=3.

So, the cola is acidic.

The value of pH for tomato juice is 14-10=4.

So, the tomato juice is acidic.

The value of pH for liquid drain cleaner is 14-0=14

So, the liquid drain cleaner is basic.

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Answer:

13 grams Na₂SO₄ (2 sig.figs.)

Explanation:

1st convert mass values to moles and solve yield using limiting reactant principles and reaction ratio of balance equation.

Determine limiting reactant and

Complete problem by converting yield into grams.

H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O

moles H₂SO₄ = 8.83g/98g·mol⁻¹ = 0.0901 mole H₂SO₄

moles NaOH = 9.1g/40g·mol⁻¹ = 0.2275 mole NaOH

Determine Limiting Reactant by dividing each mole value by the respective coefficient in the balanced equation. The smaller value is always the limiting reactant.

H₂SO₄ => (0.0901/1) = 0.0901 <= Limiting Reactant (smaller value)

NaOH =>  (0.2275/2) = 0.1138

NOTE: when working the problem use the calculated moles of reactant NOT the LR test number. In this problem, use 0.0901 mole H₂SO₄. (yeah, it is the same but this does not occur for the LR in many other problems). Anyways...

                    H₂SO₄        +        2NaOH       =>       Na₂SO₄      +      2H₂O

moles      0.0901 mole          0.2275 mol            0.0901 mol      2(0.0901 mol)

mass (g) Na₂SO₄ = 0.0901 mole x 142.04 g/mol = 12.798 grams ≅ 13 grams                                    Na₂SO₄ (2 sig.figs.)

Answer:

13 grams Na₂SO₄ (2 sig.figs.)

Explanation:

Firstly

We convert mass values to moles and solve yield using limiting reactant principles and reaction ratio of balance equation.

To determine limiting reactant and

Complete problem by converting yield into grams. We write the equations for the reaction

H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O

To give

moles H₂SO₄ = 8.83g/98g·mol⁻¹ = 0.0901 mole H₂SO₄

Then,

moles NaOH = 9.1g/40g·mol⁻¹ = 0.2275 mole NaOH

Determine Limiting Reactant by dividing each mole value by the respective coefficient in the balanced equation. The smaller value is always the limiting reactant.

H₂SO₄ => (0.0901/1) = 0.0901 <= Limiting Reactant (smaller value)

NaOH =>  (0.2275/2) = 0.1138

N/B: when working the problem use the calculated moles of reactant NOT the LR test number. In this problem, use 0.0901 mole H₂SO₄. (yeah, it is the same but this does not occur for the LR in many other problems). Anyways...

H₂SO₄        +        2NaOH       =>       Na₂SO₄      +      2H₂O

moles      0.0901 mole          0.2275 mol            0.0901 mol      2(0.0901 mol)mass (g)

Na₂SO₄ = 0.0901 mole x 142.04 g/mol = 1

2.798 grams ≅ 13 grams                                    Na₂SO₄ (2 sig.figs.)

Answer:

The temperature is 298.9K = 25.75 °C

Explanation:

Step 1: Data given

Volume of the flask = 1.00 L

Number of moles oxygen = 0.0400 moles

Pressure = 0.981 atm

Step 2: Calculate the temperature

p*V = n*R*T

⇒with p = the pressure of oxygen gas = 0.981 atm

⇒with V = the volume of the flask = 1.00 L

⇒with n = the number of moles of oxygen = 0.0400 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temeprature = TO BE DETERMINED

T = (p*V) / (n*R)

T = (0.981 atm * 1.00 L) / (0.0400 moles * 0.08206 L*atm/mol*K)

T = 298.9 K

The temperature is 298.9K = 25.75 °C

Answer:

5.0 %

Explanation:

Given data

Step 1: Calculate the volume of solution

If we assume that the volumes are additive, the volume of the solution is equal to the sum of the volume of the solute and the solvent.

V = 50 mL + 950 mL = 1000 mL

Step 2: Calculate the percent by volume

We will use the following expression.

[tex]\% v/v = \frac{volume\ of\ solute}{volume\ of\ solution} \times 100 \% = \frac{50mL}{1000mL} \times 100 \% = 5.0\%[/tex]

Answer:

hydrogen ions

Explanation:

Acids are substances that when dissolved in water release hydrogen ions, H+(aq). Bases are substances that react with and neutralise acids, producing water. When dissolved, bases release hydroxide ions, OH-(aq) into solution. Water is the product of an acid and base reacting.

When acids dissolve in water, they produce hydrogen ions (H+). This is known as the process of ionization.

The hydrogen ions are responsible for the acidic properties of the solution.When HCl is added to water, it dissociates into hydrogen ions (H+) and chloride ions (Cl-). The presence of hydrogen ions gives the solution acidic properties.

Similarly, other acids such as sulfuric acid, nitric acid, and acetic acid also produce hydrogen ions when dissolved in water.

It's important to note that not all substances that dissolve in water are acids. Acids have a pH value less than 7, and their properties can vary depending on their strength. Strong acids, like hydrochloric acid, completely dissociate in water, producing a large number of hydrogen ions. Weak acids, like acetic acid, only partially ionize, producing fewer hydrogen ions.

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Answer:

0.23 mol/L

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)

Step 2:

Determination of the number of in 12.79g of PbCl2. This is illustrated below:

Mass of PbCl2 = 12.79g

Molar Mass of PbCl2 = 207 + (2x35.5) = 207 + 71 = 278g/mol

Number of mole of PbCl2 =?

Number of mole = Mass/Molar Mass

Number of mole of PbCl2 = 12.79/278

Number of mole of PbCl2 = 0.046 mole

Step 3:

Determination of the number of mole of Pb(NO3)2 that reacted.

This is illustrated below:

From the balanced equation above,

1 mole of Pb(NO3)2 reacted to produce 1 mole of PbCl2.

Therefore, it will also take 0.046 mole of Pb(NO3)2 to react to produce 0.046 mole of PbCl2.

Step 4:

Determination of the molarity of Pb(NO3)2. This is illustrated:

Mole of Pb(NO3)2 = 0.046 mole

Volume of the solution = 200 mL = 200/1000 = 0.2 L

Molarity =?

Molarity is defined as the mole of solute per unit litre of solution. It is given by:

Molarity = mole of solute /Volume

Molarity of Pb(NO3)2 = 0.046/0.2

Molarity of Pb(NO3)2 = 0.23 mol/L

Answer:

i have an answer but i can only show you because my teacher helped my on it and wrote it down for me to remember! hope this helps!!!

Explanation:

A 1.00 g sample ofn-hexane (C6H14) under-goes complete combustion with excess O2ina bomb calorimeter. The temperature of the1502 g of water surrounding the bomb risesfrom 22.64◦C to 29.30◦C. The heat capacityof the hardware component of the calorimeter(everything that is not water) is 4042 J/◦C.What is ΔUfor the combustion ofn-C6H14?One mole ofn-C6H14is 86.1 g.The specificheat of water is 4.184 J/g·◦C.1.-9.96×103kJ/mol2.-7.40×104kJ/mol3.-1.15×104kJ/mol4.-4.52×103kJ/mol5.-5.92×103kJ/molcorrectExplanation:mC6H8= 1.00 gmwater= 1502 gSH = 4.184 J/g·◦CHC = 4042 J/◦CΔT= 29.30◦C-22.64◦C = 6.66◦CThe increase in the water temperature is29.30◦C-22.64◦C = 6.66◦C. The amount ofheat responsible for this increase in tempera-ture for 1502 g of water isq= (6.66◦C)parenleftbigg4.184Jg·◦Cparenrightbigg(1502 g)= 41854 J = 41.85 kJThe amount of heat responsible for the warm-ing of the calorimeter isq= (6.66◦C)(4042 J/◦C)= 26920 J = 26.92 kJ

Answer:

165.5 g of CO2

Explanation:

We must first put down the balanced reaction equation:

C4H10(g) + 13/2 O2(g) ------> 4CO2(g) + 5H2O(g)

From the reaction equation, one mole of butane occupies 22.4 L hence we can establish the stoichiometry of the reaction thus:

22.4 L of butane created 174 g of CO2

Therefore 21.3 L of butane will create 21.3 × 174/22.4 = 165.5 g of CO2

Answer:

A. 11.26

B. 12.37

Explanation:

A. Step 1:

Dissociation of Mg(OH)2. This is illustrated below below:

Mg(OH)2 <==> Mg2+ + 2OH-

A. Step 2:

Determination of the concentration of the OH-

From the above equation,

1 mole of Mg(OH)2 produce 2 moles of OH-

Therefore, 9.01x10^-4 M Mg(OH)2 will produce = 9.01x10^-4 x 2 = 1.802x10^-3 M of OH-

A. Step 3:

Determination of the pOH. This is illustrated below:

pOH = - Log [OH-]

[OH-] = 1.802x10^-3 M

pOH = - Log [OH-]

pOH = - Log 1.802x10^-3

pOH = 2.74

A. Step 4:

Determination of the pH.

pH + pOH = 14

pOH = 2.74

pH + 2.74 = 14

Collect like terms

pH = 14 - 2.74

pH = 11.26

B. Step 1:

Dissociation of NH4OH. This is illustrated below below:

NH4OH <==> NH4+ + OH-

B. Step 2:

Determination of the concentration of the OH-

From the above equation,

1 mole of NH4OH produce 1 moles of OH-

Therefore, 2.33x10^-2M of NH4OH will also produce 2.33x10^-2M of OH-

B. Step 3:

Determination of the pOH. This is illustrated below:

pOH = - Log [OH-]

[OH-] = 2.33x10^-2M

pOH = - Log [OH-]

pOH = - Log 2.33x10^-2M

pOH = 1.63

B. Step 4:

Determination of the pH.

pH + pOH = 14

pOH = 1.63

pH + 1.63 = 14

Collect like terms

pH = 14 - 1.63

pH = 12.37

The mass of AgI in the precipitate is approximately 38.65 grams (rounded to two significant figures) with the appropriate units.

To find the mass of AgI in the precipitate, we first need to determine the moles of AgI formed. We'll use the information provided and follow these steps:

1. Write the balanced chemical equation for the precipitation reaction:

[tex]\[ \text{Ag}^+ + \text{I}^- \rightarrow \text{AgI} \][/tex]

[tex]\[ \text{Hg}_2^{2+} + 2\text{I}^- \rightarrow \text{HgI}_2 \][/tex]

2. Determine the limiting reactant:

  - For AgI: [tex]\( \text{Ag}^+ + \text{I}^- \)[/tex] (1 mole of Ag per mole of I)

  - For HgI2: [tex]\( \frac{1}{2}\text{Hg}_2^{2+} + \text{I}^- \)[/tex] (1 mole of Hg per 2 moles of I)

  The limiting reactant is the one that produces the fewer moles of I^-, as it determines the amount of AgI formed.

  [tex]\[ \text{moles of I}^- = 0.100 \, \text{L} \times 1.71 \, \text{mol/L} = 0.171 \, \text{mol} \][/tex]

  The limiting reactant is AgI, as it requires 0.171 moles of I^-, while HgI2 would require 0.342 moles of I^-.

3. Calculate the moles of AgI formed:

  [tex]\[ \text{moles of AgI} = \text{moles of I}^- \times \frac{1 \, \text{mol AgI}}{1 \, \text{mol I}^-} = 0.171 \, \text{mol} \][/tex]

4. Calculate the mass of AgI formed:

  [tex]\[ \text{mass of AgI} = \text{moles of AgI} \times \text{molar mass of AgI} \][/tex]

  The molar mass of AgI is the sum of the atomic masses of Ag (107.87 g/mol) and I (126.904 g/mol).

  [tex]\[ \text{mass of AgI} = 0.171 \, \text{mol} \times (107.87 + 126.904) \, \text{g/mol} \][/tex]

  [tex]\[ \text{mass of AgI} \approx 38.65 \, \text{g} \][/tex]

Therefore, the mass of AgI in the precipitate is approximately 38.65 grams (rounded to two significant figures) with the appropriate units.

Answer:

Electrons

Explanation:

In an atom there would be three subatomic particles: Neutrons, electrons, protons. The smallest and lightest in terms of mass is electrons. This is because the nucleus is comprised of the protons and the neutrons, these have a greater mass than electrons as electrons has very little mass that can considered to be 0.