The Haber Process is the main industrial procedure to produce ammonia. The reaction combines nitrogen from air with hydrogen mainly from natural gas (methane) and is reversible and exothermic. The enthalpy change for this reaction is - 92 kJ mol-1. In an experiment, 1.5 moles of N2 and 4.0 moles of H2 is mixed in a 1.50 dm3 reaction vessel at 450 °C. After reaching equilibrium, the mixture contained 0.9 mole of NH3.
A) With the above information, write the reaction equilibrium equation in the Haber process. t.
B) Calculate Kc for this reaction.
C) What is the equilibrium yield of ammonia in this reaction?
D) Referring to Le Chatelier's principle and above information, suggest two ways to increase the yield of ammonia in this reaction and explain.
Answer:
A) [tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex].
B) [tex]Kc=0.0933[/tex].
C) 0.9 mol.
D) Increasing both temperature and pressure.
Explanation:
Hello,
In this case, given the information, we proceed as follows:
A)
[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]
B) For the calculation of Kc, we rate the equilibrium expression:
[tex]Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]
Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent [tex]x[/tex], we have:
[tex][NH_3]=0.6M=2*x[/tex]
[tex]x=\frac{0.6M}{2}=0.3M[/tex]
Next, the concentrations of nitrogen and hydrogen at equilibrium are:
[tex][N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M[/tex]
[tex][H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M[/tex]
Therefore, the equilibrium constant is:
[tex]Kc=\frac{(0.6M)^2}{(0.7M)*(1.77M)^3}\\ \\Kc=0.0933[/tex]
C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.
D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.
Best regards.
To determine the concentration of chloride ion (Cl-) in a 100 mL sample of ground water, a chemist adds a large enough volume of AgNO3 solution to precipitate all Cl- as AgCl. The mass of the resulting precipitate is 93.9 mg. What is the chloride ion concentration in milligrams of chloride per liter of ground water
Answer:
[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]
Explanation:
Hello,
In this case, we can represent the chemical reaction as:
[tex]Cl^-(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NO_3^-(aq)[/tex]
In such a way, since the mass of the obtained silver chloride is 93.9 mg, we can compute the chloride ions in the ground water by using the following stoichiometric procedure whereas the molar mass of chloride ions and silver chloride are 35.45 g/mol and 143.32 g/mol respectively:
[tex]m_{Cl^-}=93.3mgAgCl*\frac{1mmolAgCl}{143.32mgAgCl}*\frac{1mmolCl^-}{1mmolAgCl} *\frac{35.45mgCl^-}{1mmolCl^-} =23.23mgCl^-[/tex]
Finally, for the given volume of water in liters (0.100L), we compute the required concentration:
[tex][Cl^-]=\frac{23.2mgCl^-}{0.100L}\\[/tex]
[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]
Best regards.
The concentration of chloride ions in the groundwater sample is 230 mg/L.
Calculation of chloride ion concentration:Based on the given information,
The mass of the resulting precipitate, that is, AgCl is 93.9 mg or 0.0939 g. The molar mass of AgCl is 143.2 g/mol.Now the number of moles of AgCl precipitate can be calculated as,
n = Given mass/Molar mass
Now putting the values we get,
[tex]n = \frac{0.939 g}{143.32 g/mol} \\n = 6.5 * 10^{-4}[/tex]
Thus, 6.5 × 10⁻⁴ moles of AgCl comprises 6.5 × 10⁻⁴ chloride ions. Therefore, 6.5 × 10⁻⁴ of chloride ions are present in the sample of 100 ml.
Now the molar mass of chloride ion is 35.453 g/mol, the mass of chloride ion will be,
Mass = Mole × Molar mass
Mass = 6.5 × 10⁻⁴ moles × 35.453 g/mol
Mass = 0.0230 g or 23 mg
The volume of the groundwater sample is 100 ml or 0.1000 L.
Now the concentration of the chloride ions in the sample given is,
C = 23 mg/0.1000 L
C = 230 mg/L
Thus, the concentration of chloride ions in the groundwater sample is 230 grams per liter.
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Do you expect the compound Na4S to be a stable sulfur compound? Explain why or why not. Select the correct answer below: A. Yes, because sulfur is significantly more electronegative than sodium, so it can ionize sodium. B. Yes, because sodium is significantly more electronegative than sulfur, so it can ionize sulfur. C. No, because sulfur does not typically form negative ions or oxidation states less than 2−. The binary compound formed by sulfur and sodium is Na2S. D. No, because elemental sulfur is not a strong enough oxidation agent to oxidize sodium.
Answer:
No, because sulfur does not typically form negative ions or oxidation states less than 2−. The binary compound formed by sulfur and sodium is Na2S
Explanation:
Sulphur is a member of group 16. The oxidation states expected for sulphur in group 16 are -1, -2, +1, +2,+3,+4,+5 or +6. The elements of group 16 usually form negative ions with oxidation number of -2. They do not typically form negative ions with oxidation state less than -2.
The implication of this is that we actually do not expect the existence of a compound in which sulphur forms an S^4- anion. In reality, such an anion does not exist. Rather a binary compound of sulphur and sodium will have the formula Na2S because it contains the S^2- anion.
· A 0.100g sample of Mg when combined with O2 yields 0.166g of Mgo, a
second Mg sample with a mass of 0.144g is also combined with O2. What
mass of MgO is produced from the second sample?
Answer:
[tex]m_{MgO}=0.239gMgO[/tex]
Explanation:
Hello,
In this case, the chemical reaction between magnesium and oxygen to yield magnesium oxide is:
[tex]2Mg+O_2\rightarrow 2MgO[/tex]
In such a way, for 0.144 g of magnesium reacting with sufficient oxygen, the mass of magnesium oxide, whose molar mass is 40.3 g/mol (2:2 mole ratio) turns out:
[tex]m_{MgO}=0.144gMg*\frac{1molMg}{24.3gMg} *\frac{2molMgO}{2molMg}* \frac{40.3gMgO}{1molMgO} \\\\m_{MgO}=0.239gMgO[/tex]
Best regards.
What are representative elements"?
A. Elements in the short columns of the periodic table
B. Elements in the same row of the periodic table
C. Elements that share the same properties on the periodic table
D. Elements in the tall columns of the periodic table
The representative elements are the elements in the tall columns of the periodic table.
What are the representative elements?The representative elements that can also be referred to as the main group elements. They can be used to represent the chemistry of the group to which they belong.
Hence, the representative elements are the elements in the tall columns of the periodic table.
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A pressure cooker contains 5.68 liters of air at a temperature of 394 K. If the absolute pressure of the air in the pressure cooker is 205 pascals, how many moles of air are in the cooker? The cooker contains _______ moles of air. 1 SEE ANSWER
Answer:
Explanation:
We shall find out volume of air at NTP or at 273 K and 10⁵ Pa ( 1 atm )
Let it be V₂
[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]
[tex]\frac{2\times 10^5\times 5.68}{394} =\frac{10^5\times V_2}{273}[/tex]
V₂ = 7.87 litres
22.4 litres of any gas is equivalent to 1 mole
7.87 litres of air will be equivalent to
7.87 / 22.4 moles
= .35 moles .
a certain compound was found to contain 54.0 g of carbon and 10.5 grams of hydrogen. its relative molecular mass is 86.0. find the empirical and molecular formulas
Answer:
empirical formula = C3H7
molecular formula = C6H14
Question 7 options: The cell potential of an electrochemical cell made of an Fe, Fe2 half-cell and a Pb, Pb2 half-cell is _____ V. Enter your answer to the hundredths place and do not leave off the leading zero, if needed.
Answer: Thus the cell potential of an electrochemical cell is +0.28 V
Explanation:
The calculation of cell potential is done by :
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Fe^{2+}/Fe]}= -0.41V[/tex]
[tex]E^0_{[Pb^{2+}/Pb]}=-0.13V[/tex]
As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.
[tex]E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Fe^{2+}/Fe]}[/tex]
[tex]E^0=-0.13- (-0.41V)=0.28V[/tex]
Thus the cell potential of an electrochemical cell is +0.28 V
Calculate the frequency (Hz) and wavelength (nm)
of the emitted photon when an electron drops from
the n = 4 to the n=2 level in a hydrogen atom
Answer:
wavelength, λ = 486.6 nm
frequency, f = 6.16 * 10¹⁴ Hz
Explanation:
a. Wavelength
Using the wavelength equation; 1/λ = (1/hc) * 2.18 * 10⁻¹⁸ J * (1/nf² - 1/ni²)
Where nf is the final energy level; ni is the initial energy level; h is Planck's constant = 6.63 * 10⁻³⁴ J.s; c is velocity of light = 3 * 10⁸ m/s
1/λ = 1/(6.63 * 10⁻³⁴ J.s * 3 * 10⁸ m/s) * 2.18 * 10⁻¹⁸ J * (1/2² - 1/4²)
1/λ = 2.055 * 10⁶ m
λ = 4.866 * 10⁻⁷ m
wavelength, λ = 486.6 nm
b. Frequency
Using f = c/λ
f = (3 * 10⁸ m/s) / 4.866 * 10⁻⁷ m
frequency, f = 6.16 * 10¹⁴ Hz
Which characteristic of life best describes the process of photosynthesis?
Answer:
Using energy.
Explanation:
Which of the following best describes the interaction of the alkali metals with water? Select the correct answer below: A. They all dissolve easily in water. B. They do not react or dissolve in water. C. They react strongly with water to produce an alkaline solution and hydrogen. D. They react strongly with water to produce an alkaline solution and oxygen.
Answer:
C. They react strongly with water to produce an alkaline solution and hydrogen
Explanation:
All alkali metals react vigorously with cold water. In the reaction, hydrogen gas is given off and the metal hydroxide is produced.
Hope that helps.
The reaction, 2 SO3(g) <--> 2 SO2(g) + O2(g) is endothermic. Predict what will happen if the temperature is increased.
Explanation:
This reaction is in equilibrium and would hence obey lechatelier's principle. This principle states that whenever a system at equilibrium undergoes a change, it would react in way so as to annul that change.
Since it is an endothermic reaction, increasing the temperature would cause the reaction to shift towards the right.
This means that it favours product formation and more of the product would be formed.
The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. A(g) + 2B(g) ↔ AB2(g) Kc = 59 AB2(g) + B(g) ↔ AB3(g) Kc = ? A(g) + 3B(g) ↔ AB3(g) Kc = 478
Answer:
The correct answer is 8.10
Explanation:
Given:
A(g) + 2B(g) ↔ AB₂(g) Kc = 59 ---- Eq. 1
A(g) + 3B(g) ↔ AB₃(g) Kc = 478 ----- Eq. 2
We have to rearrange the chemical equations in order to obtain:
AB₂(g) + B(g) ↔ AB₃(g) Kc = ?
AB₂(g) is a reactant, so we have to use the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant is the same: Kc= 478. The following is the sum of rearranged chemical equations, and the compounds in bold and italic are canceled:
AB₂(g) ↔ A(g) + 2B(g) Kc₁= 1/59
A(g) + 3B(g) ↔ AB₃(g) Kc₂= 478
-----------------------------------------
AB₂(g) + B(g) ↔ AB₃(g)
If we add reactions at equilibrium, the equilibrium constants Kc are mutiplied as follows:
Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10
The value of the missing equilibrium constant is 8.10.
The value of the missing equilibrium constant is 8.10
Chemical Equations:Since
A(g) + 2B(g) ↔ AB₂(g) Kc = 59 ---- Eq. 1
A(g) + 3B(g) ↔ AB₃(g) Kc = 478 ----- Eq. 2
Now we have to rearrange the chemical equations in order to obtain:
AB₂(g) + B(g) ↔ AB₃(g) Kc = ?
Here AB₂(g) represents a reactant, so we have to applied the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant should be the same: Kc= 478.
The following is the sum of rearranged chemical equations, and the compounds in bold and italic should be canceled:
AB₂(g) ↔ A(g) + 2B(g) Kc₁= 1/59
A(g) + 3B(g) ↔ AB₃(g) Kc₂= 478
-----------------------------------------
AB₂(g) + B(g) ↔ AB₃(g)
In the case when we add reactions at equilibrium, the equilibrium constants Kc are multiplied as follows:
Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10
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The titration of 78.5 mL of an unknown concentration H3PO4 solution requires 134 mL of 0.224 M KOH solution. What is the concentration of the H3PO4 solution
Answer:
0.127 M.
Explanation:
The balanced equation for the reaction is given below:
H3PO4 + 3KOH —› K3PO4 + 3H2O
From the balanced equation above, we obtained the following data:
Mole ratio of acid, H3PO4 (nA) = 1
Mole ratio of base, KOH (nB) = 3
Data obtained from the question include:
Volume of acid, H3PO4 (Va) = 78.5 mL
Molarity of acid, H3PO4 (Ma) =...?
Volume of base, KOH (Vb) = 134 mL
Molarity of base, KOH (Mb) = 0.224 M
The concentration of the acid, H3PO4 can be obtained as follow:
MaVa / MbVb = nA/nB
Ma x 78.5 / 0.224 x 134 = 1/3
Cross multiply
Ma x 78.5 x 3 = 0.224 x 134 x 1
Divide both side by 78.5 x 3
Ma = (0.224 x 134 x 1) /(78.5 x 3)
Ma = 0.127 M
Therefore, the concentration of the acid, H3PO4 is 0.127 M.
For the following reaction at equilibrium, which gives a change that will shift the position of equilibrium to favor formation of more products? 2NOBr(g) 2NO(g) + Br 2(g), ΔHº rxn = 30 kJ/mol
Answer:
Based on the given reaction, it is evident that the reaction is endothermic as indicated by a positive sign of enthalpy of reaction. Thus, it can be stated that the favoring of the forward reaction will take place by upsurging the temperature of the reaction mixture.
Apart from this, based on Le Chatelier’s principle, any modification in the quantity of any species is performed at equilibrium and the reaction will move in such an orientation so that the effect of the change gets minimized. Therefore, a slight enhancement in the concentration of the reactant will accelerate the reaction in the forward direction and hence more formation of the product takes place.
what is the IUPAC name of KNO3, Fe(OH)3, H2SO3 and Al2(SO4)3
Answer:
KNO₃ - Potassium nitrate
Fe(OH)₃ - Iron(III) hydroxide
H₂SO3 - Sulfurous acid
Al₂(SO₄)₃ - Aluminum sulfate
Hope that helps.
Suppose of nickel(II) iodide is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of nickel(II) cation in the solution. You can assume the volume of the solution doesn't change when the nickel(II) iodide is dissolved in it. Round your answer to significant digits.
Answer:
0.619 M to 3 significant figures.
Explanation:
1 mole of [tex]NiI_{2}[/tex] - 312.5 g
? mole of [tex]NiI_{2}[/tex] - 2.9 g
= 2.9/312.5
= 0.0928 moles.
Concentration = no. of moles/vol in litres = [tex]\frac{0.0928}{0.150L}[/tex]
= 0.619 M
If the sugar concentration in a cell is 3% and the concentration in the cell’s environment is 5%, how can the cell obtain more sugar? {Hint: Sugar is polar and does not pass through the cell membrane easily.}
Select one:
a. Sugar can undergo facilitated diffusion through a channel protein, and no energy is required.
b. The cell must use active transport to pump the sugar in. Energy is required.
c. Sugar can diffuse straight through the phospholipid bilayer.
d. The cell CANNOT obtain more sugar. It is doomed.
The cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.
FACILITATED DIFFUSION:
Facilitated diffusion is the process whereby molecules move across a cell through the help of carrier/channel proteins. Facilitated diffusion is a type of passive transport and hence, does not require energy to occur. This is because movement occurs down a concentration gradient (high to low). According to this question, the sugar concentration in a cell is 3% and the concentration in the cell’s environment is 5%. This means that the sugar can travel down a concentration gradient across the cell membrane. However, because sugar is a polar molecule and does not pass through the cell membrane easily, a carrier proteins is needed to aid its movement. Therefore, the cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.Learn more at: https://brainly.com/question/18122054?referrer=searchResults
A major component of gasoline is octane, C8H18. When octane is burned in air, it chemically reacts with oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O) . What mass of carbon dioxide is produced by the reaction of 3.2g of oxygen gas? Round your answer to 2 significant digits.
Answer:
[tex]m_{CO_2}=2.8gCO_2[/tex]
Explanation:
Hello,
In this case, the combustion of octane is chemically expressed by:
[tex]C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O[/tex]
In such a way, due to the 25/2:8 molar ratio between oxygen and carbon dioxide, we can compute the yielded grams of carbon dioxide (molar mass 44 g/mol) as shown below:
[tex]m_{CO_2}=3.2gO_2*\frac{1molO_2}{32gO_2} *\frac{8molCO_2}{\frac{25}{2}molO_2 } *\frac{44gCO_2}{1molCO_2}\\ \\m_{CO_2}=2.8gCO_2[/tex]
Best regards.
the equilibrium concentrations were found to be [H2O]=0.250 M , [H2]=0.600 M , and [O2]=0.800 M . What is the equilibrium constant for this reaction?
Answer:
Keq = 0.217
Explanation:
Let's determine the equilibrium reaction.
In gaseous state, water vapor can be decomposed to hydrogen and oxygen and this is a reversible reaction.
2H₂(g) + O₂(g) ⇄ 2H₂O (g) Keq
Let's make the expression for the equilibrium constant
Products / Reactants
We elevate the concentrations, to the stoichiometry coefficients.
Keq = [H₂O]² / [O₂] . [H₂]²
Keq = 0.250² / 0.8 . 0.6² = 0.217
g A spontaneous process is one in which: A. releases a large amount of heat B. may happen (is possible) C. will rapidly approach equilibrium D. will happen quickly
Answer:
A. releases a large amount of heat
Explanation:
A reaction is said to be spontaneous if it can proceed on its own without the addition of external energy. A spontaneous reaction is not determined by the length of time, because some spontaneous reactions are completed after a long period of time. They are exothermic in nature. An example is the conversion of graphite to carbon which takes a long period of time to complete. Spontaneous reactions are known to increase entropy in a system. Entropy is the rate of disorder in a system.
In the combustion of fire, energy is released to the surroundings as there is a decrease in energy. This is an example of a spontaneous reaction because it is an exothermic reaction, which causes an increase in entropy and a decrease in energy.
The equilibrium between carbon dioxide gas and carbonic acid is very important in biology and environmental science. CO2 ( aq) + H2O ( l) H2CO3 ( aq) Which one of the following is the correct equilibrium constant expression (K c) for this reaction?
a) K =[H2CO3]/ [CO2]
b) K=[CO2]/ [H2CO3]
c) K=[H2CO3]/ [CO2][H2O]
d) K=[CO2][H2O]/ [H2CO3]
e) K=1/[H2CO3]
Answer:
Kc = [H₂CO₃] / [CO₂]
Explanation:
Equilibrium constant expression (Kc) of any reaction is defined as the ratio between molar concentrations in equilibrium of products over reactants.
Pure solids and liquids don't affect the equilibrium and you don't have to take its concentrations in the equilibrium.
Also, each specie must be powered to its reactant coefficient.
For example, for the reaction:
aA(s) + bB(aq) ⇄ cC(l) + nD(g) + xE(aq)
The equilibrium constant, kc is:
Kc = [D]ⁿ / [B]ᵇ[E]ˣ
You don't take A nor C species because are pure solids and liquids. b, n and x are the reactant coefficients of each substance. Ratio of products over reactants
Thus, for the reaction:
CO₂(aq) + H₂O(l) ⇄ H₂CO₃(aq)
The Kc is:
Kc = [H₂CO₃] / [CO₂]
30. What is the Bronsted base of H2PO4- + OH- ⟶HPO42- + H2O?
Answer:
OH⁻ is the Bronsted-Lowry base.
Explanation:
A Bronsted-Lowry base is a substance that accepts protons. In the chemical equation, OH⁻ accepts a proton from H₂PO₄⁻ to become H₂O. H₂PO₄⁻ would be a Bronsted-Lowry acid because it donates a proton to OH⁻ and becomes HPO₄²⁻.
Hope that helps.
After heating a sample of hydrated CuSO4, the mass of released H2O was found to be 2.0 g. How many moles of H2O were released if the molar mass of H2O is 18.016 g/mol
Answer:
0.1110 mol
Explanation:
Mass = 2g
Molar mass = 18.016 g/mol
moles = ?
These quantities are realted by the following equation;
Moles = Mass / Molar mass
Substituting the values of the quantities and solving for moles, we have;
Moles = 2 / 18.016 = 0.1110 mol
PLEASE HELP ME Calculate the change in boiling point when 0.402 moles of sodium chloride are added to 0.200 kilograms of water. Kf = -1.86°C/m; Kb = 0.512°C/m -2.1°C 2.1°C -7.5°C 7.5°C
Answer:
The change in the boiling point would be 2.1°C.
Explanation:
1 ) Let us first determine the molarity of this solution :
M = mol / kg,
M = 0.402 mol / 0.200 kg = 2.01 M NaCl
2 ) ΔT = i [tex]*[/tex] K [tex]*[/tex] m
ΔT = 2 [tex]*[/tex] 0.512C/m [tex]*[/tex] 2.01m
ΔT = 2.06C
As you can see, this is none of the answer choices. However the van't Hoff factor i in this case was taken to be 2, but this value is actually less than the predicted ideal solution. This is due to the ion pairing, causing i to be around 1.7 to 1.8. Therefore our solution is option b, 2.1°C.
Calculate the heat absorbed by a sample of water that has a mass of 45.00 g when the temperature increases from 21.0oC to 38.5 oC. (s=4.184 J/g.o C)
Answer:
The heat absorbed by the sample of water is 3,294.9 J
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q=?m= 45 gc= 4.184 [tex]\frac{J}{g*C}[/tex]ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 CReplacing:
Q= 4.184 [tex]\frac{J}{g*C}[/tex] * 45 g* 17.5 C
Solving:
Q=3,294.9 J
The heat absorbed by the sample of water is 3,294.9 J
Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest .
Answer:
-471 Kj/mole acrylic acid
Explanation:
THIS IS THE COMPLETE QUESTION BELOW;
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide: CaC (s) + 2 H2O(g) - CH (9) + Ca(OH),(s) AH -414. kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6 C H (9) + 3 CO2(9) + 4H2O(g) - SCH,CHCO,H) AH-132. kJ Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ?
The two equations from the reaction can be written as;
a)CaC₂(s) + 2H₂O(l) ------->C₂H₂(g) + CaOH₂(s)
Δ H= -414Kj ........................ equation (a)
b)6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj ...................... equation (b)
In equation (b)acrylic acid was produced by the reaction between Acetylene carbon dioxide and water
Then we can multiply equation(a) by factor of 6 and the ΔH Then we have (6× -414Kj)= ΔH= -2484Kj.
6CaC₂(s) + 12H₂O(l) ------->6C₂H₂(g) + 6CaOH₂(s)
Δ H= -2484Kj.................. equation (c)
6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj
Then add equation (c) and equation(b) then we have
6CaC₂(s) + 16H₂O(l)+3CO₂(g)------> 5CH₂CHCO₂H(g) + 6CaOH₂(s) ΔH= -2352Kj
ΔH(net)= -2352Kj/5moles
=-471Kj/mole
therefore, net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ? is -471Kj/mole acrylic acid
A compound decomposes with a half-life of 8.0 s and the half-life is independent of the concentration. How long does it take for the concentration to decrease to one-ninth of its initial value
Answer:
The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.
Explanation:
The decomposition of the compound has an exponential behavior and process can be represented by this linear first-order differential equation:
[tex]\frac{dc}{dt} = -\frac{1}{\tau}\cdot c(t)[/tex]
Where:
[tex]\tau[/tex] - Time constant, measured in seconds.
[tex]c(t)[/tex] - Concentration of the compound as a function of time.
The solution of the differential equation is:
[tex]c(t) = c_{o} \cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]c_{o}[/tex] is the initial concentration of the compound.
The time is now cleared in the result obtained previously:
[tex]\ln \frac{c(t)}{c_{o}} = -\frac{t}{\tau}[/tex]
[tex]t = -\tau \cdot \ln \frac{c(t)}{c_{o}}[/tex]
Time constant as a function of half-life is:
[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex]
Where [tex]t_{1/2}[/tex] is the half-life of the composite decomposition, measured in seconds.
If [tex]t_{1/2} = 8\,s[/tex], then:
[tex]\tau = \frac{8\,s}{\ln 2}[/tex]
[tex]\tau \approx 11.542\,s[/tex]
And lastly, given that [tex]\frac{c(t)}{c_{o}} = \frac{1}{9}[/tex] and [tex]\tau \approx 11.542\,s[/tex], the time taken for the concentration to decrease to one-ninth of its initial value is:
[tex]t = -(11.542\,s)\cdot \ln\frac{1}{9}[/tex]
[tex]t \approx 25.360\,s[/tex]
The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.
Which of the following properties should carbon (C) have based on its position on
the periodic table?
A. Shiny
B. Dense
C. Malleable
D. Poor conductor
Answer:
D- poor conductor
Explanation:
metallic properties decrease as we go on the right of the periodic table. Carbon is a non metal hence it is dull and a poor conductor.
it has a low density and is ductile.
Answer: Poor conductor
Explanation:
After the reaction between sodium borohydride and the ketone is complete, the reaction mixture is treated with water and H2SO4 to produce the desired alcohol. Explain the reaction by clearly indicating the source of the hydrogen atom that ends up on the oxygen
Answer:
The hydrogen can be gotten from the added Acid or water during "workup".
Explanation:
Basically we can say that the reaction describe in this question is a Reduction reaction because of the chemical compound called sodium borohydride. In the reaction described above we can see that there is a Reduction of ketone to alcohol by the compound; sodium borohydride.
For the reduction Reaction to occur the C-O bond must break so as to enable the formation of O-H bond and C-H bond.
So, "the reaction mixture is treated with water and H2SO4 to produce the desired alcohol", thus, the oxygen will definitely pick up the hydrogen from H2SO4 or H2O.