Final answer:
The maximum possible mass of NO2 that could be formed from the given amounts of reactants is 755.20 grams, and the amount of CO2 that would be produced from the combustion of 37.3 grams of C6H6 is 126.10 grams.
Explanation:
Calculating the Maximum Amount of NO2 Formed
For the balanced reaction 2 NO + O2 → 2 NO2, the stoichiometry shows a 1:1 ratio between NO and NO2. With 16.42 mol of NO, if O2 is in excess, 16.42 mol of NO2 could theoretically be formed. The molar mass of NO2 is 46.0055 g/mol, so the maximum mass of NO2 would be 16.42 mol × 46.0055 g/mol = 755.20 grams of NO2.
Amount of CO2 Produced from C6H6 Combustion
The balanced reaction for the combustion of benzene (C6H6) is 2 C6H6 + 15 O2 → 12 CO2 + 6 H2O. One mole of C6H6 (78.1134 g/mol) produces 6 moles of CO2. With 37.3 grams of C6H6, there would be 37.3 g / 78.1134 g/mol = 0.4775 mol C6H6 which would yield 0.4775 mol C6H6 × 6 mol CO2/mol C6H6 = 2.865 mol CO2.
The molar mass of CO2 is 44.0095 g/mol, so the mass of CO2 would be 2.865 mol × 44.0095 g/mol = 126.10 grams of CO2.
Final answer:
The maximum amount of NO2 that can be formed from 16.42 mol of NO and 14.47 mol of O2 is 755.1 grams. When reacting 37.3 grams of C6H6 with 126.1 grams of O2, the amount of CO2 produced is 125.96 grams.
Explanation:
Reaction Stoichiometry and Limiting Reactants:
To determine the maximum amount of NO2 that can be formed from 16.42 mol of NO and 14.47 mol of O2, we use the balanced chemical equation 2 NO + O2 → 2 NO2. This equation shows that 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. We can see that NO will be the limiting reactant because 14.47 mol of O2 would theoretically react with only 28.94 mol of NO, which is more than the 16.42 mol of NO present.
Since NO is the limiting reactant, we can form a maximum of 16.42 mol of NO2. Using the molar mass of NO2 (46.01 g/mol), this is equivalent to 755.1 grams of NO2. For the reaction C6H6 + O2 → CO2 + H2O, first, we would calculate the moles of C6H6 and O2, then use stoichiometry to determine the moles of CO2 produced. The molar mass of C6H6 (78.11 g/mol) means we start with 0.477 moles of C6H6.
From the balanced equation, we can see that for every 2 moles of NO, we need 1 mole of O2 to produce 2 moles of NO2. This gives us the mole ratio of NO to O2 as 2:1.
To find the limiting reagent (the reactant that is completely consumed and determines the maximum amount of product formed), we compare the number of moles of each reactant to the mole ratio.
For NO:
16.42 mol NO * (1 mol O2 / 2 mol NO) = 8.21 mol O2
For O2:
14.47 mol O2 * (2 mol NO / 1 mol O2) = 28.94 mol NO
Since the calculated amount of O2 (28.94 mol) is greater than the actual amount of O2 (14.47 mol), O2 is in excess and NO is the limiting reagent.
Now, we need to calculate the maximum amount of NO2 produced from the limiting reagent, which is NO.
From the balanced equation, we know that 2 moles of NO react to form 2 moles of NO2. Therefore, the mole ratio of NO to NO2 is 2:2.
Using the moles of NO (8.21 mol), we can calculate the moles of NO2:
8.21 mol NO * (2 mol NO2 / 2 mol NO) = 8.21 mol NO2
To convert moles of NO2 to grams, we need to use the molar mass of NO2, which is 46.01 g/mol.
8.21 mol NO2 * 46.01 g/mol = 377.27 g NO2
Therefore, the maximum amount of NO2 that could be formed from 16.42 mol of NO and 14.47 mol of O2 is 377.27 grams of NO2.
For the second question, we have the balanced chemical equation:
C6H6 + 15O2 → 6CO2 + 3H2O
From the balanced equation, we can see that for every mole of C6H6, we need 15 moles of O2 to produce 6 moles of CO2.
To determine the amount of CO2 produced, we first need to find the limiting reagent.
For C6H6:
37.3 g C6H6 * (1 mol C6H6 / 78.11 g C6H6) = 0.477 mol C6H6
For O2:
126.1 g O2 * (1 mol O2 / 32 g O2) = 3.94 mol O2
Since the calculated amount of C6H6 (0.477 mol) is less than the actual amount of O2 (3.94 mol), C6H6 is the limiting reagent.
Using the mole ratio from the balanced equation, we find that 0.477 mol of C6H6 will produce 6 moles of CO2.
To convert moles of CO2 to grams, we need to use the molar mass of CO2, which is 44.01 g/mol.
0.477 mol CO2 * 44.01 g/mol = 21.0 g CO2
Therefore, 37.3 grams of C6H6 reacted with 126.1 grams of O2 will produce 21.0 grams of CO2.
The oxidizing agent our bodies use to obtain energy from food is oxygen (from the air). If you breathe 17 times a minute (at rest), taking in and exhaling 0.50 L of air with each breath, what volume of air do you breathe each day?
Answer:
The volume of the air breath per day is =12240 L day ^ -1 and he volume of oxygen breathe each day is= 2570.4 L
Explanation:
Complete question: The oxidizing agent our bodies use to obtain energy from food is oxygen (from the air). If you breathe 17 times a minute (at rest), taking in and exhaling 0.50 L of air with each breath, what volume of air do you breathe each day? Air is 21% oxygen by volume. what volume of oxygen do you breathe each day?
Solution
Given that,
The volume of the air with each breath = 0.50L
The frequency of breath is = 17 times per minute
the next step is to calculate the volume of the air breath per day
The volume of the breathed air = volume of air breathed/ breath * 17 breaths/min * 60 minutes / per hour * 24 hours/ per day
= 0.50L/ breath * 17 breath /min * 60 minutes / per hour * 24 hours/ per day
= 12240 L day ^ -1
The next step is to calculate the oxygen breathed per day:
The volume of oxygen breathed = volume of air breathed * percent oxygen in air/100
= 12240 L day ^ -1 * 21 /100
= 2570.4 L day ^ -1
Therefore for each day, the volume of oxygen breathe is 2570.4 L
Answer:
Volume of oxygen we breathe each day = 2570.4 L/day
Explanation:
Given that :
volume of air used per breath = 0.5 L
Frequency of breath = 17 times per minute
We know that the percentage volume of air in the atmosphere is approximately = 21%
Hence; the volume of air breathed = [tex]\frac{volume \ of \ air breathed}{breath } *\frac{17 \ breaths}{ minute } * \frac{60 \ minutes }{hour}*\frac{24 \ hour}{day}[/tex]
the volume of air breathed = [tex]\frac{0.5 \ L }{breath } *\frac{17 \ breaths}{ minute } * \frac{60 \ minutes }{hour}*\frac{24 \ hour}{day}[/tex]
the volume of air breathed = 12,240 L/day
To calculate the volume of oxygen breathed per day; we have:
Volume of oxygen breathed = [tex]volume \ of \ air \ breathed * \frac{percentage \ of \ air }{100}[/tex]
Volume of oxygen breathed = [tex]12240* \frac{21 }{100}[/tex]
Volume of oxygen breathed = 2570.4 L/day
For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate. barium chloride (aq) potassium sulfate (aq) barium sulfate (s) potassium chloride (aq)
For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate.
The reaction consumes _____ moles of barium chloride. The reaction produces _____ moles of barium sulfate and _____ moles of potassium chloride.
Answer: a) The reaction consumes 0.365 moles of barium chloride.
b) The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of barium chloride}=\frac{76.0}{208g/mol}=0.365moles[/tex]
[tex]\text{Moles of potassium sulphate}=\frac{67.0}{174g/mol}=0.385moles[/tex]
[tex]BaCl_2(aq)+K_2SO_4(aq)\rightarrow BaSO_4(s)+2KCl(aq)[/tex]
According to stoichiometry :
1 mole of [tex]BaCl_2[/tex] require 1 mole of [tex]K_2SO_4[/tex]
Thus 0.365 moles of [tex]BaCl_2[/tex] will require=[tex]\frac{1}{1}\times 0.365=0.365moles[/tex] of [tex]K_2SO_4[/tex]
Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]K_2SO_4[/tex] is the excess reagent.
As 1 moles of [tex]BaCl_2[/tex] give = 1 moles of [tex]BaSO_4[/tex]
Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{1}{1}\times 0.365=0.365moles[/tex] of [tex]BaSO_4[/tex]
As 1 moles of [tex]BaCl_2[/tex] give = 2 moles of [tex]KCl[/tex]
Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{2}{1}\times 0.365=0.730moles[/tex] of [tex]KCl[/tex]
Thus the reaction consumes 0.365 moles of barium chloride. The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.
Which energy level has the least energy?
n=3
n=1
n=5
n=7
Answer:n=1
Explanation:
Final answer:
The energy level n=1 has the least energy.
Explanation:
The energy levels of electrons in an atom are determined by the principal quantum number, denoted as n. The higher the value of n, the higher the energy level. Therefore, the energy level n=7 has the highest energy. On the other hand, the energy level n=1 has the lowest energy, which makes it the answer to your question.
What is the difference between reactants and products in a chemical equation (Explain)? In the equation “carbon plus oxygen makes carbon dioxide”, what substance(s) is/are the reactant(s)? What substance(s) is/are the product(s) (Explain).
Explanation:
Reactants can be referred to as starting materials for a chemical reaction. They undergo a change to form products. Reactants are usuallyconsumed in the reaction process.
Reactants are written at the left side before the arrow sign in a reaction.
Products are the ending materials of a reaction. They are what is left after the reactants has been consumed.
Prpducts are written on the right side after the arrow sign in a reaction
Basically, a reaction is given as;
Reactants --> Products
A + B --> C + D
A and B are reactants and C and D are products
In the reaction;
“carbon plus oxygen makes carbon dioxide”
C + O2 --> CO2
The reactants are; C and O2.
The product is CO2
Answer:
The answer to your question is given below
Explanation:
A. The difference between reactants and products in a chemical equation is that the reactants are located on the left hand side of the equation while the products are located on the right hand side of the equation.
B. When carbon (C) combine with oxygen (O2) to produce carbon dioxide (CO2), the details of the reaction are given below:
Reactants => C and O2
Product => CO2.
The balanced equation is given below:
C + O2 —> CO2
C and O2 are located on the left side which indicates that they are the reactants while CO2 is located at the right side which indicates that it is the product.
Sulfurous Acid, H2SO3, is a diprotic acid, with the following acid dissociation constants: Ka1 = 1.4 × 10-2 Ka2 = 6.7 × 10-8 What is the Kb for SO32- at 25ºC? Group of answer choices 9.4 × 10-10 1.5 × 10-7 6.7 × 10-8 7.2 × 10-13
Answer:
Kb = 7.1 x 10⁻¹³
Explanation:
Ka x Kb = Kw => Kb = 1 x 10⁻¹⁴/1.4 x 10⁻² = 7.1 x 10⁻¹³
The Kb for SO32- at 25ºC is approximately 1.5 x 10^-7.
therefore correct option (b).
Explanation:To find the Kb for SO32-, we can use the relationship between Ka and Kb. Since the diprotic acid, H2SO3, forms SO32- in the second dissociation step, we can use the formula Kb = Kw/Ka. Kb is the equilibrium constant for the reaction of a base with water to form the hydroxide ion. Kw is the ion product of water, which is 1.0 x 10^-14 at 25ºC. So, using the given Ka2 value of 6.7 x 10^-8, we can calculate the Kb for SO32-:
Kb = Kw/Ka = (1.0 x 10^-14) / (6.7 x 10^-8)
Kb ≈ 1.5 x 10^-7
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What mass of helium is in a 2.00L ballon at STP
Answer:
0.357 g He
Explanation:
1 mol of gas at STP = 22.4 L
2.00 L * 1 mol/22.4 L = 0.08929 mol He
M(He) = 4.00 g/mol
0.08929 mol He * 4.00 g/ 1 mol He = 0.357 g He
(45 pts) What is the theoretical yield (in g) of iron(III) carbonate that can be produced from 1.72 g of iron(III) nitrate and an excess of sodium carbonate? The molar mass of iron(III) carbonate (Fe2(CO3)3) is 291.73 g∙mol–1 and the molar mass of iron(III) nitrate (Fe(NO3)3) is 241.88 g∙mol–1.
2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)
Answer:
1.04g of iron III carbonate
Explanation:
First, we must put down the equation of reaction because it must guide our work.
2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)
From the question, we can see that sodium carbonate is in excess while sodium nitrate is the limiting reactant.
Number of moles of iron III nitrate= mass of iron III nitrate reacted/ molar mass of iron III nitrate
Mass of iron III nitrate reacted= 1.72g
Molar mass of iron III nitrate= 241.88 g∙mol–1
Number of moles of iron III nitrate= 1.72g/241.88 g∙mol–1= 7.11×10^-3 moles
From the equation of the reaction;
2 moles of iron III nitrate yields 1 mole of iron III carbonate
7.11×10^-3 moles moles of iron III nitrate yields 7.11×10^-3 × 1/ 2= 3.56×10^-3 moles of iron III carbonate
Theoretical mass yield of iron III carbonate = number of moles of iron III carbonate × molar mass
Theoretical mass yield of iron III carbonate = 3.56×10^-3 moles ×291.73 g∙mol–1 = 1.04g of iron III carbonate
The theoretical yield of Fe₂(CO₃)₃ obtained from the reaction is 1.04 g
We'll begin by calculating the mass of Fe(NO₃)₃ that reacted and the mass of Fe₂(CO₃)₃ produced from the balanced equation.
2Fe(NO₃)₃ + 3Na₂CO₃ —> Fe₂(CO₃)₃ + 6NaNO₃
Molar mass of Fe(NO₃)₃ = 241.88 g/mol
Mass of Fe(NO₃)₃ from the balanced equation = 2 × 241.88 = 483.4 g
Molar mass of Fe₂(CO₃)₃ = 291.73 g/mol
Mass of Fe₂(CO₃)₃ from the balanced equation = 1 × 291.73 = 291.73 g
From the balanced equation above,
483.4 g of Fe(NO₃)₃ reacted to produce 291.73 g of Fe₂(CO₃)₃
Finally, we shall determine the theoretical yield of Fe₂(CO₃)₃. This can be obtained as follow:From the balanced equation above,
483.4 g of Fe(NO₃)₃ reacted to produce 291.73 g of Fe₂(CO₃)₃
Therefore,
1.72 g of Fe(NO₃)₃ will react to produce = (1.72 × 291.73) / 483.4 = 1.04 g of Fe₂(CO₃)₃
Thus, the theoretical yield of Fe₂(CO₃)₃ is 1.04 g
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A student in lab titrated a diprotic weak acid with 0.100 M NaOH. To reach the equivalence point required 22.58 mL of NaOH solution. How many moles of diprotic acid were present in the titration
Answer:
0.0023 moles of H₂A
Explanation:
moles H₂A in soln = moles NaOH used in titration
moles = Molarity x Volume in Liters
moles H₂A = moles NaOH used
Which is = (0.100M)(0.02258L) = 0.0023 mol H₂A
The number of moles in a diprotic acid is 0.0023 moles of H₂A. This can be identified using law of dilution.
Law of dilution:While performing titrations, the law of dilution is used.
Molarity is defined as the quantity of moles of solute partitioned by the volume of the arrangement in liters.
Moles of acid = Moles of base
n= M /V
Moles H₂A = moles NaOH used
= (0.100M)(0.02258L)
= 0.0023 mol H₂A
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What mass of potassium bromide (in grams) do you need to make 250.0 mL of a 1.50 M potassium bromide solution?
Answer:
44.63g
Explanation:
First, let us calculate the number of mole of KBr in 1.50M KBr solution.
This is illustrated below:
Data obtained from the question include:
Volume of solution = 250mL = 250/1000 = 0.25L
Molarity of solution = 1.50M
Mole of solute (KBr) =.?
Molarity is simply mole of solute per unit litre of solution
Molarity = mole /Volume
Mole = Molarity x Volume
Mole of solute (KBr) = 1.50 x 0.25
Mole of solute (KBr) = 0.375 mole
Now, we calculate the mass of KBr needed to make the solution as follow:
Molar Mass of KBr = 39 + 80 = 119g/mol
Mole of KBr = 0.375 mole
Mass of KBr =?
Mass = number of mole x molar Mass
Mass of KBr = 0.375 x 119
Mass of KBr = 44.63g
Therefore, 44.63g of KBr is needed to make 250.0mL of 1.50 M potassium bromide (KBr) solution
Final answer:
To make a 1.50 M potassium bromide solution with a volume of 250.0 mL, you would need approximately 44.625 grams of potassium bromide.
Explanation:
To calculate the mass of potassium bromide required, we can use the formula:
mass (g) = concentration (M) x volume (L) x molar mass (g/mol)
The concentration is given as 1.50 M, the volume is 250.0 mL (which can be converted to 0.250 L), and the molar mass of potassium bromide (KBr) is 119.0 g/mol. Plugging in these values, we get:
mass (g) = 1.50 M x 0.250 L x 119.0 g/mol = 44.625 g
Therefore, you would need approximately 44.625 grams of potassium bromide to make 250.0 mL of a 1.50 M potassium bromide solution.
what is the ratio of the atoms in molecular formulae (NCl3) nitrogen trichloride
Answer:
3:1 M ratio at pH 3-4
Explanation:
Be sure to answer all parts. A freshly isolated sample of 90Y was found to have an activity of 2.2 × 105 disintegrations per minute at 1:00 p.m. on December 3, 2006. At 2:15 p.m. on December 17, 2006, its activity was measured again and found to be 5.8 × 103 disintegrations per minute. Calculate the half-life of 90Y. Enter your answer in scientific notation.
Answer:
The half-life is [tex]t_h = 3.856*10^{3} minute[/tex]
Explanation:
From the question we are told that
The sample is 90 Y
The first activity is [tex]A_1 = 2.2 *10^5[/tex] per minute
The second activity is [tex]A_2 = 5.8 *10^3[/tex] per minute
The duration from 1:00 p.m. on December 3, 2006 to 2:15 p.m. on December 17, 2006 is
[tex]t = 14 \ days \ 1 hr \ 15 min[/tex]
Converting to minutes we have
[tex]t = (14 * 24 * 60) + (1* 60) + 15[/tex]
[tex]t = 20235 \ minutes[/tex]
The first order rate constant for this disintegrations can be mathematically represented as
[tex]ln \frac{A_2}{A_1} = - \lambda t[/tex]
Where [tex]\lambda[/tex] is the rate constant
Substituting values
[tex]ln [\frac{5.8 * 10^{3}}{2.2 *10^{5}} ] = - \lambda * 20235[/tex]
[tex]-3.6358 = - \lambda * 20235[/tex]
So
[tex]\lambda = \frac{3.6358}{20235}[/tex]
[tex]\lambda = 1.7968 *10^{-4} minute^{-1}[/tex]
The half life is mathematically represented as
[tex]t_{h} = \frac{0.693}{\lambda }[/tex]
So [tex]t_h = \frac{0.693}{1.7968 *10^{-4}}[/tex]
[tex]t_h = 3.856*10^{3} minute[/tex]
The half-life of the isolated sample of 90Y is calculated using the formula for exponential decay and the given information. The decay constant (lambda) is found and then used to find the half-life, which is approximately 2.67 days.
Explanation:The subject of this question is the half-life of a isolated sample of 90Y. Half-life is the rate at which radioactive substances decay. The half-life can be calculated using the formula for exponential decay, N = N0e-lambda*t. We are given N0 (the initial amount of material, 2.2 x 10^5 disintegrations per minute), N (the remaining amount of material, 5.8 x 10^3 disintegrations per minute), and t (the time elapsed, 14.25 days). We can use these values to find lambda (the decay constant). Lambda is equal to the natural log of N0/N divided by t. To find the half-life, we use the formula, T = ln(2)/lambda.
Performing the appropriate calculations, we find that the half-life of 90Y is approximately 2.67 days (or 2.67 x 10^0 days in scientific notation).
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Switching one substance for another is called..?
a) displacement
b) synthesis
c) decomposition
d) oxidation
Switching one substance for another is called displacement.
What is chemical reaction?Chemical reactions are those reactions in which reactants are react with each other for the formation of product.
Displacement reactions are those reactions in which one molecule of the reactant get displaced by another molecule.Synthesis reactions are those reactions in which two reactants are combined with each other for the formation of product.Decomposition reactions are those reactions in which one reactant molecule will convert into two small molecules.Oxidation reactions are those in which addition of oxygen atom or removal of hydrogen atom or electrons takes place.Hence, given reaction is the displacement reaction.
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A voltaic cell that uses the reaction PdCl42−(aq)+Cd(s) → Pd(s)+4Cl−(aq)+Cd2+(aq) has a measured standard cell potential of +1.03 V. You may want to reference (Pages 860 - 867) Section 20.4 while completing this problem. Part A Write the half-cell reaction at the cathode.
Answer: The half-cell reaction occurring at cathode is [tex]PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)[/tex]
Explanation:
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
Reduction reaction is defined as the reaction in which a chemical specie accepts electrons. The oxidation state of the chemical specie reduces.
The given balanced chemical equation is:
[tex]PdCl_4^{2-}(aq)+Cd(s)\rightarrow Pd(s)+4Cl^-(aq)+Cd^{2+}(aq)[/tex]
The half cell reaction occurring at cathode follows:
[tex]PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)[/tex]
Hence, the half-cell reaction occurring at cathode is given above.
According to the experimental procedure of Experiment F1, 135 microliters of acetophenone (120.15 g/mol, 1.03 g/mL) was reacted with 127 mg of 4-nitrobenzaldehyde (151.12 g/mol). What is the theoretical yield, in milligrams (mg), of trans-4-nitrochalcone (253.25 g/mol)? Enter your answer as digits only (no units), using the proper number of significant figures.
Answer: The theoretical yield of 4-nitrochalcone is, [tex]2.13\times 10^2[/tex]
Explanation : Given,
Volume of acetophenone = 135 microliters = 135 × 10⁻⁶ L = 0.135 mL
conversion used : (1 microliter = 10⁻⁶ L) and (1 L = 1000 mL)
Density of acetophenone = 1.03 g/mL
Mass of acetophenone = Density × Volume = 1.03 g/mL × 0.135 mL = 0.139 g
Mass of 4-nitrobenzaldehyde = 127 mg = 0.127 g
Conversion used : (1 mg = 0.001 g)
First we have to calculate the moles of acetophenone and 4-nitrobenzaldehyde
[tex]\text{Moles of acetophenone}=\frac{\text{Given mass acetophenone}}{\text{Molar mass acetophenone}}[/tex]
[tex]\text{Moles of acetophenone}=\frac{0.139g}{120.15g/mol}=0.00116mol[/tex]
and,
[tex]\text{Moles of 4-nitrobenzaldehyde}=\frac{\text{Given mass 4-nitrobenzaldehyde}}{\text{Molar mass 4-nitrobenzaldehyde}}[/tex]
[tex]\text{Moles of 4-nitrobenzaldehyde}=\frac{0.127g}{151.12g/mol}=0.000840mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]C_8H_8O+C_7H_5NO_3\rightarrow C_{15}H_{11}NO_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of 4-nitrobenzaldehyde react with 1 mole of acetophenone
So, 0.000840 mole of 4-nitrobenzaldehyde react with 0.000840 mole of acetophenone
From this we conclude that, acetophenone is an excess reagent because the given moles are greater than the required moles and 4-nitrobenzaldehyde is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of 4-nitrochalcone
From the reaction, we conclude that
As, 1 mole of 4-nitrobenzaldehyde react to give 1 mole of 4-nitrochalcone
So, 0.000840 mole of 4-nitrobenzaldehyde react to give 0.000840 mole of 4-nitrochalcone
Now we have to calculate the mass of 4-nitrochalcone
[tex]\text{ Mass of 4-nitrochalcone}=\text{ Moles of 4-nitrochalcone}\times \text{ Molar mass of 4-nitrochalcone}[/tex]
Molar mass of 4-nitrochalcone = 253.25 g/mole
[tex]\text{ Mass of 4-nitrochalcone}=(0.000840moles)\times (253.25g/mole)=0.21273g=212.73mg=2.13\times 10^2mg[/tex]
(1 g = 1000 g)
Therefore, the theoretical yield of 4-nitrochalcone is, [tex]2.13\times 10^2mg[/tex]
At a certain temperature, the K p for the decomposition of H 2 S is 0.739 . H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g ) Initially, only H 2 S is present at a pressure of 0.215 atm in a closed container. What is the total pressure in the container at equilibrium?
Answer:
The total pressure in the container is 0.389 atm
Explanation:
Step 1: Data given
Kp = 0.739
The initial pressure of H2S = 0.215 atm
Step 2: The balanced equation
H2S(g) ⇆ H2(g) + S(g)
Step 3: The initial pressures
pH2S = 0.215 atm
pH2 = 0 atm
pS = 0 atm
Step 4: The pressures at the equilibrium
pH2S = 0.215 - X atm
pH2 = X atm
pS = X atm
Step 5:
Kp = 0.739 = (pS)*(pH2) / (pH2S)
0.739 = X*X / (0.215 - X)
0.739 = X² / (0.215 - X)
X² = 0.739*(0.215-X)
X² = 0.1589 - 0.739X
X² +0.739X - 0.1589 = 0
X = 0.174
pH2S = 0.215 - 0.174 atm = 0.041 atm
pH2 = 0.174 atm
pS = 0.174 atm
Step 6: Calculate the total pressure in the container
Total pressure = 0.041 atm + 0.174 atm + 0.174 atm
Total pressure = 0.389 atm
The total pressure in the container is 0.389 atm
Soaps feel slippery on our hands because they dissolve the oil on our skin decreasing friction this is because soaps contain
Answer:
.
Explanation:
Oil or Lipid + Base lead to Glycerol and Soap
The combustion of 1.771 g of propanol ( C 3 H 7 OH ) increases the temperature of a bomb calorimeter from 298.00 K to 302.34 K . The heat capacity of the bomb calorimeter is 13.70 kJ/K . Determine Δ H for the combustion of propanol to carbon dioxide gas and liquid water.
Answer: ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ
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=C\times \Delta T[/tex]
Q = Heat absorbed by calorimeter =?
C = heat capacity of calorimeter = 13.70 kJ/K
Initial temperature of the calorimeter = [tex]T_i[/tex] = 298.00 K
Final temperature of the calorimeter = [tex]T_f[/tex] = 302.34 K
Change in temperature ,[tex]\Delta T=T_f-T_i=(302.34-298.00)K=4.34K[/tex]
Putting in the values, we get:
[tex]Q=13.70kJ/K\times 4.34K=59.4kJ[/tex]
As heat absorbed by calorimeter is equal to heat released by combustion of propanol
[tex]Q=q[/tex]
[tex]\text{Moles of propanol}=\frac{\text{given mass}}{\text{Molar Mass}}=\frac{1.771g}{60g/mol}=0.030mol[/tex]
Heat released by 0.030 moles of propanol = 59.4 kJ
Heat released by 1 mole of propanol = [tex]\frac{59.4}{0.030}\times 1=1980kJ[/tex]
ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ/mol
The pH of a 1.1M solution of hexanoic acid HC6H11O2 is measured to be 2.40. Calculate the acid dissociation constant Ka of hexanoic acid. Be sure your answer has the correct number of significant digits.
Answer:
Ka = 1.4 x 10⁻⁴
Explanation:
HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻
At equilibrium 1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M
Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵
Answer:
Ka = 1.4 x 10⁻⁴
Explanation:
HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻
At equilibrium
1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M
Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵
How much energy is evolved during the formation of 98.7 g of Fe, according to the reaction below? Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) ΔH°rxn = -852 kJ How much energy is evolved during the formation of 98.7 g of Fe, according to the reaction below? Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) ΔH°rxn = -852 kJ 753 kJ 482 kJ 1.51 x 103 kJ 4.20 x 103 kJ 241 kJ
The formation of 98.7 grams of Fe from the reaction Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) with an enthalpy change of -852 kJ per 112 grams of Fe formed, would result in approximately -753 kJ of energy being released. The reaction is exothermic, meaning it releases energy as heat.
Explanation:The question asks about the energy evolved during the formation of 98.7 grams of iron (Fe) from the reaction given: Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) with a enthalpy change (ΔH°rxn) of -852 kJ. The reaction is exothermic, meaning it releases energy as heat. The ΔH for the reaction is -852 kJ per 112 grams of Fe formed (the molar mass of Fe is 56 g/mol, so 2 mol of Fe is 112 g).
To find out how much energy is released in forming 98.7 g of Fe, you can use simple proportion: (98.7 g Fe / 112 g Fe) x -852 kJ = approx. -753 kJ. So, the formation of 98.7 grams of Fe would result in approximately -753 kJ of energy being released, pointing to answer 753 kJ among the options given.
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The energy evolved during the formation of 98.7 g of Fe is 753 kJ.
Explanation:The given reaction is:
Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)
The enthalpy change for the reaction is -852 kJ.
To calculate the energy evolved during the formation of 98.7 g of Fe, we need to use the molar mass of Fe.
The molar mass of Fe is 55.845 g/mol.
By converting the given mass of Fe to moles and using the stoichiometry of the balanced equation, we can calculate the energy evolved.
98.7 g Fe * (1 mol Fe / 55.845 g Fe) * (-852 kJ / 2 mol Fe) = -753 kJ
Therefore, the energy evolved during the formation of 98.7 g of Fe is 753 kJ.
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A chemist prepares hydrogen fluoride by means of the following reaction:
CaF2 + H2SO4 --> CaSO4 + 2HF
The chemist uses 11 g of CaF2 and an excess of H2SO4, and the reaction produces 2.2 g of HF.
(a) Calculate the theoretical yield of HF.
(b) Calculate the percent yield of HF.
Answer:
39.3%
Explanation:
Our guide in solving the problem must be the reaction equation hence it is pertinent to put down first:
CaF2 + H2SO4 --> CaSO4 + 2HF
We have a very important information in the question, sulphuric acid is present in excess. This implies that calcium fluoride is the limiting reactant.
Number of moles of calcium fluoride reacted= mass of calcium fluoride reacted/ molar mass of calcium fluoride
Molar mass of calcium fluoride= 78.07 g/mol
Number of moles of calcium fluoride= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of 0.28 moles of hydrogen fluoride = number of moles× molar mass
Molar mass of hydrogen flouride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g this is the theoretical yield of HF
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%
Answer:
A. The theoretical yield of HF is 5.64g
B. The percentage yield of HF is 39%
Explanation:
Step 1:
The balanced equation for the reaction:
CaF2 + H2SO4 --> CaSO4 + 2HF
Step 2:
Determination of the mass of CaF2 that reacted and the mass of HF produced from the balanced equation. This is illustrated below:
Molar Mass of CaF2 = 40 + (19x2) = 40 + 38 = 78g/mol
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF from the balanced equation = 2 x 20 = 40g.
From the balanced equation above,
78g of CaF2 reacted and 40g of HF were produced.
A. Determination of the theoretical yield of HF.
This is illustrated below:
From the balanced equation above,
78g of CaF2 reacted to produce 40g of HF.
Therefore, 11g of CaF2 will react to produce = (11 x 40)/78 = 5.64g of HF.
The theoretical yield of HF is 5.64g
B. Determination of the percentage yield.
The percentage yield of HF can be obtained as follow:
Actual yield = 2.2g
Theoretical yield = 5.64g
Percentage yield =?
Percentage yield = Actual yield/Theoretical yield x100
Percentage yield = 2.2/5.64 x 100
Percentage yield = 39%
The percentage yield of HF is 39%
Which of the following statements about hybridization are true? Select all that apply: Hybrid orbitals exist in isolated atoms. Hybrid orbitals within the same atom have the same energy and shape. Hybrid orbitals are described mathematically as a linear combination of atomic orbitals. An atom can have both hybridized and unhybridized orbitals at the same time.
Answer:
True:
Hybrid orbitals within the same atom have the same energy and shape. Hybrid orbitals are described mathematically as a linear combination of atomic orbitals. An atom can have both hybridized and unhybridized orbitals at the same time.Explanation:
Hybrid orbitals do not exist in isolated atoms. They form only in covalently bonded atoms.
Hybridization happens when several atomic orbitals combine to form other orbitals with the same energy and greater stability.
A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that combined to produce the set.
Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds, and both can appear in an atom at the same time.
The statements about hybridization that are true include:
B. Hybrid orbitals within the same atom have the same energy and shape.
C. Hybrid orbitals are described mathematically as a linear combination of atomic orbitals.
D. An atom can have both hybridized and unhybridized orbitals at the same time.
A sublevel is an energy level that is typically associated with the valence electrons found outside an atomic nucleus.
In Chemistry, there are four (4) types of sublevel and these includes:
I. s orbital (sublevel): it has one (1) orbital i.e 1s.
II. p orbital (sublevel): it has three (3) orbitals.
III. d orbital (sublevel): it has five (5) orbitals.
IV. f orbital (sublevel): it has seven (7) orbitals.
Hybridization can be defined as a phenomenon which involves the linear combination of two or more atomic orbitals of a molecule, so as to form the same number of hybrid orbitals, with each of the orbital having the same energy and shape.
Generally, the two types of hybridization an atom can have include:
1. Hybridized orbitals.
2. Unhybridized orbitals
Hence, we can deduce the following points from the above:
Hybrid orbitals of the same atom of a chemical molecule have the same energy and shape.Hybrid orbitals can be described mathematically as a linear combination of multiple atomic orbitals.An atom of a chemical element can have both hybridized and unhybridized orbitals at the same time.Read more: https://brainly.com/question/24403297
Almost all energy on Earth came from the blank
Answer:
sun
Explanation:
Answer:
The sun is where almost all energy on Earth came from.
9. How many molecules are contained in 6.44 moles of oxygen gas?
Answer:
3.88 × 10²⁴ molecules
Explanation:
In order to solve this question, we need to consider the Avogadro's number. know that 1 mole of particles contains 6.02 × 10²³ particles. This applies to different kinds of particles: atoms, molecules, electrons.
In this case, 1 mole of molecules of oxygen gas contains 6.02 × 10²³ molecules of oxygen gas. We will use this relation to find the number of molecules of oxygen gas in 6.44 moles of oxygen gas.
[tex]6.44mol \times \frac{6.02 \times 10^{23}molecules }{1mol} = 3.88 \times 10^{24}molecules[/tex]
What is the volume, in liters, occupied by 2.79 moles of N2 gas at 439 mmHg pressure and a
temperature of 64°C?
Answer:
132.93 L
Explanation:
Step 1:
Data obtained from the question:
Volume (V) =?
Number of mole (n) = 2.79 moles
Pressure (P) = 439 mmHg
Temperature (T) = 64°C
Step 2:
Conversion to appropriate unit
For pressure:
760mmHg = 1atm
Therefore, 439 mmHg = 439/760 = 0.58 atm
For temperature:
Temperature (Kelvin) = temperature (celsius) + 273
temperature (celsius) = 64°C
Temperature (Kelvin) = 64°C + 273 = 337 K
Step 3:
Determination of the volume.
The volume occupied by N2 can be obtained by using the ideal gas equation as follow:
Volume (V) =?
Number of mole (n) = 2.79 moles
Pressure (P) = 0.58 atm
Temperature (T) = 337 K
Gas constant (R) = 0.082atm.L/Kmol
PV = nRT
0.58 x V = 2.79 x 0.082 x 337
Divide both side by 0.58
V = (2.79 x 0.082 x 337)/0.58
V = 132.93 L
Therefore, the volume occupied by N2 is 132.93 L
Consider the mechanism. Step 1: A + B ⟶ C A+B⟶C slow Step 2: A + C ⟶ D A+C⟶D fast Overall: 2 A + B ⟶ D 2A+B⟶D Determine the rate law for the overall reaction, where the overall rate constant is represented as k .
The rate law is defined as the rate of reaction in which reactants are expressed in their molar concentration raised to the power of their stoichiometric coefficient.
In the given reaction, the slow step determines the rate of reaction.
The chemical reaction is:
2A + B [tex]\rightarrow[/tex] D
The intermediate reaction of the mechanism follows:
Step 1: A + B [tex]\rightleftharpoons[/tex] C (slow)
Step 2: A + C [tex]\rightleftharpoons[/tex] D (fast)
In the given reaction, the first step is a slow step, which determines the rate of the reaction. The rate for the equation can be given as:
Rate = k [A]² [B]
Therefore, the rate law expression for the overall reaction is Rate = k [A]² [B].
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The rate law for the given overall reaction, considering that the first step is the slow step, would be rate = k[A][B]. The reaction is first order with respect to both A and B, making it an overall second-order reaction.
Explanation:The question requires determining the rate law for an overall reaction from a given two-step reaction mechanism. In the provided reaction, step 1: A + B ⟶ C is the rate-determining (or slow) step while step 2: A + C ⟶ D is a fast step.
Given this scenario, the rate-determining step dictates the rate law of the overall reaction. The rate law for a reaction where the slow step is the first step can be written as , where
[tex]rate = k[A]^m[B]^n[/tex]is the rate constant and m and n are the orders of the reaction with respect to reactants A and B respectively.
In this case, since both A and B are involved in the slow step, the rate law would be rate = k[A][B], where the reaction is first-order with respect to both A and B, making it an overall second-order reaction.
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If 25.21 ml of NaOH solution is required to react completely with 0.550 g KHP, what is the molarity of the NaOH solution? Answer MUST be in Mol NaOH/L solution. Do not put in scientific notation.
Answer:
[NaOH] = 0.107M
Explanation:
25.21ml of NaOH(aq) soln + 0.550g KHPh => NaKPh + H₂O
The reaction is 1 to 1 NaOH to KHPh => moles NaOH neutralized = moles KHPh used. (KHPh => Potassium Hydrogen Phthalate f.wt. = 204.22 g/mol)
moles KHPh = 0.550g / 204.22g·mol⁻¹ = 0.0027 mole
moles NaOH neutralized = moles KHPh used = 0.0027 mole NaOH in 25.21ml aqueous solution ...
Molar concentration of NaOH solution = moles NaOH/Liters solution = 0.0027mol NaOH/0.0251L = 0.107M NaOH solution.
Answer:
[NaOH] = 0.107M
Explanation:
25.21ml of NaOH(aq) soln + 0.550g KHPh => NaKPh + H₂O
The reaction is 1 to ratio 1 NaOH to KHPh => moles NaOH neutralized = moles KHPh used. (KHPh => Potassium Hydrogen Phthalate f.wt. = 204.22 g/mol)
moles KHPh = 0.550g / 204.22g·mol⁻¹ = 0.0027 mole
moles NaOH neutralized = moles KHPh used = 0.0027 mole NaOH in 25.21ml aqueous solution ...
Molar concentration of NaOH solution = moles NaOH/Liters solution = 0.0027mol NaOH/0.0251L = 0.107M NaOH solution.
A technician plates a faucet with 2.68 g of Cr metal by electrolysis of aqueous Cr2(SO4)3. If 15.2 min is allowed for the plating, what current is needed? Use 96500 C/mol e- for the Faraday constant. Enter a number to 2 decimal places.
Answer:
16.36A
Explanation:
We'll begin by writing a balanced dissociation equation of aqueous Cr2(SO4)3. This is illustrated below:
Cr2(SO4)3 —> 2Cr^3+ 3(SO4)^2-
From the above, we can see that Cr is trivalent.
Next, let us determine the number of faraday needed to deposit metallic Cr. This is illustrated below:
Cr^3+ 3e- —> Cr
From the above equation, 3 faradays are needed to deposit metallic Cr
1 faraday = 96500C
Therefore, 3 faraday = 3 x 96500C = 289500C.
Molar Mass of Cr = 52g/mol
Now let us determine the quantity of electricity needed for 2.68g of Cr metal
This is shown below:
52g of Cr required 289500C.
Therefore, 2.68g of Cr will require = (2.68 x 289500)/52 = 14920.38C
Now, with this quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process. This is illustrated below:
Q (quantity of electricity) = 14920.38C
t (time) = 15.2mins = 15.2 x 60 = 912secs
I (current) =?
Apply the equation Q = It
Q = It
14920.38 = I x 912
Divide both side by 912
I = 14920.38/912
I = 16.36A
Therefore, a current of 16.36A is needed for the process.
The current needed is 16.36 A. The quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process.
Balanced dissociation equation:[tex]Cr_2(SO_4)_3 ---- > 2Cr^{3+}+ 3SO_4^{2-}[/tex]
The number of faradays needed to deposit metallic Cr. This is illustrated below:
[tex]Cr^{3+}+ 3e^- ---- > Cr[/tex]
Given:
3 faradays are needed to deposit metallic Cr
1 faraday = 96500C
Therefore, 3 faraday = 3 * 96500C = 289500C.
Molar Mass of Cr = 52g/mol
52g of Cr required 289500C.
Therefore, 2.68g of Cr will require = (2.68 * 289500)/52 = 14920.38C
Now, with this quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process. This is illustrated below:
Q (quantity of electricity) = 14920.38C
t (time) = 15.2mins = 15.2 x 60 = 912secs
To find:
I (current) =?
Apply the equation,
Q = It
14920.38 = I * 912
I = 14920.38/912
I = 16.36A
Therefore, a current of 16.36A is needed for the process.
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What is the maximum mass in grams of NH3 that can be produced by the reaction of of 2.5 g N2 with 2.5 g of H2 via the equation below?
N2 (g) + 3 H2 (g) → 2 NH3 (g)
Answer: The mass of [tex]NH_3[/tex] produced is, 3.03 grams.
Explanation : Given,
Mass of [tex]N_2[/tex] = 2.5 g
Mass of [tex]H_2[/tex] = 2.5 g
Molar mass of [tex]N_2[/tex] = 28 g/mol
Molar mass of [tex]H_2[/tex] = 2 g/mol
First we have to calculate the moles of [tex]N_2[/tex] and [tex]H_2[/tex].
[tex]\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}=\frac{2.5g}{28g/mol}=0.089mol[/tex]
and,
[tex]\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}=\frac{2.5g}{2g/mol}=1.25mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]N_2[/tex] react with 3 mole of [tex]H_2[/tex]
So, 0.089 moles of [tex]N_2[/tex] react with [tex]0.089\times 3=0.267[/tex] moles of [tex]H_2[/tex]
From this we conclude that, [tex]H_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]N_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]NH_3[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]N_2[/tex] react to give 2 mole of [tex]NH_3[/tex]
So, 0.089 mole of [tex]N_2[/tex] react to give [tex]0.089\times 2=0.178[/tex] mole of [tex]NH_3[/tex]
Now we have to calculate the mass of [tex]NH_3[/tex]
[tex]\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3[/tex]
Molar mass of [tex]NH_3[/tex] = 17 g/mole
[tex]\text{ Mass of }NH_3=(0.178moles)\times (17g/mole)=3.03g[/tex]
Therefore, the mass of [tex]NH_3[/tex] produced is, 3.03 grams.
What is the molarity of a solution made by dissolving 14.8 g of ammonium hydroxide NH4OH, in enough water to make 250.0 mL of solution
Answer:
Molarity= 1.69M
Explanation:
m= 14.8, Mm= 35, V= 0.25dm3, C= ?
Moles = m/M= C×V
Substitute and Simplify
m/M= C×V
14.8/35= C×0.25
C= 1.69M
Final answer:
The molarity of the ammonium hydroxide solution is 1.684 M.
Explanation:
The molarity of a solution can be calculated using the formula:
Molarity (M) = moles of solute / liters of solution
First, we need to convert the mass of ammonium hydroxide (NH4OH) to moles:
1. Calculate the molar mass of NH4OH:
Molar mass of N = 14.01 g/molMolar mass of H = 1.01 g/molMolar mass of O = 16.00 g/molTotal molar mass of NH4OH = (14.01 x 1) + (1.01 x 4) + 16.00 = 35.05 g/mol
2. Calculate the number of moles:
moles = mass / molar mass = 14.8 g / 35.05 g/mol = 0.421 mol
Next, we need to convert the volume of the solution from milliliters to liters:
250.0 mL = 0.250 L
Finally, we can calculate the molarity:
Molarity (M) = 0.421 mol / 0.250 L = 1.684 M
As an approximation, we can assume that proteins exist either in the native (or physiologically functioning) state or the denatured state. The standard molar enthalpy and entropy of the denaturation of a certain protein are 545 kJ·mol−1 and 1.55 kJ·K−1·mol−1, respectively. Comment on the signs and magnitudes of these quantities.
Answer:
Reaction is spontaneous at high temperature and nonspontaneous at low temperature
Explanation:
Given:
Enthalpy change [tex]\Delta H= 545 \frac{KJ}{mol}[/tex]
Entropy change [tex]\Delta S = 1.55[/tex] [tex]\frac{KJ }{K. mol}[/tex]
From the formula of change in free energy,
[tex]\Delta G = \Delta H - T\Delta S[/tex]
But for spontaneous process the values of quantities are given below
For spontaneous process value of [tex]\Delta G[/tex] is negative
For nonspontaneous process value of [tex]\Delta G[/tex] is positive
Here values of [tex]\Delta H[/tex] and [tex]\Delta S[/tex] are positive, so reaction is spontaneous at
high temperature and nonspontaneous at low temperature