Answer:
The answer is lithium(LI)
Which of the following aqueous solutions are good buffer systems?
0.25 M ammonium nitrate + 0.40 M ammonia
0.23 M nitric acid + 0.20 M sodium nitrate
0.16 M potassium hydroxide + 0.22 M potassium chloride
0.39 M hypochlorous acid + 0.25 M potassium hypochlorite
0.12 M hydrofluoric acid + 0.14 M sodium fluoride
Answer:
0.25 M ammonium nitrate + 0.40 M ammonia
0.39 M hypochlorous acid + 0.25 M potassium hypochlorite
0.12 M hydrofluoric acid + 0.14 M sodium fluoride
Explanation:
A buffer consists of a weak acid and its salt or a weak base and its salt.
NH₃ is a weak base, and HClO and HF are weak acids.
B is wrong. HNO₃ is a strong acid.
C is wrong. KOH is a strong base.
When 26.0 mL of 0.500 M H2SO4 is added to 26.0 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.)
The enthalpy of reaction is -55.8 KJ/mol.
From the equation of the reaction;
2KOH(aq) + H2SO4(aq) ------> K2SO4(aq) + 2H2O(l)
Number of moles of H2SO4 = 26.0/1000 × 0.500 M = 0.013 moles
Number of moles of KOH = 26.0/1000 × 1.00 M = 0.026 moles
2 moles of KOH produces 2 moles of water
Hence 0.0026 moles of KOH produces 0.026 moles of water.
Total volume of solution = 26.0 mL + 26.0 mL = 52 mL
Mass of water = density × volume = 1.00 g/mL × 52 mL = 52 g
Using the formula;
ΔH = mcθ
Mass of solution (m) = 52 g
Specific heat capacity of solution (c) = 4.184 J/g·°C
Temperature difference(θ) = 30.17°C - 23.50°C = 6.67°C
Substituting values;
ΔH = -() 52 g × 4.184 J/g·°C × 6.67°C/ 0.026 moles
ΔH = -(1.45 KJ/0.026 moles)
ΔH = -55.8 KJ/mol
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What material would be the best for a hip implant?
Fiber glass
Titanium
ABS
Steel
Rubber
Answer:
titanium
Explanation:
On another planet, the isotopes of titanium have the given natural abundances. Isotope Abundance Mass (amu) 46Ti 72.000% 45.95263 48Ti 14.300% 47.94795 50Ti 13.700% 49.94479 What is the average atomic mass of titanium on that planet?
The average atomic mass of titanium on the given planet is 46.800 amu.
Explanation:In order to calculate the average atomic mass of titanium on the given planet, we multiply the mass of each isotope by its natural abundance and sum them up.
For 46Ti: 72.000% x 45.95263 amu = 33.105 amuFor 48Ti: 14.300% x 47.94795 amu = 6.839 amuFor 50Ti: 13.700% x 49.94479 amu = 6.856 amuAdd the results together: 33.105 amu + 6.839 amu + 6.856 amu = 46.800 amuTherefore, the average atomic mass of titanium on the planet is 46.800 amu.
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To avoid dehydration during or after a long distance athletic event many athletes suggest sponging down with isopropanol. Why would this procedure help avoid the loss of water?
Final answer:
Sponging down with isopropanol can help lower the body temperature through rapid evaporation, which can assist in preventing dehydration. Nevertheless, athletes should prioritize proper hydration by consuming adequate fluids, particularly sports drinks that contain a balance of water, electrolytes, and sugar, to maintain endurance and performance.
Explanation:
To avoid dehydration during or after a long-distance athletic event, athletes may sponge down with isopropanol because it accelerates the evaporation process when applied to the skin. This rapid evaporation causes a cooling effect which can help to lower the body temperature. However, it is important to note that the best way to prevent dehydration is through proper hydration, which involves drinking enough fluids, such as water or sports drinks, before, during, and after exercise. These fluids should contain the necessary electrolytes like sodium, which is crucial for fluid absorption and replacing what is lost through sweat.
The American College of Sports Medicine recommends that the goal of drinking fluids during exercise is to prevent dehydration without over-hydrating. Replacing lost fluids without overconsumption is key to maintaining optimal endurance and performance. To ensure proper rehydration, it's essential to consume sports drinks that contain the correct proportions of sugar, water, and sodium. Sodium in sports drinks not only enhances fluid absorption but also aids in maintaining blood-glucose levels for muscle fuel and replaces some of the electrolytes lost in sweat.
3-methyl-1-hexene is reacted with HCl. Select ALL statements below that are true for this reaction. a) Reaction is stereoselective b) Reaction involves carbocation intermediate c) A 1,2-shift is expected in reaction mechanism d) Reaction is regioselective
Answer:The options b, c and d are True.
Explanation:
The above reaction is an example of electrophilic addition to alkenes and is a typical reaction known as Markownikoffs addition reaction.
In the above reaction we are using Hydrogen chloride as an acid and as HCl is a strong acid so it will provide acidic protons.
The acidic protons (H⁺) acts as electrophile and the the pi-bond in 3-methyl-1-hexene has sufficient electron density to attack the electrophilic protons. As a result of the attack hydrogen is added on one of the carbon atoms across the pi bond and a generation of carbocation takes place.
Once the carboation is formed then it rearranges into a more stable carbocation if it can and hence a stable carbocation is generated. The chloride anion can attack this stable carbocation leading to a regio-selective product.
The carbocation formed is the reaction intermediate in this case.
The statement (a) is incorrect as the above reaction involves carbocation as intermediate and as carbocation is a planar molecule so there are two faces available for the chloride ion to attack hence a racemic product would be formed resulting from the attack on both the sides. So the re action is not stereoselective as not a specific isomer is formed.
The statement (b) is correct as the intermediate formed in the reaction is a carbocation.
The statement (c) is correct as carbocation is formed and we know that carbocations undergo rearrangements to form more stable carbocations so 1,2 shifts can occur in the reaction. As 1,2 shift is a way through rearrangements can occur in carbocation.
The statement (d) is correct as the reaction would lead to formation of a stable carbocation and hence the chloride ion will only attack the stable carbocation so the reaction would be regioselective in nature.
How many ATP are produced from a fatty acid that is 14 carbons long?
Answer:
92 ATP
Explanation:
Fatty acid oxidation results in the formation of large number of ATP molecules. Three important process of fatty acid are activation of the fatty acid, beta oxidation and entry of acetyl CoA in Krebs cycle.
14 carbon fatty acid is Miristic acid. The complete oxidation of Miristic acid results in the formation of 7 acetyl CoA + 6NADH and [tex]6FADH_2[/tex]
1 Acetyl CoA gives 10 molecules of ATP then 7 acetyl CoA gives 70 molecules of ATP.
1NADH = 2.5 ATP, 6NADH = 15 ATP.
[tex]1FADH_2[/tex] = 1.5 ATP, [tex]6FADH_2[/tex] = 9 ATP.
2 ATP has been consumed in the activation of fatty acid.
Total ATP = 70+15+9-2
=92 ATP.
Thus, the total ATP generated from the oxidation of 14 carbon fatty acid is 92.
For the reaction H2(g) + CO2(g) ⇌ H2O(g) + CO(g) at 700ºC, Kc = 0.534. Calculate the number of moles of H2 that are present at equilibrium if a mixture of 0.300 mole of CO and 0.300 mole of H2O is heated to 700ºC in a 10.0-L container.
Answer:
[tex][H_2]_{eq}=0.0173M[/tex]
Explanation:
Hello,
In this case, for the given reaction, the law of mass action turns out:
[tex]Kc=\frac{[H_2O][CO]}{[H_2][CO_2]}[/tex]
In such a way, the reactants initial concentrations are:
[tex][CO_2]_0=[H_2]_0=\frac{0.300mol}{10.0L} =0.030M[/tex]
And considering the change [tex]x[/tex] due to the equilibrium, the law of mass action takes the following form:
[tex]Kc=\frac{(x)(x)}{(0.030M-x)(0.030M-x)} =\frac{x^2}{0.0009-0.06x+x^2} \\Kc(0.0009-0.06x+x^2)=x^2\\0.0004806-0.03204x+0.534x^2-x^2=0\\0.466x^2+0.03204x-0.0004806=0\\x_1=-0.0814M\\x_2=0.0127M[/tex]
The feasible answer is [tex]x=0.0127M[/tex]
Therefore, the hydrogen moles at equilibrium are:
[tex][H_2]_{eq}=0.030M-0.0127M=0.0173M[/tex]
Best regards.
The number of moles of H2 present at equilibrium is 0.04moles
Data;
Kc = 0.534CO = 0.3molH2O = 0.30molV = 10LT = 700ºCEquilibrium Concentrationlet's find the concentration of the species in the reaction.
The concentration of CO is
[tex][CO] = \frac{0.3}{10} = 0.03M[/tex]
The concentration of H2O is
[tex][H_2O] = \frac{0.3}{10} 0.03M[/tex]
The equation of this reaction is
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
final -x -x 0.03 + x 0.03 + x
The Kc i.e the equilibrium constant of this reaction is
[tex]K_c = \frac{[product]}{[reactant]}\\[/tex]
let's substitute the values and solve.
[tex]K_c = \frac{product}{reactant} \\0.534 = \frac{(0.03+x)(0.03+x)}{x * x} \\x= 0.04M[/tex]
The number of moles of H2 present at equilibrium can be calculated as
[tex]0.04 = \frac{0.04}{1}[/tex]
The number of moles of H2 present at equilibrium is 0.04moles
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Given the recipe: 2 cups flour+ 1 egg+3 oz blueberries 4 muffins. Assuming you have sufficient flour and blueberries, you can make 16 muffins from 3 eggs (T/F)
Answer:
False, we cannot made 16 muffins from 3 eggs.
Explanation:
2 cups flour + 1 egg + 3 oz blueberries → 4 muffins
According top recipe, 2 cups of flour, 1 egg and 3 oz of blueberries are made into batter to give 4 muffins.
Then 16 muffins will made from:
For each muffin we need =[tex]\frac{1}{4} egg[/tex]
Then for 16 muffins we will need:
[tex]\frac{1}{4}\times 14 eggs=4 eggs[/tex]
4 eggs will be required to make 16 muffins
From 1 egg we can made 4 muffins
Then from 3 eggs we can made = 3 × 4 muffins = 12 muffins
12 muffins will made from 3 eggs.
We cannot make 16 muffins from 3 eggs.
Consider the neutralization reaction 2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq) A 0.105 L sample of an unknown HNO3 solution required 46.5 mL of 0.100 M Ba(OH)2 for complete neutralization. What is the concentration of the HNO3 solution?
Answer: The concentration of [tex]HNO_3[/tex] comes out to be 0.088 M.
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]
We are given:
Conversion factor: 1 L = 1000 mL
[tex]n_1=1\\M_1=?M\\V_1=0.105L=105mL\\n_2=2\\M_2=0.100M\\V_2=46.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 105=2\times 0.100\times 46.5\\\\M_1=0.088M[/tex]
Hence, the concentration of [tex]HNO_3[/tex] comes out to be 0.088 M.
The concentration of the HNO3 solution is 0.443 M.
Explanation:To determine the concentration of the HNO3 solution, we can use the equation, concentration = (volume of solution titrated)/(volume of solution required for complete neutralization). In this case, the volume of solution titrated is 0.105 L and the volume of 0.100 M Ba(OH)2 solution required for complete neutralization is 46.5 mL (or 0.0465 L). Plugging these values into the equation, we get: concentration = 0.0465 L / 0.105 L = 0.443 M.
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Calculate the standard potential for the following galvanic cell: Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s) which has the overall balanced equation: Ni(s)+2Ag+(aq)→Ni2+(aq)+2Ag(s) Express your answer to three significant figures and include the appropriate units.
Answer:
1.06 V
Explanation:
The standard reduction potentials are:
Ag^+/Ag E° = 0.7996 V
Ni^2+/Ni E° = -0.257 V
The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are
Ni → Ni^2+ + 2e- E° = 0.257 V
2Ag^+ 2e- → 2Ag E° = 0.7996 V
Ni + 2Ag^+ → Ni^2+ + 2Ag E° = 1.0566 V
To three significant figures, the standard potential for the cell is 1.06 V .
The standard cell potential for the galvanic cell is 1.05 V.
The overall balanced equation of the reaction is;
Ni(s) + 2Ag+(aq) →Ni2+(aq) + 2Ag(s)
Since it is a galvanic cell, Nickel is the anode and silver is the cathode.
We know that;
E°cell = E°cathode - E°anode
E°cathode = -0.25 V
E°anode = 0.80 V
E°cell = 0.80 V - (-0.25 V)
E°cell = 1.05 V
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Consider the following elementary reaction: NO2(g) + F2(g) --> NO2F(g) + F (g) Suppose we let k1 stand for the rate constant of this reaction, and k -1 stand for the rate constant of the reverse reaction. Write an expression that gives the equilibrium concentration of F2 in terms of k1, k -1, and the equilibrium concentrations of NO2, NO2F, and F.
To express the equilibrium concentration of F2, the equation [F2] = (k -1[NO2F][F]) / (k1[NO2]) is used, based on the principle that the forward reaction rate equals the reverse reaction rate at equilibrium.
Explanation:To write an expression for the equilibrium concentration of F2 in terms of the rate constants k1 and k -1, and the equilibrium concentrations of NO2, NO2F, and F, one can use the principle that at equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. For the given reaction NO2(g) + F2(g) → NO2F(g) + F(g), the forward rate is given by k1[NO2][F2] and the reverse rate by k -1[NO2F][F]. Setting these equal to each other gives us:
k1[NO2][F2] = k -1[NO2F][F]
Solving for the equilibrium concentration of F2, we have:
[F2] = (k -1[NO2F][F]) / (k1[NO2])
This expression relates the equilibrium concentration of F2 to the rate constants and the equilibrium concentrations of NO2, NO2F, and F.
The compound AX2 decomposes according to the equation, 2 AX2(g) => 2 AX(g) + X2(g). In one experiment, AX2 was measured at various times, and these data recorded: Time (s) {AX2} mol/L 0 0.0500 2.0 0.0448 6.0 0.0300 8.0 0.0249 10.0 0.0209 20.0 0.0088 What is the average rate from 8.0 to 20.0 seconds in mol / L s? Use 4 decimal places.
Answer:
0.001341 mol/L s is the average rate from 8.0 to 20.0 seconds.
Explanation:
[tex]2 AX_2(g)\rightarrow 2 AX(g) + X_2(g)[/tex]
Average rate of the reaction =[tex]R_a[/tex]
[tex]R_a=-\frac{\Delta [x]}{\Delta T}=-\frac{x_2-x_1}{t_2-t_1}[/tex]
[tex]R_a[/tex] =Average rate of the reaction during the given time interval.
[tex]\Delta [x][/tex] = Change in concentration of reactant with respect to time.
[tex]\Delta T[/tex] = Change in time.
[tex]x_1[/tex]=Concentration of reactant at time[tex]t_1[/tex]
[tex]x_2[/tex]=Concentration of reactant at time[tex]t_2[/tex]
So, at [tex]t_1=8.0 sec[/tex] the concentration of [tex]AX_2[/tex] :
[tex]x_1=0.0249 mol/L[/tex]
And at [tex]t_2=2.0 sec[/tex] the concentration of [tex]AX_2[/tex] :
[tex]x_2=0.0088 mol/L[/tex]
The average rate of the reaction at given interval will be given as:
[tex]R_a=-\frac{x_2-x_1}{t_2-t_1}=-\frac{0.0088 mol/L-0.0249mol/L}{20.0s-8.0 s}=0.001341 mol/L s[/tex]
0.001341 mol/L s is the average rate from 8.0 to 20.0 seconds.
Now let M react, and at time = 0, [M] = 0.102 M, A(545nm) = 0.491; at time = 30 seconds, A(545nm) = 0.404. What is the [M] at 30 seconds? What is the slope of the Absorbance vs. [M] function? What is the (average) rate of the reaction during this time period.
If a new reaction was run leaving [N] the same, but beginning (time = 0) with [M] = 0.204 M, A(545nm) = 0.981. After 30 seconds, A (545 nm) is 0.807. What is the order of the reaction with respect to [M] ?
Answer:
election of Cell conduct of your daily India Cases, itu is'n give you some reason=✓
Explanation:
[tex] \geqslant { \sqrt[ \geqslant \sqrt[ \geqslant \leqslant \sqrt[ \geqslant ]{0.time} \times \frac{ log_{30.0m}(3.71m) }{204.9m} ]{3528t} ]{3} }^{8} \times \frac{4}{3} [/tex]
Find the fugacity coefficient of a gaseous species of mole fraction 0.4 and fugacity 25 psia in a mixture at a total pressure of 50 psia
Answer: The fugacity coefficient of a gaseous species is 1.25
Explanation:
Fugacity coefficient is defined as the ratio of fugacity and the partial pressure of the gas. It is expressed as [tex]\bar{\phi}[/tex]
Mathematically,
[tex]\bar{\phi}_i=\frac{\bar{f_i}}{p_i}[/tex]
Partial pressure of the gas is expressed as:
[tex]p_i=\chi_iP[/tex]
Putting this expression is above equation, we get:
[tex]\bar{\phi}_i=\frac{\bar{f_i}}{\chi_iP}[/tex]
where,
[tex]\bar{\phi}_i[/tex] = fugacity coefficient of the gas
[tex]\bar{f_i}[/tex] = fugacity of the gas = 25 psia
[tex]\chi_i[/tex] = mole fraction of the gas = 0.4
P = total pressure = 50 psia
Putting values in above equation, we get:
[tex]\bar{\phi}_i=\frac{25}{0.4\times 50}\\\\\bar{\phi}_i=1.25[/tex]
Hence, the fugacity coefficient of a gaseous species is 1.25
The lowest pressure attainable using the best available vacuum techniques is about 10−12 N/m2. Part A At such a pressure, how many molecules are there per cm3 at 2 ∘C?
Hey there!:
We are given P = 10⁻²² N/m2 ,
Volume, V = 1 cm3= 10⁻⁶ m³ and
T= 13ºC= 13+273= 286 K
Using gas law, P*V = n R T, where n is number of moles and R=8.314JK⁻¹.
n = PV/(RT) => n = 10-12*10⁻⁶/ (8.314*286) = 4.205*10₋²² moles
Hence number of molecules = number of moles * Avogadro's number
= 4.205*10-22 moles * 6.023*1023 molecules/mole
= 253
Number of molecules = 250 (upto 2 significant figures)
Number of molecules: N = 263.31
Further explanationSome of the laws regarding gas can apply to ideal gas (volume expansion does not occur when the gas is heated),:
Boyle's law at constant T,
[tex] \displaystyle P = \dfrac {1} {V} [/tex]
Charles's law, at constant P,[tex] \displaystyle V = T [/tex]
Avogadro's law, at constant P and T,[tex] \displaystyle V = n [/tex]
So that the three laws can be combined into a single gas equation, the ideal gas equation
In general, the gas equation can be written
[tex] \large {\boxed {\bold {PV = nRT}}} [/tex]
where
P = pressure, atm , N/m²
V = volume, liter
n = number of moles
R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)
T = temperature, Kelvin
n = N / No
n = mole
No = Avogadro number (6.02.10²³)
n = m / m
m = mass
M = relative molecular mass
Known
P = 10−12 N / m2
V = 1 cm3 = 10-6 m3
T = 2 ºC = 2 + 273 = 275 K
R = 8,314 J / mol. K
[tex]\rm n=\dfrac{PV}{RT}\\\\n=\dfrac{10^{-12}\times 10^{-6}}{8.314\times 275}\\\\n=\boxed{\bold{4.374\times 10^{-22}}}[/tex]
then the number of molecules (N):
N = n x No
N = 4,374.10⁻²² x 6.02.10²³
N = 263.31
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Magnesium and nitrogen react in a combination reaction to produce magnesium nitride: 3 Mg N2 → Mg3N2 In a particular experiment, a 8.33-g sample of N2 reacts completely. The mass of Mg consumed is ________ g.
Answer:
The mass of Mg consumed is 21.42g
Explanation:
The reaction is
[tex]3Mg+N_{2}-->Mg_{3}N_{2}[/tex]
As per balanced equation, three moles of Mg will react with one mole of nitrogen to give one mole of magnesium nitride.
as given that mass of nitrogen reacted = 8.33g
So moles of nitrogen reacted = [tex]\frac{mass}{molarmass}=\frac{8.33}{28}=0.2975mol[/tex]
moles of Mg required = 3 X moles of nitrogen taken = 3X0.2975 = 0.8925mol
Mass of Mg required = moles X molar mass = 0.8925 X 24 = 21.42 g
The stoichiometric reaction between magnesium and nitrogen to produce magnesium nitride requires 3 moles of magnesium for each mole of nitrogen. Therefore, for an 8.33g sample of nitrogen, approximately 21.66g of magnesium will be consumed.
Explanation:The subject question involves a stoichiometric calculation regarding a reaction between magnesium (Mg) and nitrogen (N2) to produce magnesium nitride (Mg3N2). The balanced equation for the reaction is: 3 Mg + N2 → Mg3N2. This ratio tells us that for every 1 mole of N2, 3 moles of Mg are needed.
The molar mass of N2 is 28.014 g/mol. In an 8.33-g sample, there are (8.33 g)/(28.014 g/mol) = 0.297 mol of N2.
As per the reaction's stoichiometry, three times this amount in moles of Mg are needed, so 0.297 mol x 3 = 0.891 mol of Mg are required.
The molar mass of Mg is 24.305 g/mol. Thus, the mass of Mg consumed is (0.891 mol) x (24.305 g/mol) = 21.66 g.
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The feed to an ammonia synthesis reactor contains 25 mole% nitrogen and the balance hydrogen. The fow rate of the stream is 3000 kg/h. Calculate the rate of flow of nitrogen into the reactor in kg/h. (Suggestion: First calculate the average molecular weight of the mixture.)
Answer:
The rate of flow of nitrogen into the reactor is 2,470.588 kg/h.
Explanation:
Mole percentage of nitrogen gas = 25 mole%
Mole percentage of hydrogen gas = 75 mole%
Average molecular weight of the mixture:
[tex](0.25\times 28 g/mol)+(0.75\times 2 g/mol)=8.5 g/mol[/tex]
Rate of flow of the stream = 3000 kg/h
Mass of stream in 1 hour = 3000 kg = 3,000,000 g
Moles of stream :
[tex]\frac{3000 g}{8.5 g/mol}=352,941.17 mol[/tex]
Moles of nitrogen gas in 352,941.17 moles of stream be x
[tex]25\%=\frac{\text{Moles of nitrogen gas}}{\text{Moles of stream}}\times 100[/tex]
[tex]25=\frac{x}{352,941.17mol}\times 100[/tex]
x = 88,235.294 mol
Mass of 8,823,529.4 mole of nitrogen gas:
[tex]88235.294 mol\times 28 g/mol=2.470588.2 g=2,470.588 kg[/tex]
The rate of flow of nitrogen into the reactor is 2,470.588 kg/h.
To calculate the rate of flow of nitrogen into the reactor, you need to calculate the average molecular weight of the mixture and use the ideal gas law.
Explanation:To calculate the rate of flow of nitrogen into the reactor, we first need to calculate the average molecular weight of the mixture. The average molecular weight (MW) is calculated by summing the product of the mole fraction of each component and its molecular weight:
MW = (mole fraction of N2 * molecular weight of N2) + (mole fraction of H2 * molecular weight of H2)
Now, we can calculate the rate of flow of nitrogen using the ideal gas law:
Rate of flow of nitrogen = (mole fraction of N2 * flow rate of the stream * MW of N2) / molecular weight of N2
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which type of heat exchanger shell and tube or plate heat exchanger is suitable for evaporation?
Answer:
shell and tube type heat exchanger
Explanation:
for evaporation the shell and tube type heat exchanger is best suited.
in the plate heat exchanger there is gaskets in between every part so this part become weak part in heat echanger and there is possibilities of leakage through this part, there is no such problem in shell and tube type.the plate type cant be used when there is high temperature and high pressure drop but shell and tube type can be usedin evaporation there the liquids change into vapors due to which there is sudden change in pressure and in which plate type is not used because there is chances of leakageThe suitable type of heat exchanger for evaporation is the shell and tube heat exchanger.
Evaporation processes typically involve the boiling of a liquid to produce vapor, which requires a heat exchanger that can handle phase change and the associated volume changes. Shell and tube heat exchangers are well-suited for this application due to their ability to withstand high pressures and temperatures, as well as their design flexibility to accommodate the two-phase flow of vapor and liquid.
In a shell and tube heat exchanger, the tubes can be designed to allow for the expansion of vapor and the efficient separation of the vapor phase from the liquid phase. The tubes also provide a large surface area for heat transfer, which is necessary for the evaporation process. Additionally, shell and tube heat exchangers can be designed with multiple passes to increase the time of contact between the heating or cooling medium and the product, thereby improving the efficiency of the evaporation process.
On the other hand, plate heat exchangers are generally not suitable for evaporation processes that involve boiling and phase change. This is because the plates in a plate heat exchanger are not designed to handle the high pressure differentials and volume changes associated with the production of vapor. Plate heat exchangers are more suitable for applications involving low to medium pressure and temperature, and where the fluids remain in a single phase.
In summary, for processes involving evaporation, the robust design and flexibility of the shell and tube heat exchanger make it the more suitable choice compared to the plate heat exchanger."
Consider the reaction of 2-methyl-1,3-cyclohexadiene with HCl. Assuming that the reaction takes place at a very low temperature: (a) draw the intermediate and product structures, including any formal charges
Answer:.The product formed in this reaction would be a kinetic product formed from 1,2-addition.
Kindly refer attachments for mechanism and structures of product.
Explanation:
In organic chemistry once the intermediate is generated the reaction can go in two ways. In one way product formed would be rate dependent and would be known as kinetic product and in the other way product formed would be stable and would be know as thermodynamic product.
A kinetic product is formed faster but it is generally not that stable as a thermodynamic product so a kinetic product is formed at lower temperatures where the molecular energy is very less.
A thermodynamic product formation takes time to form and the reaction is carried out at higher temperatures where the molecules have energy. The thermodynamic product is relatively stable.
In this case since we are doing our reaction at very low temperature so the major product formed in the reaction would be under kinetic control and hence product formed would be rate dependent.
The reaction of 2-methyl-1,3-cyclohexadiene with HCl would be a electrophilic addition reaction in which the pi bond would attack HCl to generate a carbocation intermediate. After the formation of carbocation chloride anion can attack the carbocation and can form the product.
Once the carbocation is formed it can be stabilised by rearrangement or other stabilizing mechanisms.
In this case initially 1-methylcyclohex-2-en-1-ylium carbocation is generated which is at tertiary center as well as allylic position. This carbocation formed initially can stabilize itself through resonance as the charge can be delocalised with the allyl group.
The reaction can happen in two ways :
In the first way the initially formed 1-methylcyclohex-2-en-1-ylium carbocation is been attacked by the chloride anion and this leads to the 1,2 addition product.
In the other way the initially formed 1-methylcyclohex-2-en-1-ylium carbocation delocalizes its positive charge with the allyl group as it is in conjugation with the allyl group thereby generating a positive charge at 4 postion and form 3-methylcyclohex-2-en-1-ylium. Now this carbocation is attacked by the chloride anion and this leads to the 1,4 addition product.
The 1,2 addition product is a Kinetic product as it can quickly lead to products wheras the 1,4 addition product is a thermodynamic product.
The product formed in this reaction would be a kinetic product formed form 1,2-addition.
Kindly refer the attachments for reaction mechanism.
1) For this reaction,
SiCl4 (l)+ 2H2O(g) <-> SiO2 (s) + 4HCl (g) DeltaH = -127 KJ
Predict the effect increase (i), decrease(d) or no change(nc) that each change has on the QUANTITY for the equilibrium mixture. Assume that each change is applied separately to the system.
Change Quantity Effect
increase in pressure amount of H2O
i, d or nc
decrease temp amount of SiCl4 i d or nc
increase temp Kc i d or nc
increase temp amount of H2O i d or nc
add H2O amount of SiCl4 i d or nc
add SiO2 amount of HCl i d or nc
add HCl amount of SiO2 i d or nc
Answer:
a) Increase
b) the amount of SiCl₄ will decrease.
c) K will increases.
d) H₂O will increase.
e) SiCl₄ will decrease
f) HCl will decrease.
g) SiO₂ will decrease
Explanation:
The equilibrium changes are explained by Le-Chatelier's Principle.
If we apply a change to an equilibrium system then it shifts to the side where the effect of stress can be released.
a) Change : Increase in pressure
Quantity : amount of water
Change: We are increasing the pressure it will decrease the volume and hence will increase the moles per unit volume. so the system will move in the direction where the number of gaseous moles are less. It will move towards reactant side. thus the amount of water will increase.
b) Change : decrease in temperature
Quantity : amount of SiCl₄
As this is an exothermic reaction, the decrease in temperature will shift the reaction towards product side and hence the amount of SiCl₄ will decrease.
c)
Change : Increase in Temperature
Quantity: Kc
The equilibrium constant will increase as it increases with temperature.
d) Change : increase in temperature
Quantity : amount of H₂O
As this is an exothermic reaction, the increase in temperature will shift the reaction towards reactant side and hence the amount of H₂O will increase.
e) Change : Add H₂O
Quantity: Amount of SiCl₄
As we are adding reactant to the equilibrium, the equilibrium will shift towards product side thus the amount of SiCl₄ will decrease
f) Change : Add SiO₂
Quantity: Amount of HCl
As we are adding product, the equilibrium will shift in reactant side and thus there will be decrease in amount of HCl.
g) Change : Add HCl
Quantity: Amount of SiO₂
As we are adding product to the equilibrium the equilibrium will shift in reactant side and thus there will be decrease in amount of SiO₂
Give the structure of the expected organic product in the reaction of 3−phenylpropanal with sodium hydroxide in ethanol at 70°C. If your structure contains an aldehyde do NOT use the condensed formula (−CHO), draw it out.
Answer:The expected product formed is 2-benzyl-5-phenylpent-2-enal .
Kindly refer the attachment for structure.
Explanation:
3-phenylpropanal on reaction with sodium hydroxide undergoes an self- aldol condensation reaction and leads to formation of 2-benzyl-5-phenylpent-2-enal as final product.
In the first step of the reaction the highly basic hydroxide anions abstract a acidic hydrogen available at the carbon next to carbonyl carbon.These hydrogens are acidic because of the electron withdrawing effect of carbonyl group.
The proton abstraction leads to the generation of a carbanion and it further delocalizes forming an enolate anion.
The carbanion can behave as a nucleophile and can attack at the electrophilic carbon centers.
The carbonyl carbon is electrophilic in nature due to the electron withdrawl from oxygen which generates a partial positive charge on the carbonyl carbon.
The carbanion further reacts with another molecule of 3-phenylpropanal at its electrophilic carbonyl center and forms 2-benzyl-3-hydroxy-5-phenylpentanal. This reaction is known as self- aldol condensation reaction
2-benzyl-3-hydroxy-5-phenylpentanal has a OH group and this OH group is protonated through the available acidic protons from the solvent .
As OH is protonated it easily leaves leading to the formation of C=C double bond.
This reaction is one of the examples of Carbon-Carbon bond forming reaction.
Kindly refer attachments for reaction, structure and mechanism.
Which of the following is the correct balanced equation for the neutralization of barium hydroxide with hydrochloric acid? Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O BaOH2 + 2 HCl → BaCl2 + H2O BaOH + HCl → BaCl + H2O Ba(OH)2 + HCl → BaCl2 + H2O
Answer: The correct answer is [tex]Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O[/tex]
Explanation:
Neutralization reaction is defined as the reaction in which an acid reacts with a base to produce a salt and also releases water molecule.
The general equation for this reaction follows:
[tex]BOH+HX\rightarrow BX+H_2O[/tex]
The balanced chemical equation for the reaction of barium hydroxide and hydrochloric acid follows:
[tex]Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of barium hydroxide reacts with 2 moles of hydrochloric acid to produce 1 mole of barium chloride and 2 moles of water molecule.
The balanced chemical equation for the neutralization of barium hydroxide with hydrochloric acid is Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O, leading to the formation of water and a salt, BaCl2.
Explanation:The correct balanced equation for the neutralization of barium hydroxide with hydrochloric acid is Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O. In this equation, Ba(OH)2 is the base and HCl is the acid. A neutralization reaction between an acid and a base results in the formation of water and salt. In this case, the salt is BaCl2. Each reactant and product in the equation is balanced, with the same number of each type of atom appearing on both sides of the equation, hence it's a balanced chemical equation.
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How many grams of Br are required to react completely with 29.5 g of NO? 2NO+Br2- 2NOBr
Answer: 78.54 grams
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of molesof} NO=\frac{29.5g}{30.01g/mol}=0.98moles[/tex]
[tex]2NO+Br_2\rightarrow 2NOBr[/tex]
According to stoichiometry :
2 moles of [tex]NO[/tex] react with 1 mole of [tex]Br_2[/tex]
0.98 moles of [tex]NO[/tex] react with =[tex]\frac{1}{2}\times 0.98=0.49[/tex] moles of [tex]Br_2[/tex]
Mass of [tex]Br_2=moles\times {\text{Molar mass}=0.49\times 159.8=78.54g[/tex]
Thus 78.54 g of [tex]Br_2[/tex] are required to react completely with 29.5 g of NO.
Select the false statement below : Fe2+ is more paramagnetic (has more unpaired electrons) than Fe3+ Fe3+ is isoelectronic with Mn2+ Fe has 2 outer electrons, 8 valence electrons, and 18 core electrons Fe3+ is predicted to be a stronger potential oxidizing agent (can be reduced more) than Fe2+ none of these ions (Fe2+, Fe3+, Mn2+) have any electrons in the 4s subshell
hey there!:
1) none of these ions ( Fe²⁺ , Fe³⁺ , Mn²⁺ 0 have any electrons in the 4s subshell . ( TRUE )
2) Fe²⁺ is more paramagnetic ( has more unpaired electrons ) than Fe³⁺.
( FALSE ) explanation :
* Fe²⁺ has no free electrons and is diamagnetic
* Fe³⁺ has one unpaired electron and is paramagnetic
3) Fe³⁺ is isoeletronic with Mn²⁺ . ( TRUE )
4) Fe has 2 outer electrons , 8 valence electrons , and 18 core eletrons. (TRUE )
electron configuration Fe = 1s², 2s², 2p⁶, 3s² , 3p⁶, 4s² , 3d⁶ or :
[Ar] = 4s² , 3d⁶
5) Fe³⁺ is predicted to be a stronger potencial oxidizing agent ( can be reduced more ) than Fe²⁺ ( TRUE )
Hope this helps!
The false statement is that; "Fe2+ is more paramagnetic (has more unpaired electrons) than Fe3+."
Iron is a transition element that has 26 valence electrons in the ground state. The ground state electron configuration of iron is [Ar] 3d6 4s2. The 4s electrons are the outer electrons hence Fe2+ is a d6 specie while Fe3+ is a d5 specie.
This means that Fe3+ has five unpaired electrons while Fe2+ has only four unpaired electrons. Hence, the false statement is that; "Fe2+ is more paramagnetic (has more unpaired electrons) than Fe3+."
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For all gas law calculations, the temperature must be in kelvins. (T/F)
Answer:
True
Explanation:
For, all the gas law calculations, Temperature must be used in Kelvins. This statement is true.
The reason is:
Ideal gas equation is made up of gas laws like P/T (Amonton's Law), V/T (Charle's Law) and combined gas laws like PV/T ( Boyle's Laws, Amonton's Law and Charle's Law).
Charle's Law and Amonton's Law are only valid when temperature values are put in Kelvin and only then V/T or P/T will be constant.
Also, in all the cases, Temperature occurs in the denominator. If we measure temperature values in Celcius, then it will lead to wrong calculations. Also, if we put [tex]0^0C[/tex] in the equation, then the equation will have zero in the denominator which will solve as no solution.
But, If we will put 0K in the equation, then it will achieve absolute state where all the things stop and there will be zero entropy (Third law of thermodynamics). In that case, we dont have to think about any of the parameters to be calculated.
Calculate the vapor pressure of a solution containing 24.6 g of glycerin (C3H8O3) in 134 mL of water at 30.0 ∘C. The vapor pressure of pure water at this temperature is 31.8 torr. Assume that glycerin is not volatile and dissolves molecularly (i.e., it is not ionic) and use a density of 1.00 g/mL for the water.
Answer : The vapor pressure of a solution is, 30.687 torr
Explanation :
First we have to calculate the moles of glycerin.
[tex]\text{Moles of }C_3H_8O_3=\frac{\text{Mass of }C_3H_8O_3}{\text{Molar mass of }C_3H_8O_3}=\frac{24.6g}{92.09g/mole}=0.267moles[/tex]
Now we have to calculate the mass of water.
[tex]\text{Mass of }H_2O=\text{Density of }H_2O\times \text{Volume of }H_2O[/tex]
[tex]\text{Mass of }H_2O=(1.00g/ml)\times (134ml)=134g[/tex]
Now we have to calculate the moles of water.
[tex]\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{134g}{18g/mole}=7.444moles[/tex]
Now we have to calculate the mole fraction of water.
[tex]\text{Mole fraction of water}=\frac{\text{Moles of water}}{\text{Moles of water}+\text{Moles of glycerin}}[/tex]
[tex]\text{Mole fraction of water}=\frac{7.444mole}{7.444mole+0.267mole}=0.965[/tex]
Now we have to calculate the vapor pressure of the solution.
According to the Raoult's law,
[tex]p_A=X_A\times p^o_A[/tex]
where,
[tex]p_A[/tex] = vapor pressure of solution = ?
[tex]p^o_A[/tex] = vapor pressure of pure water= 31.8 torr
[tex]X_A[/tex] = mole fraction of water = 0.965
Now put all the given values in this formula, we get the vapor pressure of solution.
[tex]p_A=0.965\times 31.8\text{ torr}[/tex]
[tex]p_A=30.687\text{ torr}[/tex]
Therefore, the vapor pressure of a solution is, 30.687 torr
Considering the Raoult's Law, the vapor pressure of a solution that containing 24.6 g of glycerin in 134 mL of water at 30.0 ∘C is 30.687 torr.
Raoult's LawRaoult's Law is a gas law that relates the vapor pressure and mole fraction of each gas in a solution (solution).
If a solute has a measurable vapor pressure, the vapor pressure of its solution is always less than that of the pure solvent. Thus, the relationship between the vapor pressure of the solution and the vapor pressure of the solvent depends on the concentration of the solute in the solution.
Raoult's law allows us to calculate the vapor pressure of a substance when it is part of an ideal solution, knowing its vapor pressure when it is pure (at the same temperature) and the composition of the ideal solution in terms of molar fraction.
Then, this law establishes as a conclusion that: “In an ideal solution, the partial pressures of each component in the vapor are directly proportional to their respective mole fractions in the solution”.
Mathematically, Raoult's law states that the partial pressure of a solvent above a solution P₁ is given by the vapor pressure of the pure solvent P₁⁰ multiplied by the mole fraction of the solvent in the solution X₁:
P₁= P₁⁰ × X₁
Vapor pressure in this caseThe mole fraction is defined as the ratio of moles of solute to total moles of solution.
So, the moles of glycerin (solute) is calculated as follow:
[tex]moles of glycerin=\frac{mass of glycerin}{molar mass of glycerin}=\frac{24.6 g}{92.09 \frac{g}{mole} } =[/tex] 0.267 moles
Being the density the measure of the amount of mass in a certain volume of a substance, the mass of water (solvent) in the solution is calculated as:
mass of water= volume of water× density of water
mass of water= 134 mL× 1 [tex]\frac{g}{mL}[/tex]= 134 g
So, the moles of water (solvent) is calculated as follow:
[tex]moles of water=\frac{mass of water}{molar mass of water}=\frac{134 g}{18 \frac{g}{mole} } =[/tex] 7.44 moles
Then, the mole fraction of water can be calculated as:
[tex]Mole fraction of water=\frac{moles of water}{moles of glycerin + moles of water}[/tex]
[tex]Mole fraction of water=\frac{7.44 moles}{0.267 moles + 7.44 moles}[/tex]
Solving:
mole fraction of water= 0.965
Then, you know that:
[tex]P_{water} =[/tex] ?[tex]P_{water} ^{0}=[/tex] 31.8 torr[tex]x_{water}=[/tex] 0.965Replacing in Raoult's Law:
[tex]P_{water} =[/tex] 31.8 torr × 0965
Solving:
[tex]P_{water} =[/tex] 30.687 torr
In summary, the vapor pressure of a solution that containing 24.6 g of glycerin in 134 mL of water at 30.0 ∘C is 30.687 torr.
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Question 5 A solution is prepared at 25°C that is initially 0.35M in chlorous acid HClO2, a weak acid with =Ka×1.110−2, and 0.29M in sodium chlorite NaClO2. Calculate the pH of the solution. Round your answer to 2 decimal places.
Answer : The pH of the solution is, 1.88
Explanation : Given,
[tex]K_a=1.1\times 10^{-2}[/tex]
Concentration of [tex]HClO_2[/tex] = 0.35 M
Concentration of [tex]NaClO_2[/tex] = 0.29 M
First we have to calculate the value of [tex]pK_a[/tex].
The expression used for the calculation of [tex]pK_a[/tex] is,
[tex]pK_a=-\log [K_a][/tex]
Now put the value of [tex]K_a[/tex] in this expression, we get:
[tex]pK_a=-\log (1.1\times 1-^{-2})[/tex]
[tex]pK_a=2-\log (1.1)[/tex]
[tex]pK_a=1.96[/tex]
Now we have to calculate the pH of the solution.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[NaClO_2]}{[HClO_2]}[/tex]
Now put all the given values in this expression, we get:
[tex]pH=1.96+\log (\frac{0.29}{0.35})[/tex]
[tex]pH=1.88[/tex]
Therefore, the pH of the solution is, 1.88
The pH of a buffer solution consisting of a weak acid and its conjugate base can be calculated using the Henderson–Hasselbalch equation, with the concentrations of the weak acid and base, and the Ka or pKa of the weak acid.
Explanation:In this scenario, we are dealing with a weak acid, chlorous acid (HClO2), and the salt of its conjugate base, sodium chlorite (NaClO2). When a weak acid is in solution with the salt of its conjugate base, a buffer solution is formed. Buffer solutions resist significant changes in pH upon the addition of small quantities of acid or base. The pH of the buffer solution can be calculated using the Henderson–Hasselbalch equation:
pH = pKa + log([A-]/[HA]).
Where,
pKa is the negative log of the acid dissociation constant Ka,[A-] is the concentration of the base form, here it is sodium chlorite NaClO2,[HA] is the concentration of the weak acid, here it is chlorous acid HClO2.Using the given concentrations and Ka or pKa, substitute them into the formula to calculate the pH of the solution.
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The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from the high-altitude jet plane, the SST. The reaction is
O3 + NO ---> O2 + NO2
If 0.827 g of O3 reacts with 0.635 g of NO, how many grams of NO2 will be produced? g NO2 Which compound is the limiting reagent? ozone (O3) nitric oxide (NO) Calculate the number of moles of the excess reagent remaining at the end of the reaction.
Answer: Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.
Explanation:
To calculate the number of moles, we use the equation
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
For ozone:Given mass of ozone = 0.827 g
Molar mass of ozone = 48 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol[/tex]
For nitric oxide:Given mass of nitric oxide = 0.635 g
Molar mass of nitric oxide = 30.01 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol[/tex]
For the given chemical equation:
[tex]O_3+NO\rightarrow O_2+NO_2[/tex]
By Stoichiometry of the reaction:
1 mole of ozone reacts with 1 mole of nitric oxide.
So, 0.0172 moles of ozone will react with = [tex]\frac{1}{1}\times 0.0172=0.0172moles[/tex] of nitric oxide
As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.
Thus, ozone is considered as a limiting reagent because it limits the formation of product.
Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 molesBy Stoichiometry of the reaction:
1 mole of ozone produces 1 mole of nitrogen dioxide.
So, 0.0172 moles of ozone will react with = [tex]\frac{1}{1}\times 0.0172=0.0172moles[/tex] of nitrogen dioxide
Now, calculating the mass of nitrogen dioxide from equation 1, we get:
Molar mass of nitrogen dioxide = 46 g/mol
Moles of nitrogen dioxide = 0.0172 moles
Putting values in equation 1, we get:
[tex]0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g[/tex]
Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.
Final answer:
To find the mass of NO₂ produced in the reaction between O₃ and NO, and to identify the limiting reagent, one must convert the mass of each reactant to moles, compare these to the reaction stoichiometry, and perform subsequent calculations.
Explanation:
The concern over the depletion of ozone (O3) in the stratosphere due to reactions with nitric oxide (NO) from high-altitude jet planes is a crucial environmental issue. The reaction of interest is O3 + NO → O2 + NO2. To calculate the amount of NO2 produced from 0.827 grams of O₃ and 0.635 grams of NO, we must first determine the limiting reagent, which will dictate the maximum amount of product formed. This involves converting the masses of the reactants to moles, using their molar masses (O₃ = 48.00 g/mol and NO = 30.01 g/mol), and comparing the mole ratios to the stoichiometry of the reaction.
Calculations show that NO is the limiting reagent. From the stoichiometry of the reaction, it's clear that 1 mole of NO produces 1 mole of NO2. Thus, the mass of NO₂ produced can be directly calculated from the moles of NO reacted, taking into account the molar mass of NO2 (46.01 g/mol). The remaining amount of the excess reagent, O3, can also be determined by subtracting the moles of O₃ that reacted (equal to the moles of NO reacted) from the initial moles of O₃.
An exothermic reaction has a positive enthalpy (heat) of reaction.(T/F)
Answer:
True.
Explanation:
An exothermic reaction has a positive enthalpy (heat) of reaction. However, it can be negative in some circumstances.