Answer: The value of equilibrium constant for reverse reaction is [tex](\frac{1}{95})^2[/tex]
Explanation:
The given chemical equation follows:
[tex]N_2O_5(g)\rightarrow 2NO_2(g)+\frac{1}{2}O_2(g)[/tex]
The equilibrium constant for the above equation is 95.
We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:
[tex]O_2(g)+4NO_2(g)\rightarrow 2N_2O_5(g)[/tex]
The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.
If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant of initial reaction.
The value of equilibrium constant for reverse reaction is:
[tex]K_{eq}'=(\frac{1}{95})^2[/tex]
Hence, the value of equilibrium constant for reverse reaction is [tex](\frac{1}{95})^2[/tex]
The equilibrium constant for the reaction O2(g) + 4 NO2(g) --> 2 N2O5(g) at 25ºC is (1/95)^2. This is due to Le Chatelier's principle stating that K for the reverse reaction is the reciprocal of the original one and it's squared when the reaction is doubled.
Explanation:The second reaction in your question is the reverse reaction of the first one, but also multiplied by two. According to Le Chatelier's principle, the equilibrium constant (K) of the reverse reaction is the reciprocal of the original one. So the K of the reverse of 1: N2O5(g) --> 2 NO2(g) + ½ O2(g) would be 1/95. However, the actual reaction to the question is twice that.
When you double a reaction, you square its equilibrium constant. Thus, the K for the reaction: O2(g) + 4 NO2(g) --> 2 N2O5(g) would be (1/95)^2.
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How much heat is absorbed/released when 25.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(l) according to the following chemical equation? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) ΔH° = 1168 kJ
428.7 kJ of heat are absorbed when 25.00 g of NH₃(g) reacts in the presence of excess O₂(g) to produce NO(g) and H₂O(l).
Let's consider the following thermochemical equation.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(l) ΔH° = 1168 kJ
The standard enthalpy of the reaction is positive, which means that the reaction is endothermic, that is, heat is absorbed.
We will convert 25.00 g of NH₃ to moles using its molar mass (17.03 g/mol).
[tex]25.00 g \times \frac{1mol}{17.03g} = 1.468 mol[/tex]
1168 kJ of heat are absorbed when 4 moles of NH₃ react. The heat absorbed when 1.468 moles of NH₃ react is:
[tex]1.468 mol \times \frac{1168kJ}{4mol} = 428.7 kJ[/tex]
428.7 kJ of heat are absorbed when 25.00 g of NH₃(g) reacts in the presence of excess O₂(g) to produce NO(g) and H₂O(l).
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Chemical reactions are defined as the reaction in which two or more reactants combine to form single or more products. The heat can be absorbed or evolved in the reaction. The heat absorbed is 428.7kJ when ammonia reacts with excess oxygen.
Given that,
Mass of ammonia = 25 gMolar mass of ammonia = 17.03 g/molThe chemical equation between ammonia and oxygen is:
[tex]\text {4 NH}_3 \;+\; \text {5 O}_2\;\rightarrow\; \text {4 NO + 6 H}_2\text {O}\;\;\;\;\;\Delta\text H&=1168 \text {kJ}[/tex]Now, the enthalpy of the reaction is positive, such that the reaction is endothermic. In endothermic reaction heat is absorbed.
Now, converting the mass of ammonia into moles, we get:
[tex]25.00\text {g} \;\times\;\dfrac{1 \text{mol}}{17.03\text{g}} &= 1.468\; \text {mol}[/tex]1168 kJ of heat is absorbed when 4 moles of ammonia react with oxygen. The heat absorbed in 1.468 moles, will be:
[tex]1.468\;\text{mol}\;\times\;\dfrac{1168\;\text{kJ}}{4\;\text{mol}}&=428.7\;\text{kJ}[/tex]Therefore, when 25 grams of ammonia reacts in the presence of oxygen 428.7 kJ will be absorbed. Hence, it is an endothermic reaction.
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Almost all amino acids are chiral. The body only uses L amino acids, and proteins are chains of amino acids that in the most simple sense fold up into a single structure by themselves when they are formed. Do you think a cell could live with all D (the opposite of L) amino acids, explain your logic. When I ask 'live', I mean could the organism/cell be composed completely of D amino acids and live (This is just a thought question, don't bother trying to find the answer anywhere…)
Answer:
An organism that is completely composed D amino acids cannot survive.
Explanation:
Most of amino acids in all organism are present in L conformation.As result in all organism the enzymes are specific for L amino acids but not for D amino acids.
Bacterial cell wall contain some D amino acids such as D glutamate,D alanine but not entirely composed of D amino acids.
A solution of the primary standard potassium hydrogen phthalate, (204.22 g/mol), was prepared by dissolving 0.4877 g of in about 50 mL of water. The solution was titrated with an solution and mL were needed to reach the phenolphthalein end point. What is the molarity of the solution?
Answer: The molarity of KOH solution is 0.0663 M.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of KHP = 0.4877 g
Molar mass of KHP = 204.22 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of KHP}=\frac{0.4877g}{204.22g/mol}=0.0024mol[/tex]
The chemical reaction for the formation of chromium oxide follows the equation:
[tex]KHC_8H_4O_4(aq.)+KOH\rightarrow K_2C_8H_4O_4(aq.)+H_2O(l)[/tex]
By Stoichiometry of the reaction:
1 mole of KHP reacts with 1 mole of KOH.
So, 0.0024 moles of KHP will react with = [tex]\frac{1}{1}\times 0.0024=0.0024mol[/tex] of KOH.
To calculate the molarity of KOH, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
Moles of KOH = 0.0024 moles
Volume of solution = 36.21 mL = 0.03621 L (Assuming) (Conversion factor: 1L = 1000 mL)
Putting values in above equation, we get:
[tex]\text{Molarity of KOH }=\frac{0.0024mol}{0.03621L}=0.0663M[/tex]
Hence, the molarity of KOH solution is 0.0663 M.
Arizona was the site of a 400,000-acre wildfire in June 2002. How much carbon dioxide (CO2) was produced into the atmosphere by that fire? [Hints: Assume that the density of carbon on the acreage was 10 kg/m2 and that 50% of the biomass burned. In addition, 10,000 m2 = 2.47 acre].
Answer:
2.97 × 10¹³ g
Explanation:
First, we have to calculate the biomass the is burned. We can establish the following relations:
2.47 acre = 10,000 m² 10 kg of C occupy an area of 1 m²50% of the biomass is burnedThe biomass burned in the site of 400,000 acre is:
[tex]400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC[/tex]
Let's consider the combustion of carbon.
C(s) + O₂(g) ⇒ CO₂(g)
We can establish the following relations:
The molar mass of C is 12.01 g/mol1 mole of C produces 1 mole of CO₂The molar mass of CO₂ is 44.01 g/molThe mass of produced is CO₂:
[tex]8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}[/tex]
For fatty acids with the same number of carbon atoms, how does the melting point change as the number of double bonds in the fatty acid changes?
a. The melting point of the fatty acid decreases as the number of double bonds increases.
b. There is no relationship between the melting point of a fatty acid and the number of double bonds.
c. The melting point of the fatty acid is unchanged as the number of double bonds changes.
d. The melting point of the fatty acid decreases as the number of double bonds decreases.
The answer is a. "The melting point of the fatty acid decreases as the number of double bonds increases" because when there are double bonded carbons it cancels the polarity of the carboxyl in the opposite direction, increasing that effect as double bonded carbons add to the molecule, giving the net dipole moment to the molecule. As the net dipole moment of the molecule decrease, the melting point of the fatty acid also decreases.
In addition, when the fatty acid is major than 8 carbons and have a double bond that prevents the fatty acid from crystal lattice formation, so the melting point will be lower also for this reason in this case.
Final answer:
The melting point of a fatty acid decreases as the number of double bonds increases because double bonds create kinks in the fatty acid chains, preventing tight packing and weakening intermolecular forces. The correct answer is option a.
Explanation:
For fatty acids with the same number of carbon atoms, the melting point changes as the number of double bonds in the fatty acid changes. The correct option is (a) - the melting point of the fatty acid decreases as the number of double bonds increases. This decrease in melting point occurs because unsaturated fatty acids, which have one or more double bonds, tend to have kinks or bends at the location of these bonds.
These kinks inhibit the fatty acids from packing closely together, resulting in weaker intermolecular Van der Waals forces (specifically London dispersion forces) and thus a lower melting point. In contrast, saturated fatty acids without double bonds have straight chains and can pack tightly together, leading to stronger intermolecular forces and higher melting points.
A good illustration of this concept is the substantial difference in melting points between palmitoleic acid, which contains a cis-double bond and melts over 60℃ lower than its saturated counterpart, palmitic acid. When considering the physical properties of fatty acids, it's important to note that more saturated fatty acids are typically more solid at room temperature due to their higher melting points, while unsaturated fatty acids are usually liquid.
At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 is present. Calculate Kc.
Answer:
[tex]K_c=0.0867[/tex]
Explanation:
Moles of SO₃ = 0.760 mol
Volume = 1.50 L
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity=\frac{0.760}{1.50\ L}[/tex]
[SO₃] = 0.5067 M
Considering the ICE table for the equilibrium as:
[tex]\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}[/tex]
Given:
Equilibrium concentration of O₂ = 0.130 mol
Volume = 1.50 L
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity=\frac{0.130}{1.50\ L}[/tex]
[O₂] = x = 0.0867 M
[SO₂] = 2x = 0.1733 M
[SO₃] = 0.5067-2x = 0.3334 M
The expression for the equilibrium constant is:
[tex]K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}[/tex]
[tex]K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}[/tex]
[tex]K_c=0.0867[/tex]
(a)-Use Lewis symbol store present there action that occurs between Ca and F atoms. (b)-What is the chemical formula of the most likely product? (c)-How many electrons are transferred? (d)-Which atom loses electrons in the reaction?
The Lewis symbol for Ca is Ca²⁺, while F is F⁻. The Ca atom loses two electrons, and the F atom gains one electron. The most likely product is CaF₂.
Explanation:(a) The Lewis symbol for Ca is Ca2+, while the Lewis symbol for F is F-. The action that occurs between Ca and F atoms is a redox reaction, where one atom loses electrons (oxidation) and the other gains electrons (reduction).
(b) The chemical formula of the most likely product is CaF2 since one Ca atom can bond with two F atoms.
(c) In the reaction, two electrons are transferred, with Ca losing two electrons and each F atom gaining one electron.
(d) Ca atom loses electrons in the reaction.
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When 1.98g of a hydrocarbon is burned in a bomb calorimeter, the temperature increases by 2.06∘C. If the heat capacity of the calorimeter is 69.6 J∘C and it is submerged in 944mL of water, how much heat (in kJ) was produced by the hydrocarbon combustion?
Answer:
8.3 kJ
Explanation:
In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:
q for water:
q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g
c: specific heat of water = 4.186 J/gºC
ΔT : change in temperature = 2.06 ºC
so solving for q :
q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J
For calorimeter
q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC
ΔT : change in temperature = 2.06 ºC
q calorimeter = 69.60J x 2.06 ºC = 143.4 J
Total heat released = 8,140 J + 143.4 J = 8,2836 J
Converting into kilojoules by dividing by 1000 we will have answered the question:
8,2836 J x 1 kJ/J = 8.3 kJ
The heat produced by the combustion of 1.98g of the hydrocarbon is 0.143 kJ.
The volume of water was not needed for this particular calculation.
To determine the heat produced by burning 1.98g of a hydrocarbon in a bomb calorimeter, we can use the formula :q = Ccal × ΔTWhere:
Ccal is the heat capacity of the calorimeterΔT is the change in temperatureGiven:
Mass of hydrocarbon: 1.98gTemperature increase: 2.06°CHeat capacity of calorimeter: 69.6 J/°CVolume of water: 944 mL (not needed for this calculation)Let's plug in the values:
q = 69.6 J/°C × 2.06°Cq = 143.376 JSince we need the heat in kJ :q = 143.376 J / 1000 = 0.143376 kJThe heat produced by the hydrocarbon combustion is 0.143 kJ.Glucose is present in blood plasma at a concentration of approximately 80 mg/dL. A typical 70 kg human has approximately 5 L blood, which is 55% plasma. Using the 44.96 mg/mL glucose concentration of Sprite that you calculated in the previous question, how many mL of Sprite does a human need to drink to reach a plasma glucose concentration of 80 mg/dL, assuming none is mobilized? (Note: all extracellular glucose is about 10 times this level, which is not important for this calculation.(Please pay attention to the units mg/dL)
Answer:
A human need to drink 48.93 mL of sprite to reach a plasma glucose concentration of 80 mg/dL.
Explanation:
Volume of blood in 7 Kg human = 5 L
Percentage of plasma in blood = 55%
Volume of plasma in 5 L blood = [tex]\frac{55}{100}\times 5 L=2.75 L=27.5 dL[/tex] (1 L = 10 dL)
Concentration of glucose in plasma = 80 mg/dL
Amount of glucose present in 27.5 dL : 80 mg /dL × 27.5 dL= 2200 mg
Let the volume of sprite with 2200 mg glucose be x
Concentration of glucose in sprite = 44.96 mg/mL
[tex]x\times 44.96 mg/mL=2200 mg[/tex]
[tex]x=\frac{2200 mg}{44.96 mg/mL}=48.93 mL[/tex]
A human need to drink 48.93 mL of sprite to reach a plasma glucose concentration of 80 mg/dL.
A saturated solution of lead(II) chloride, PbCl2, was prepared by dissolving solid PbCl2 in water. The concentration of Pb2+ ion in the solution was found to be 1.62×10−2 M . Calculate Ksp for PbCl2.
Answer:
Ksp for PbCl2 is 1.7 *10^-5
Explanation:
Step 1: Data given
The concentration of Pb2+ ion in the solution was found to be 1.62 *10^−2 M
Step 2: The balanced equation
PbCl2 ⇔Pb^2+ + 2Cl -
Step 2: ICE chart
The initial concentration of Pb2+ is 0 M
There will react X M and 2X of Cl-
At the equillibrium there is X M of Pb^2+ and 2X M of Cl-
The concentration of Pb2+ ion in the solution was found to be 1.62 *10^−2 M
Step 3: Calculate Ksp
Since PbCl2 is solid, it doesn't aply for Ksp
Ksp = [Pb^2+][Cl-]²
Ksp = X*(2X)²
Ksp = 4X³
⇒ X = 1.62 *10^-2 M
Ksp = 4*( 1.62 *10^-2)³
Ksp =1.7 *10^-5
Ksp for PbCl2 is 1.7 *10^-5
A gas has a sample initial pressure of 1.24 atm and an initial volume of 0.671 L. What is the pressure (in torr) if the final volume of the gas is changed to 583. mL? Assume constant temperature and amount of gas.
Answer:
1086.8 torr
Explanation:
Using Boyle's law
[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]
Given ,
V₁ = 0.671 L
V₂ = 583 mL = 0.583 L ( 1 mL = 0.001 L )
P₁ = 1.24 atm
P₂ = ?
Using above equation as:
[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]
[tex]{1.24\ atm}\times {0.671\ L}={P_2}\times {0.583\ L}[/tex]
[tex]{P_2}=1.43\ atm[/tex]
The conversion of P(atm) to P(torr) is shown below:
[tex]P(atm)={760}\times P(torr)[/tex]
So,
Pressure = 1.43*760 torr = 1086.8 torr
"In the absence of an adequate supply of oxygen, yeasts obtains metabolic energy by fermentation of glucose to produce ethanol. C6H12O6(s) LaTeX: \longrightarrow⟶ 2 C2H5OH(l) + 2 CO2(g) Use the standard enthalpies of formation to calculate ΔH for this reaction" Substance ΔHo glucose(s) -304.5 kcal/mol CO2(g) -93.9 kcal/mol C2H5OH(l) -66.4 kcal/mol
Answer: [tex]\Delta H=-16.5 kcal[/tex]
Explanation:
The balanced chemical reaction is,
[tex]C_6H_{12}O_6(s)\longrightarrow 2C_2H_5OH(l)+2CO_2(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{C_2H_5OH}\times \Delta H_{C_2H_5OH})+(n_{CO_2}\times \Delta H_{CO_2})]-[(n_{C_6H_{12}O_6}\times \Delta H_{C_6H_{12}O_6})][/tex]
where,
n = number of moles
Now put all the given values in this expression, we get
[tex]\Delta H=[(2\times -66.4)+(2\times -93.9)]-[(1\times -304.5)][/tex]
[tex]\Delta H=-16.5kcal[/tex]
Therefore, the enthalpy change for this reaction is -16.5 kcal
An air/gasoline vapor mix in an automobile cylinder has an initial temperature of 180 ∘C and a volume of 13 cm3 . If the mixture is heated to 587 ∘C with the pressure and amount held constant, what will be the final volume of the gas in cubic centimeters? Express your answer in cubic centimeters to three significant figures.
Answer:
The final volume will be 24.7 cm³
Explanation:
Step 1: Data given:
Initial temperature = 180 °C
initial volume = 13 cm³ = 13 mL
The mixture is heated to a fina,l temperature of 587 °C
Pressure and amount = constant
Step 2: Calculate final volume
V1/T1 = V2/T2
with V1 = the initial volume V1 = 13 mL = 13*10^-3
with T1 = the initial temperature = 180 °C = 453 Kelvin
with V2 = the final volume = TO BE DETERMINED
with T2 = the final temperature = 587 °C = 860 Kelvin
V2 = (V1*T2)/T1
V2 = (13 mL *860 Kelvin) /453 Kelvin
V2 = 24.68 mL = 24.7 cm³
The final volume will be 24.7 cm³
A sample of NI3 is contained in a piston and cylinder. The samples rapidly decomposes to form nitrogen gas and iodine gas, and releases 3.30 kJ of heat and does
950 J of work. What is ΔE?
Answer:
ΔE is -4250 J
Explanation:
Step1: Data given
3.30 kJ of heat is released
There is 950 J of work done
Step 2: Calculate ΔE
ΔE = q + w
⇒ with q = heat of energy released = 3.3 kJ = 3300 J ( negative because the heat is released)
⇒ work done = 950 J ( also negative )
ΔE = -3300J-950J= -4250 J
ΔE is -4250 J
The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed rate of disappearance of NO is 9.3×10−5M/s.
(a) What is the rate of disappearance of O2 at this moment?
(b) What is the value of the rate constant?
(c) What are the units of the rate constant?
(d) What would happen to the rate if the concentration of NO were increased by a factor of 1.8?
Answer:
(a) The rate of disappearance of [tex]O_{2}[/tex] is: [tex]4.65*10^{-5}[/tex] M/s
(b) The value of rate constant is: 0.83036 [tex]M^{-2}s^{-1}[/tex]
(c) The units of rate constant is: [tex]M^{-2}s^{-1}[/tex]
(d) The rate will increase by a factor of 3.24
Explanation:
The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.
For the given reaction:
[tex]2NO(g)+O_{2}->2NO_{2}[/tex]
rate = [tex]-\frac{1}{2} \frac{d}{dt}[NO][/tex] = [tex]-\frac{d}{dt}[O_{2}][/tex] = [tex]\frac{1}{2}\frac{d}{dt}[NO_{2}][/tex] -----(1)
According to the question, the reaction is second order in NO and first order in [tex]O_{2}[/tex].
Then we can say that, rate = k[tex][NO]^{2}[O_{2}][/tex] -----(2)
where k is the rate constant.
The rate of disappearance of NO is given:
[tex]-\frac{d}{dt}[NO][/tex] = [tex]9.3*10^{-5}[/tex] M/s.
(a) From (1), we can get the rate of disappearance of [tex]O_{2}[/tex].
Rate of disappearance of [tex]O_{2}[/tex] = [tex]-\frac{d}{dt}[O_{2}][/tex] = (0.5)*([tex]9.3*10^{-5}[/tex]) M/s = [tex]4.65*10^{-5}[/tex] M/s.
(b) The rate of the reaction can be obtained from (1).
rate = [tex]-\frac{1}{2} \frac{d}{dt}[NO][/tex] = (0.5)*([tex]9.3*10^{-5}[/tex])
rate = [tex]4.65*10^{-5}[/tex] M/s
The value of rate constant can be obtained by using (2).
rate constant = k = [tex]\frac{rate}{[NO]^{2}[O_{2}]}[/tex]
k = [tex]\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}[/tex] = 0.83036 [tex]M^{-2}s^{-1}[/tex]
(c) The units of the rate constant can be obtained from (2).
k = [tex]\frac{rate}{[NO]^{2}[O_{2}]}[/tex]
Substituting the units of rate as M/s and concentrations as M, we get:
[tex]\frac{Ms^{-1} }{M^{3}}[/tex] = [tex]M^{-2}s^{-1}[/tex]
(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.
[tex]rate\alpha [NO]^{2}[/tex]
If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of [tex](1.8)^{2}[/tex] = 3.24
The rate of disappearance of O2 is 4.65×10−5 M/s, the rate constant is 0.19875 M^-1s^-1. The units of the rate constant for a third-order reaction are M^-1s^-1. If NO concentration were to increase by a factor 1.8, the reaction rate would increase by a factor of 3.24.
Explanation:In a chemical reaction, the rate of disappearance of a reactant matches the stoichiometric ratio. Here, the ratio of NO to O2 is 2:1, therefore, the rate of disappearance of O2 is half that of NO. So, (a) the rate of disappearance of O2 is 9.3×10−5 M/s ÷ 2 = 4.65×10−5 M/s.
From the rate law, rate = k[NO]^2[O2], we can determine the rate constant (k) by substituting the known values into the equation. (b) the value of the rate constant k is rate ÷ ([NO]^2[O2]) = 9.3×10−5 M/s ÷ ((0.040 M)^2(0.035 M)) = 0.19875 M^-1s^-1.
As for (c) the units of the rate constant for a third-order reaction are M^-1s^-1. (d) If the concentration of NO were increased by a factor of 1.8, the rate would increase by a factor of (1.8)^2 due to the reaction being second order in NO. Hence the rate would increase by a factor of 3.24 times.
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Part A Add single electrons and/or electron pairs as needed to complete the electron-dot symbol for astatine, At. To return the atom to its original state, use the reset button. Click to select an electron-pair or single electron, then click on the element symbol to add electrons. Click the reset button to clear all electrons and start over. View Available Hint(s) AANNN
Answer:
7 valence electrons
Explanation:
Astatine has the atomic number 85. Thus, its electron configuration is:
[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁵
As we can see, in the last level (6) it has 2 + 5 = 7 electrons, that is, astatine has 7 electrons in its valence shell. In the Lewis dot structure (attached) we write 3 pairs of electrons and 1 unpaired electron around the symbol of At.
What is the concentration in molarity of a solution made using 10.0 grams of KCl in 300.0 mL of water?
Please help immediately!!! :(
Answer:
From Molarity=concentration/molar mass
Concentration=10/0.3dm³
33.33g/dm³
Molar mass=H2O=18g/mol
Molarity=1.852mol/dm³
Answer:
0. 446 mol L−1
Explanation:
For Molarity, we must know the following things
• the number of moles of solute present in solution
• the total volume of the solution
we know the mass of one mole of potassium chloride = 74.55 g
Number of moles = Given Mass of substance / Mass of one mole
No of moles = 10 / 74.55
= 0.134 moles
Now we know that molarity is expressed per liter of solution. Since you dissolve 0.134 moles of potassium chloride in 300. mL of solution, you can say that 1.0 L will contain
For 300 ml of solution, no of moles are = 0.134 moles
For 1 ml of solution, no of moles are = 0.134/300 moles
For 1(1000) ml of solution, no of moles are= 0.134/300 x 1000
= 0.446 moles/ L
Answer is =0. 446 mol L−1
The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes,
(1) NO(g) + NO
2
(g)
→
N
2
O
3
(g) ;
Δ
H
o
r
x
n
= -39.kJ
(2) NO(g) + NO
2
(g) + O
2
(g)
→
N
2
O
5
(g) ;
Δ
H
o
r
x
n
= -112.5 kJ
(3) 2NO
2
(g)
→
N
2
O
4
(g) ;
Δ
H
o
r
x
n
= -57.2 kJ
(4) 2NO(g) + O
2
(g)
→
2NO
2
(g) ;
Δ
H
o
r
x
n
= -114.2 kJ
(5) N
2
O
5
(s)
→
N
2
O
5
(g) ;
Δ
H
o
s
u
b
l
= 54.1 kJ
Calculate the heat of reaction for N
2
O
3
(g) + N
2
O
5
(s)
→
2N
2
O
4
(g)
Δ
H = _ _ _ _ _ kJ
Answer:
The heat of the given reaction is -23.0 kJ.
Explanation:
We are given with ;
[tex]NO(g) + NO_2(g)\rightarrow N_2O_3(g) \Delta H_{1,rxn}=-39.kJ[/tex]..[1]
[tex]NO(g) + NO_2(g) + O_2(g)\rightarrow N_2O_5(g), \Delta H_{2,rxn} = -112.5 kJ [/tex]..[2]
[tex]2NO_2(g) \rightarrow N_2O_4(g) ,\Delta H_{3,rxn} = -57.2 kJ [/tex]..[3]
[tex]2NO(g) + O_2(g)\rightarrow 2NO_2(g), \Delta H_{4,rxn} = -114.2 kJ[/tex]..[4]
[tex]N_2O_5(s)\rightarrow N_2O_5(g) ,\Delta H_{5,sub}= 54.1 kJ[/tex]..[5]
To find heat of reaction:
[tex]N_2O_3(g) + N_2O_5(s)\rightarrow 2N_2O_4(g),\Delta H_{6,rxn} = ?[/tex]..[6]
Using Hess's law:
[5] +2 × [3] + [4] - [2] - [1] = [6]
[tex]\Delta H_{6,rxn}=\Delta H_{5,sub}+2\times \Delta H_{3,rxn}+\Delta H_{4,rxn}-\Delta H_{2,rxn}-\Delta H_{1,rxn}[/tex]
[tex]\Delta H_{6,rxn}=54.1 kJ+(2\times (-57.2 kJ))+(-114.2 kJ)-(-112.5 kJ)-(-39.0 kJ)[/tex]
[tex]\Delta H_{6,rxn}=-23.0 kJ[/tex]
The heat of the given reaction is -23.0 kJ.
A heat flux meter attached to the inner surface of a 3-cm-thick refrigerator door indicates a heat flux of 25 W/m² through the door. Also, the temperatures of the inner and the outer surfaces of the door are measured to be 7°C and 15°C, respectively. Determine the average thermal conductivity of the refrigerator door. Answer: 0.0938 W/m. C
Answer:
0.0938 W/m.°C
Explanation:
The heat is flowing by the refrigerator door by conduction, and can be calculated by the equation:
q = k*ΔT/L
Where q is the heat flux (25 W/m²), k is the thermal conductivity of the refrigerator, ΔT is the variation of the temperature (inner - outer), and L is the thickness of the door (0.03 m).
25 = k*(15-7)/0.03
8k = 0.75
k = 0.0938 W/m.°C
Complete and balance the following reaction by filling in the missing coefficients. Assume the reaction is occurring in a basic, aqueous solution.CH3CH2OH(aq)+MnO−4(aq)⟶CH3COO−(aq)+MnO2(s) Include all coefficients, even those equal to 1.
Answer:
4 MnO₄⁻(aq) + 3 CH₃CH₂OH(aq) ⟶ 4 MnO₂(s) + 1 OH⁻(aq) + 3 CH₃COO⁻(aq) + 4 H₂O(l)
Explanation:
To balance a redox reaction we use the ion-electron method.
Step 1: Identify both half-reactions
Reduction: MnO₄⁻(aq) ⟶ MnO₂(s)
Oxidation: CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq)
Step 2: Balance the mass adding H₂O and OH⁻ where necessary.
2 H₂O(l) + MnO₄⁻(aq) ⟶ MnO₂(s) + 4 OH⁻(aq)
5 OH⁻(aq) + CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq) + 4 H₂O(l)
Step 3: Balance the charge adding eelctrons where necessary.
2 H₂O(l) + MnO₄⁻(aq) + 3 e⁻ ⟶ MnO₂(s) + 4 OH⁻(aq)
5 OH⁻(aq) + CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq) + 4 H₂O(l) + 4 e⁻
Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost are the same.
4 × (2 H₂O(l) + MnO₄⁻(aq) + 3 e⁻ ⟶ MnO₂(s) + 4 OH⁻(aq))
3 × (5 OH⁻(aq) + CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq) + 4 H₂O(l) + 4 e⁻)
Step 5: Add both half-reactions and cancel what is repeated.
8 H₂O(l) + 4 MnO₄⁻(aq) + 12 e⁻ + 15 OH⁻(aq) + 3 CH₃CH₂OH(aq) ⟶ 4 MnO₂(s) + 16 OH⁻(aq) + 3 CH₃COO⁻(aq) + 12 H₂O(l) + 12 e⁻
4 MnO₄⁻(aq) + 3 CH₃CH₂OH(aq) ⟶ 4 MnO₂(s) + 1 OH⁻(aq) + 3 CH₃COO⁻(aq) + 4 H₂O(l)
To balance the given reaction: CH3CH2OH(aq)+MnO−4(aq)⟶CH3COO−(aq)+MnO2(s), in a basic, aqueous solution, the balanced equation is 8CH3CH2OH(aq) + MnO−4(aq) + 8OH-(aq) ⟶ 8CH3COO-(aq) + MnO2(s) + 4H2O(l).
Explanation:To complete and balance the given reaction: CH3CH2OH(aq)+MnO−4(aq)⟶CH3COO−(aq)+MnO2(s) in a basic, aqueous solution, we need to assign appropriate coefficients to each compound to ensure that the reaction is balanced. The balanced equation is:
8CH3CH2OH(aq) + MnO−4(aq) + 8OH-(aq) ⟶ 8CH3COO-(aq) + MnO2(s) + 4H2O(l)
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In the absence of an electric field, a radioactive beam strikes a fluorescent screen at a single point. When an electric field is applied, the radioactive beam is separated into three different components. One of the components is deflected toward the positive electrode because it is negatively charged, one of the components is deflected toward the negative electrode because it is positively charged, and one component is not deflected in any direction; instead, it moves along a straight path.
Identify the charges possessed by the different components of the radioactive beam by observing their behavior under the influence of an electric field.
Sort the appropriate items into Negatively Charged, Neutral, or Positively Charged:
1._____________ ? Rays
2. __________? Rays
3. _____________? Rays
Answer:
Negatively Charged: Beta - radiation
Neutral: Gamma radiation
Positively Charged: Beta + radiation.
Explanation:
The radioactive beam described is formed by three different types of particules:
The Negatively Charged are formed by electrons and known as Beta - radiationThe Positively Charged are formed by positrons and known as Beta + radiation. The other alternative to a positively charged radiation is Alpha radiation: formed by two positrons and two neutrines-The Neutral ones are formed by Gamma radiationIn Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, what is the mass of nitrogen remaining after the reaction?
Answer:
Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams
Explanation:
Step 1: Data given
Number of moles hydrogen = 30 moles
Number of moles nitrogen = 30 moles
Yield = 50 %
Molar mass of N2 = 28 g/mol
Molar mass of H2 = 2.02 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate limiting reactant
For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3
Hydrogen is the limiting reactant.
The 30 moles will be completely be consumed.
N2 is in excess. There will react 30/3 =10 moles
There will remain 30 -10 = 20 moles (this in the case of a 100% yield)
In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.
Step 4: Calculate moles of NH3
There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)
For a 50% yield there will be produced, 10 moles of NH3
Step 5: Calculate the mass of NH3
Mass of NH3 = mol NH3 * Molar mass NH3
Mass of NH3 = 20 moles * 17.03
Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)
Step 6: Calculate actual mass
50% yield = actual mass / theoretical mass
actual mass = 0.5 * 340.6
actual mass = 170.3 grams
Step 7: The mass of nitrogen remaining
There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain
Mass of nitrogen = 25 moles * 28 g/mol
Mass of nitrogen = 700 grams
Answer:
25 moles
Explanation:
1 mole of nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia. 30 moles of hydrogen will react with 10 moles of nitrogen to give 20 moles of ammonia. As the actual yield is 50%, ammonia formed is 10 moles, the amount of nitrogen reacted is 5 moles, and the amount of hydrogen reacted is 15 moles. The mass of the remaining hydrogen is 15 moles and of the remaining nitrogen is 25 moles.
Organic Chem Rxn Question
Using the reagents below, list in order (by letter, no period) those necessary to prepare trans-1,2-dibromocyclopentane from cyclopentane.
Note: Not all spaces provided may be needed. Type "na" in any space where you have no reagent. (There are 3 spaces)
a. Br2, heat, light
b. HBr
c. Br2
d. H2O
e. HBr, HOOH
f. EtOH
g. NaOEt, EtOH, heat
Answer:
a, g, c
Explanation:
The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.
The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.
Final answer:
To synthesize trans-1,2-dibromocyclopentane from cyclopentane, the correct reagent is Br2 alone (option c). Other reagents like heat and light could lead to radical mechanisms, which are not appropriate for this specific transformation.
Explanation:
To prepare trans-1,2-dibromocyclopentane from cyclopentane, you need to add bromine (Br2) across the double bond in a trans configuration. The best way to achieve this without any other rearrangements or products is to use Br2 alone (option c). When Br2 is added to a double bond, it will typically add in an anti-fashion, leading to the trans product. Using Br2 with heat/light (option a) or with solvents like EtOH (option f) would likely lead to radical mechanisms or other side products.
Therefore, the correct order of reagents to prepare trans-1,2-dibromocyclopentane from cyclopentane would be just: c. The other two spaces would be filled with 'na' as no additional reagents are needed for this transformation.
Thus, the list in order would be: c, na, na.
Identify two structural features of purines and Pyrimidines
Purines
A. contain only three ring nitrogen atoms.
B. contain one heterocyclic ring.
C. contain four ring nitrogen atoms.
D. contain only two ring nitrogen atoms.
E.contain two heterocyclic rings.
Pyrimidines
A.contain two heterocyclic rings.
B. contain only three ring nitrogen atoms.
C. contain only two ring nitrogen atoms.
D. contain four ring nitrogen atoms
E. contain one heterocyclic ring.
Answer:
The answers are:
Purines:
C. contain four ring nitrogen atoms.
E. contain two heterocyclic rings.
Pyrimidines:
C. contain only two ring nitrogen atoms.
E. contain one heterocyclic ring.
Explanation:
Purines and Pyrimidines are nitrogenous bases which are the building blocks of nucleic acids (DNA and RNA).
Purines are composed by two fused heterocyclic rings, one of them is a 6-ring and the other is a 5-ring. Each ring contains two nitrogen atoms which form part of the ring. Thus, the nitrogen positions in purines are: 1', 3', 7' and 9'. Depending on the functional groups bonded to the two-ring structure, a purine base can be Guanidine (G) or Adenine (A).
The structure of Pyrimidines is a single heterocycle ring wich contains two nitrogen atoms in positions 1' and 3'. Depending of the functional groups, they can be: Cytosine (C), Thymidine (T) and Uracil (U, which is found in RNA).
The purpose of the salt bridge in an electrochemical cell is to ____.
A. maintain electrical neutrality in the half-cells via migration of ions
B. provide a source of ions to react at the anode and cathode
C. provide oxygen to facilitate oxidation at the anode
D. provide a means for electrons to travel from the anode to the cathode
E. provide a means for electrons to travel from the cathode to the anode
Answer:
A. maintain electrical neutrality in the half-cells via migration of ions
Explanation:
Salt bridge -
For an electrochemical reaction , involving an anode and a cathode , both the electrodes are connect via a salt bridge to complete the circuit for the reaction .
One of the very important use of a salt bridge is to maintain the electrical neutrality of the respective half cells , which is achieved by the movement of ions .
Hence , from the given options , the correct option is ( a ) .
A salt bridge in an electrochemical cell helps maintain electrical neutrality. It does so by providing a pathway for ions to flow between the half-cells, balancing out the charge imbalance resulting from the reactions, thereby allowing the cell to continue to operate.
Explanation:The purpose of the salt bridge in an electrochemical cell is to maintain electrical neutrality in the half-cells via the migration of ions. A salt bridge, often saturated with a salt solution, provides a pathway for ions to flow between the two half-cells. During the operation of the cell, the reactions produce or use ions in the solutions, which could result in a charge imbalance. The salt bridge helps to balance out these charges, keeping the solutions electrically neutral. This flow of ions helps complete the circuit, allowing the cell to continue to operate.
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A stock solution contains a mixture of ~100 ppm chloride, fluoride, nitrite, bromide, nitrate and phosphate anions. In order to prepare 1 L of 100 ppm nitrite stock solution, you weigh out 150.0 mg of NaNO2. The actual concentration of nitrite would be:______.
Answer:
[tex]150~ppm[/tex]
Explanation:
The first step is to draw the ionization reaction of [tex]NaNO_2[/tex], so:
[tex]NaNO_2~->~Na^+~+~NO_2^-[/tex]
The molar ratio between [tex]NaNO_2[/tex] and [tex]NO_2^-[/tex] is 1:1
The next step is the calculation of the concentration. We have to remember that the formula of ppm is:
[tex]ppm=\frac{mg}{L}[/tex]
In this case we will have a mass of 150 mg and a volume of 1 L so:
[tex]ppm=\frac{150~mg}{1~L}=~150ppm[/tex]
Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole of C6H12O6. C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)
Answer : The enthalpy of combustion per mole of [tex]C_6H_{12}O_6[/tex] is -2815.8 kJ/mol
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equilibrium reaction follows:
[tex]C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})][/tex]
We are given:
[tex]\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol[/tex]
Therefore, the enthalpy of combustion per mole of [tex]C_6H_{12}O_6[/tex] is -2815.8 kJ/mol
Answer:
muahhh
Explanation:
Match each weather phenomena listed below to its appropriate scale of motion. longwave ridges and troughs microscale a high pressure system persists over the central plains for 4 days synoptic scale a supercell thunderstorm that lasts for over an hour global scale A turbulent eddy mesoscale
Answer:
Longwave ridges = Global scale
High pressure system persists over the central plains for 4 days = Synoptic scale
Supercell thunderstorm that lasts for over an hour = mesoscale
A turbulent eddy = microscale
Explanation:
Global scale has a range of the entire earth
synoptic scale has a range of about 100-1000km and can last fro days to weeks
mesoscale has a range of 4-100km and lasts for a day maximum
microscale is the least of all atmospheric motions.
A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −320.1 kJ/mol and ΔS = −86.00 J/K · mol.
Determine the temperature (in °C) below which the reaction is spontaneous.
Answer:
This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C
Explanation:
Step 1: Data given
ΔH = −320.1 kJ/mol
ΔS = −86.00 J/K · mol.
Step 2: Calculate the temperature
ΔG<0 = spontaneous
ΔG= ΔH - TΔS
ΔH - TΔS <0
-320100 - T*(-86) <0
-320100 +86T < 0
-320100 < -86T
320100/86 > T
3722.1 > T
The temperature should be lower than 3722.1 Kelvin (= 3448.9 °C)
We can prove this with Temperature T = 3730 K
-320100 -3730*(-86) <0
-320100 + 320780 = 680 this is greater than 0 so it's non spontaneous
T = 3700 K
-320100 -3700*(-86) <0
-320100 + 318200 = -1900 this is lower than 0 so it's spontaneous
The temperature is quite high because of the big difference between ΔH and ΔS.
This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C
What evidence is there to suggest that the Earth is composed of tectonic plates?
Question 1 options:
The presence of a Ring of “Fire, hot spots, and the locations of earthquakes are some kinds of evidence that indicate plates exist and move. Data from seafloor spreading is another kind of data. The presence of fossils indicates that two land masses that once were close are now far apart. Mountain ranges that once were part of the same range are now on different continents. All of these indicates plates exist and move.
The tectonic plates are cracked up due to the frequency of earthquakes and scientists can see and measure these cracks. The cracks are the proof that the earth is made up of tectonic plates.
Answer:
All of these indicate plates exist and move.
Explanation:
The Ring of Fire, hot spots, and the locations of earthquakes draw a pattern on the surface of geologic activity underground. Seafloor spreading is proof of tectonic plates because it increases the area of some oceans and decreases the area of others. Magma from the mantle pushes the plates apart. The case of matching fossils in parts of South America and Africa is consistent with the theory that at one point the continents were joined together. According to seafloor spreading, the Atlantic Ocean is young and is still growing. The same principle applies to mountain ranges now found on two continents that used to be on the same one. It is thought that the Appalachian mountains were connected to Europe and Africa.Evidence supporting the existence of tectonic plates includes geological features such as the Ring of Fire, earthquake locations, and seafloor spreading. Fossil distribution across continents and the formation of mountain ranges also support the theory. Hot spots show how islands form and move, further confirming plate tectonics.
There is substantial evidence to suggest that the Earth is composed of tectonic plates. Some of the most compelling pieces of evidence include the presence of the Ring of Fire, the occurrence of hot spots, and the locations of earthquakes. These features highlight areas where plates interact. Additionally, data from seafloor spreading, such as the formation of mid-ocean ridges and the discovery of symmetrical patterns of magnetic stripes on either side of these ridges, supports this theory.
The presence of fossils provides further proof. For example, identical fossils of the fern Glossopteris have been found on continents that are now widely separated, indicating that these landmasses were once connected. Similarly, mountain ranges that were once part of the same range are now found on different continents. These phenomena can be explained by the movement of tectonic plates.
Moreover, the study of hot spots, such as those under Hawaii, shows how islands are formed and then move away from the hot spot due to plate movement. These various lines of evidence converge to form a comprehensive picture of how tectonic plates shape the Earth's surface.