Answer:
B
Explanation:
Firstly, we will need to calculate the number of moles. To do this, we make use of the ideal gas equation
PV = nRT
n = PV/RT
The parameters have the following values according to the question:
P = 780mmHg, we convert this to pascal.
760mHG = 101325pa
780mmHg = xpa
x = (780 * 101325)/760 = 103,991 Pa
V= 400ml = 0.4L
T = 135C = 135 + 273.15 = 408.15K
n = ?
R = 8314.463LPa/K.mol
Substituting these values into the equation yields the following:
n = (103991 * 0.4)/(8314.463 * 408.15)
= 0.012 moles
Now we know 1 mole contains 6.02 * 10^23 molecules, hence, 0.012moles will contain = 0.012 * 6.02 * 10^23 = 7.38 * 10^21 molecules
Answer: The number of nitrogen molecules in the container are [tex]7.38\times 10^{21}[/tex]
Explanation:
To calculate the moles of gas, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 780 mmHg
V = Volume of the gas = 400.0 mL = 0.4 L (Conversion factor: 1 L = 1000 mL)
T = Temperature of the gas = [tex]135^oC=[135+273]K=408K[/tex]
R = Gas constant = [tex]62.364\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
n = number of moles of nitrogen gas = ?
Putting values in above equation, we get:
[tex]780mmHg\times 0.4L=n\times 62.364\text{ L. mmHg }mol^{-1}K^{-1}\times 408K\\\\n=\frac{780\times 0.4}{62.364\times 408}=0.01226mol[/tex]
According to mole concept:
1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules
So, 0.01226 moles of nitrogen gas will contain = [tex](0.012\times 6.022\times 10^{23})=7.38\times 10^{21}[/tex] number of molecules
Hence, the number of nitrogen molecules in the container are [tex]7.38\times 10^{21}[/tex]
Name the structure. CHC H.CH2CH2i CHC CH3
A) 7-chloro-3-ethyl-4-methyl-3-heptene
B) 1-chloro-5-ethyl-4-methyl-3-heptene x
C) 1-chloro-3-pentenyl-2-pentene x
D) cis-7-chloro-3-ethyl-4-methyl-3-heptene
E) trans-7-chloro-3-ethyl-4-methyl-3-heptene
The question is incomplete. The complete question is attached below.
Answer : The correct option is, (A) 7-chloro-3-ethyl-4-methyl-3-heptene
Explanation :
The rules for naming of alkene are :
First select the longest possible carbon chain.
The longest possible carbon chain should include the carbons of double bonds.
The naming of alkene by adding the suffix -ene.
The numbering is done in such a way that first carbon of double bond gets the lowest number.
The carbon atoms of the double bond get the preference over the other substituents present in the parent chain.
If two or more similar alkyl groups are present in a compound, the words di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.
The given compound name will be, 7-chloro-3-ethyl-4-methyl-3-heptene.
The structure of given compound is shown below.
The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene.
Explanation:The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene. The name is determined by identifying the longest carbon chain, the substituents attached to it, and their positions. In this case, the longest carbon chain has 7 carbons, with a chlorine atom attached at position 7. There is an ethyl group at position 3 and a methyl group at position 4. The presence of double bonds is indicated by the -ene ending.
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the reaction of 50 mL of gas with 50 mL of gas via the equation: Cl2(g) + C2H4(g) ➔ C2H4Cl2 (g) will produce a total of __________ mL of products if pressure and temperature are kept constant.
Explanation:
The given data is as follows.
50 ml of [tex]Cl_{2}[/tex], 50 ml of [tex]C_{2}H_{4}[/tex]
And, it is known that at STP 1 mole of a gas occupies 22.4 L. Hence, moles present in 50 ml of gas are as follows.
[tex]\frac{50}{22.4 \times 1000}[/tex] (As 1 L = 1000 ml)
= [tex]2.23 \times 10^{-3}[/tex] moles
So, according to the given equation [tex]2.23 \times 10^{-3}[/tex] moles of [tex]Cl_{2}[/tex] reacts with [tex]2.23 \times 10^{-3}[/tex] moles of [tex]C_{2}H_{4}[/tex].
Hence, moles of [tex]C_{2}H_{4}Cl_{2}[/tex] is equal to the moles of [tex]C_{2}H_{4}[/tex] and [tex]Cl_{2}[/tex].
Therefore, moles of [tex]C_{2}H_{4}Cl_{2}[/tex] = [tex]2.23 \times 10^{-3}[/tex] moles
1 mole of [tex]C_{2}H_{4}Cl_{2}[/tex] = 22.4 L
[tex]2.23 \times 10^{-3}[/tex] moles = [tex]22.4 \times 2.23 \times 10^{-3} moles[/tex]
= 50 ml of product
Thus, we can conclude that 50 ml of products if pressure and temperature are kept constant.
Consider the following reaction. N2(g) + O2(g) equilibrium reaction arrow 2 NO(g) If the equilibrium partial pressures of N2, O2, and NO are 0.15 atm, 0.33 atm, and 0.050 atm, respectively, at 2200°C, what is KP?
Explanation:
As the given reaction equation is as follows.
[tex]N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)[/tex]
Hence, expression for [tex]K_{p}[/tex] of this reaction is as follows.
[tex]K_{p} = \frac{P^{2}_{NO}}{P_{N_{2}} \times P_{O_{2}}}[/tex]
Now, putting the given values into the above expression as follows.
[tex]K_{p} = \frac{P^{2}_{NO}}{P_{N_{2}} \times P_{O_{2}}}[/tex]
= [tex]\frac{(0.050)^{2} atm}{(0.15 atm) \times (0.33 atm)}[/tex]
= [tex]5.05 \times 10^{-2}[/tex]
Thus, we can conclude that the value of [tex]K_{p}[/tex] is [tex]5.05 \times 10^{-2}[/tex].
Final answer:
The equilibrium constant, Kp, for the reaction N2(g) + O2(g) = 2 NO(g) at the given conditions is approximately 0.1010 when calculated using the given equilibrium partial pressures.
Explanation:
The question is asking us to calculate the equilibrium constant, KP, for the reaction given the equilibrium partial pressures of the reactants and products at a certain temperature. Using the reaction N2(g) + O2(g) = 2 NO(g), we can express the equilibrium constant KP in terms of the partial pressures of the gases:
KP = (PNO)2 / (PN2 * PO2)
Substituting in the given partial pressures:
KP = (0.050 ATM)2 / (0.15 ATM * 0.33 ATM) = 0.0050 / 0.0495 = 0.1010
So, the equilibrium constant KP at 2200°C for this reaction is approximately 0.1010.
The energy change, ∆H, associated with the following reaction is +81 kJ. NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g) What is the expected energy change for the reverse reaction of nine moles of HOBr and two moles of NH3?
Answer:
162 kJ
Explanation:
The reaction given by the problem is:
NBr₃(g) + 3 H₂O(g) → 3 HOBr(g) + NH₃(g) ∆H = +81 kJIf we turn it around, we have:
3 HOBr(g) + NH₃(g) → NBr₃(g) + 3 H₂O(g) ∆H = -81 kJIf we think now of HOBr and NH₃ as our reactants, then now we need to find out which one will be the limiting reactant when we have 9 moles of HOBr and 2 moles of NH₃:
When we have 1 mol NH₃, we need 3 mol HOBr. So when we have 2 moles NH₃, we need 6 moles HOBr. We have more than 6 moles HOBr so that's our reactant in excess, thus NH₃ is our limiting reactant.-81 kJ is our energy change when there's one mol of NH₃ reacting, so we multiply that value by two when there's two moles of NH₃ reacting. The answer is 81*2 = 162 kJ.
Final answer:
The energy change for the reverse reaction of NBr3 and H2O to form HOBr and NH3 is -81 kJ. When considering nine moles of HOBr and two moles of NH3, thrice the stoichiometric amounts, the expected energy change would be three times as much, resulting in -243 kJ.
Explanation:
The energy change, ΔH, for a reaction indicates how much energy is absorbed or released during the reaction. In the given reaction, NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g), the energy change is +81 kJ, meaning 81 kJ of energy is absorbed in the forward reaction. Since energy is a state function and the magnitude of energy change is the same for a reaction whether it goes in the forward or reverse direction, the energy change for the reverse reaction would be -81 kJ. The reaction in reverse would release 81 kJ of energy.
However, the question asks for the expected energy change for reversing the original reaction with nine moles of HOBr and two moles of NH3. The energy change provided (+81 kJ) is for the stoichiometric amounts as per the original equation. For nine moles of HOBr and two moles of NH3, which are thrice the stoichiometric amounts indicated by the original reaction, the energy released would be three times as much. Therefore, the expected energy change for the reverse reaction under these conditions would be 3 × (-81 kJ) = -243 kJ.
Draw the intermediate (in the first box) that forms in the reaction of 1-methylcyclopentene with br2 in ccl4, then draw the expected product (in the second box). do not provide stereochemistry. interactive 3d display mode
Answer:
The product of the reaction is 1,2 di-bromo-1-methylcyclopentane.
Explanation:
In the reaction of 1-methylcyclopentene with Br2 in CCl4, a bromonium ion intermediate forms, followed by nucleophilic attack by Cl-, resulting in the product 1-chloromethylcyclopentane.
In the reaction of 1-methylcyclopentene with Br2 in CCl4, a bromonium ion intermediate is formed. The bromonium ion is a three-membered ring with a bromine atom attached to two adjacent carbon atoms. Due to the lack of stereochemistry information, the structure of the intermediate is a general representation without specifying the stereochemical arrangement.
Following the formation of the bromonium ion, it undergoes a nucleophilic attack by the chloride ion (Cl-) from the solvent CCl4. The chloride ion opens the bromonium ring, leading to the substitution of bromine by chloride on one of the adjacent carbons. The product of this reaction is 1-chloromethylcyclopentane. The chlorine atom is now attached to the carbon that originally carried the methyl group.
This reaction mechanism is a classic example of halogenation of alkenes, where the bromine adds to the alkene in a concerted manner, forming a cyclic intermediate. The subsequent nucleophilic attack by the chloride ion results in the replacement of bromine with chlorine, leading to the final halogenated product.
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Consider the titration of 30.0 ml of 0.050 m nh3 with 0.025 mhcl. Calculate the ph after the following volumes of titrant have been added.
Explanation of pH changes in a titration between NH₃ and HCl.
The pH in a titration is determined by the amount of acid and base before and after the equivalence point. In the given scenario, titrating NH₃ with HCl, the initial pH is determined by the NH₃ concentration before any HCl is added.
(a) 0.00 mL: Before any HCl is added, NH₃ is in excess, making the solution basic with a pH >7.
(b) 25.00 mL: At the equivalence point, the pH is determined by the salt formed during the reaction. For NH₃ titrated with HCl, the salt formed is NH₄Cl, making the pH acidic.
A 27.5 −g aluminum block is warmed to 65.6 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.0 ∘C. The aluminum and the water are allowed to come to thermal equilibrium.
Answer:
Final temperature T is 26.197 degrees celcius.
Explanation:
As the system is in thermal equilibrium, the heat lost by aluminium is the heat gained by water.
specific heat capacity of aluminium = 0.9 J/g per degree celcius
specific heat capacity of water = 4.186 J/g per degree celcius
Let the final common temperature attained be "T".
Heat lost by aluminium = [tex](m)(c)(65.6-T)[/tex]
Heat gained by water = [tex](m)(c)(T-22)[/tex]
by law of conservation of energy,
Heat lost by hot body = Heat gained by cold body
[tex](m)(c)(65.6-T)[/tex] = [tex](m)(c)(T-22)[/tex]
[tex](27.5)(0.9)(65.6 - T) = (55.5)(4.186)(T - 22)[/tex]
calculating the value of T comes out as 26.197 degrees celcius.
Light with a wavelength of 190 nm is incident on a metal surface. The most energetic electrons emitted from the surface are measured to have 4.0 ev of kinetic energy. Which of the metals in the table is the surface most likely to be made of?
The question deals with the photoelectric effect, where light ejects electrons from a metal's surface. To identify the metal, one must calculate the work function using the kinetic energy of the emitted electrons and the frequency of the incident light.
Explanation:The question concerns the photoelectric effect, a phenomenon in physics where electrons (called photoelectrons) are ejected from a metal surface when it is exposed to light of a certain frequency or wavelength that is above the metal's threshold frequency. You've given a light with a wavelength of 190 nm incident on an unknown metal, and the most energetic electrons emitted have 4.0 eV of kinetic energy.
To identify which metal this could be, one needs to calculate the work function of the metal. This involves using the equation for the photoelectric effect: KE = hf - Φ, where KE is the kinetic energy of the ejected electrons, h is Planck's constant, f is the frequency of the incident light, and Φ represents the work function (also known as the binding energy) of the metal.
First, we need to convert the wavelength of 190 nm into frequency using the equation c = λf, where c is the speed of light, λ is the wavelength, and f is the frequency. Once we have the frequency, we can plug it into Planck's equation to find the energy of the photons. Subsequently, we'll use the initial energy of the photons to determine the work function. By comparing this calculated work function with the known work functions of various metals, we can identify the most likely metal of the surface.
Which biogeochemical cycle is the only one that does not include the atmosphere?
Answer:
Phosphorus cycle
Explanation:
Biogeochemical cycle, any of the natural pathways by which essential elements of living matter are circlated.
There are four types of biogeochemical cycle, they are ; water cycle,carbon cycle,nitrogen cycle and phosphorous cycle
Carbon cycle is the cycle in which photosynthesis and cellular respiration take place.
Water cycle involves transpiration.
Nitrogen cycle Is the cycle that is dependent upon bacteria for nitrogen fixation and denitrification.
Phosphorus cycle is one of the slowest biogeochemical cycle. It does not stay in the atmosphere, because it is normally in a liquid state at room temperature. It does not include the atmosphere.
Final answer:
The phosphorus cycle is the biogeochemical cycle that does not include an atmospheric component, differentiating it from cycles like water, carbon, and nitrogen.
Explanation:
The biogeochemical cycle that does not include the atmosphere is the phosphorus cycle. Unlike the water, carbon, and nitrogen cycles, the phosphorus cycle does not involve the atmospheric pathway. Phosphorus mainly cycles through rocks, soil, water, and living organisms. It is essential for life as part of DNA, RNA, and ATP molecules, which are vital for genetic information and energy transfer in cells. However, phosphorus doesn't enter the atmosphere in a gaseous phase; hence, it is distinct amongst biogeochemical cycles.
A student carried out a titration using HC2H3O2(aq) and NaOH(aq). The net ionic equation for the neutralization reaction that occurs during the titration is represented above. The NaOH(aq) was added from a buret to the HC2H3O2(aq) in a flask. The equivalence point was reached when a total of 20.0mL of NaOH(aq) had been added to the flask. How does the amount of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compare to the amount of HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq), and what is the reason for this result?
With every addition of NaOH, the concentration of HC2H3O2 in the solution decreases due to the neutralization reaction. Thus, the quantity of HC2H3O2 after adding 5.0 mL of NaOH will be less than the amount after adding 1.0 mL of NaOH. This is due to the stoichiometry of the reaction, where one mole of added NaOH neutralizes one mole of acetic acid.
Explanation:In this titration analysis, the neutralization reaction involves acetic acid (HC2H3O2) and Sodium Hydroxide (NaOH), creating sodium acetate and water as products. When 1.0 mL of NaOH is added, it reacts with the acetic acid present in the flask, reducing its quantity. When 5.0 mL of NaOH is added, more acetic acid reacts, and therefore, the amount of HC2H3O2 in the flask decreases more significantly compared to when only 1 mL was added. The principle of this result is based upon the stoichiometry of the neutralization reaction, indicating that for each mole of NaOH added, one mole of acetic acid is neutralized. Therefore, as more NaOH is added, more HC2H3O2 is neutralized, hence contributing to a lower concentration of HC2H3O2 in the solution.
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The amount of HC2H3O2(aq) in the flask will be less after the addition of 5.0mL NaOH than after the addition of 1.0mL NaOH. This happens because the NaOH neutralizes the acid directly proportional to its quantity added. More the NaOH added, more the acidic substance HC2H3O2 gets neutralized.
Explanation:In a titration process involving HC2H3O2(aq) and NaOH(aq), following the neutralization equation, each acid molecule reacts with one base molecule. Hence, the quantity of HC2H3O2 will reduce in direct proportion to the amount of NaOH added. So if you add 5.0mL of NaOH, you will neutralize 5 times as much HC2H3O2 as when you add 1.0mL of NaOH.
Therefore, the quantity of HC2H3O2(aq) remaining in the flask after the addition of 5.0mL NaOH will be less than that after the addition of 1.0mL NaOH. The reason is that more NaOH means more HC2H3O2(aq) has been neutralized into water and the salt NaC2H3O2.
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Both covalent-network solids and ionic solids can have melting points well in excess of room temperature, and both can be poor conductors of electricity in their pure form. However, in other ways their properties are quite different.1.Which type of solid is more likely to dissolve in water?2.Which type of solid can become a considerably better conductor of electricity via chemical substitution?
Ionic solids are more likely to dissolve in water and can become better conductors of electricity via chemical substitution.
Explanation:Both covalent-network solids and ionic solids have different properties. Ionic solids dissolve readily in water and can conduct electricity when melted or dissolved because their ions are free to move. Covalent-network solids, on the other hand, are insoluble in water and are poor conductors of electricity in any state because their particles are electrically neutral.
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Assuming each solution to be 0.10 M , rank the following aqueous solutions in order of decreasing pH.
Rank the solutions from the highest to lowest pH. To rank items as equivalent, overlap them
a.N2H2
b.Ba(OH)2
c.HOCLd.NAOH
e.HCL
Final answer:
The aqueous solutions ranked from highest to lowest pH are: NaOH (most basic), Ba(OH)2, N2H2, HOCl, and HCl (most acidic).
Explanation:
To rank the following aqueous solutions in order of decreasing pH, from highest pH (most basic) to lowest pH (most acidic), we need to understand the nature of each compound in water. The ions or molecules that each solution releases in water determine their pH.
NaOH - Sodium hydroxide is a strong base, and will completely dissociate in water, releasing OH- ions, which leads to a high pH.
Ba(OH)2 - Barium hydroxide is another strong base, and it also fully dissociates in water, but being a dibasic base, it releases twice as many OH- ions per formula unit as NaOH, potentially leading to an even higher pH.
N2H2 - Hydrazine is a weak base, it does not completely dissociate in water, but will still increase the pH to some extent.
HOCl - Hypochlorous acid is a weak acid, so it only partially dissociates in water.
HCl - Hydrochloric acid is a strong acid and fully dissociates in water, releasing H+ ions and resulting in a low pH.
The solutions in order of decreasing pH (highest pH to lowest pH) are: NaOH, Ba(OH)2, N2H2, HOCl, HCl.
A 100 W light bulb is placed in a cylinder equipped with a moveable piston. The light bulb is turned on for 2.0×10−2 hour, and the assembly expands from an initial volume of 0.85 L to a final volume of 5.88 L against an external pressure of 1.0 atm.
a)Use the wattage of the light bulb and the time it is on to calculate ΔU in joules (assume that the cylinder and light bulb assembly is the system and assume two significant figures). Express your answer using two significant figures.
b) Calculate w. Express your answer using two significant figures.
c) Calculate q. Express your answer using two significant figures.
Answer:
(a) ΔU = 7.2x10²
(b) W = -5.1x10²
(c) q = 5.2x10²
Explanation:
From the definition of power (p), we have:
[tex] p = \frac {\Delta W}{\Delta t} = \frac {\Delta U}{\Delta t} [/tex] (1)
where, p: is power (J/s = W (watt)) W: is work = ΔU (J) and t: is time (s)
(a) We can calculate the energy (ΔU) using equation (1):
[tex] \Delta U = p \cdot \Delta t = 100 \frac{J}{s} \cdot 2.0\cdot 10^{-2} h \cdot \frac{3600s}{1h} = 7.2 \cdot 10^{2} J [/tex]
(b) The work is related to pressure and volume by:
[tex] \Delta W = -p \Delta V [/tex]
where p: pressure and ΔV: change in volume = V final - V initial
[tex] \Delta W = - p \cdot (V_{fin} - V_{ini}) = - 1.0 atm (5.88L - 0.85L) = - 5.03 L \cdot atm \cdot \frac{101.33J}{1 L\cdot atm} = -5.1 \cdot 10^{2} J [/tex]
(c) By the definition of Energy, we can calculate q:
[tex] \Delta U = \Delta W + \Delta q [/tex]
where Δq: is the heat transfer
[tex] \Delta q = \Delta U - \Delta W = 7.2 J - (-5.1 \cdot 10^{2} J) = 5.2 \cdot 10^{2} J [/tex]
I hope it helps you!
It takes 208.4 kJ of energy to remove 1 mole of electrons from 1 mole of atoms on the surface of rubidium metal. How much energy does it take to remove a single electron from an atom on the surface of solid rubidium?
Answer:
3.46*10⁻²² kJ
Explanation:
By the Avogadro's number, 1 mole of electrons at 1 mole of atoms correspond to 6.02x10²³ electrons. So it's necessary 208.4 kJ to remove 6.02x10²³ electrons. To remove a single electron:
6.02*10²³ electrons ---------------- 208.4 kJ
1 electron ---------------- x
By a simple direct three rule:
6.02*10²³ x = 208.4
x = 3.46*10⁻²² kJ
List the 4 major groups of inorganic chemicals common in cells:
Water, Carbon Dioxide, Acids, and Oxygen are the inorganic chemicals common in cells.
What are inorganic chemicals?A substance without both carbon and hydrogen is said to be an inorganic compound.Many inorganic substances, like water (H₂O) and the hydrochloric acid (HCl) secreted by your stomach, do contain hydrogen atoms. In contrast, carbon atoms are only present in a small number of inorganic molecules. An example of one of the few is carbon dioxide (CO₂).Water makes up as much as 70% of an adult's body weight. The cells that make up tissues and organs have water both inside of them and in the spaces between them.A substance that releases hydrogen ions (H⁺) in solution is referred to be an acid. Because it releases all of its H⁺ in the stomach's watery environment, hydrochloric acid (HCl), which is secreted by cells in the lining of the stomach, is a powerful acid. This strong acid helps with digestion and destroys ingested microorganisms.Carbon dioxide (CO₂) is expelled during respiration in exchange for oxygen (O₂) from the surrounding air. At the cellular level, O₂ is utilized to produce CO₂ while converting metabolic energy from foods into adenosine triphosphate (ATP).
Thus, inorganic chemicals like Water, Carbon Dioxide, Acids, and Oxygen
are essentail for cells.
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The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell:________
Pt(s)
|H2(g,1atm)
|H+(aq,1.0M)
|Au3+(aq,?M)
|Au(s).
What is the concentration of Au3+ in the solution if Ecell is 1.23 V ?
Answer: The concentration of [tex]Au^{3+}[/tex] is [tex]1.096\times 10^{-6}[/tex]
Explanation:
The given chemical cell follows:
[tex]Pt(s)|H_2(g,1atm)|H^+(aq,1.0M)||Au^{3+}(aq,?M)|Au(s)[/tex]
Oxidation half reaction: [tex]H_2(g,1atm)\rightarrow 2H^{+}(aq,1.0M)+2e^-;E^o_{2H^{+}/H_2}=0.0V[/tex] ( × 3)
Reduction half reaction: [tex]Au^{3+}(aq,?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V[/tex] ( × 2)
Net cell reaction: [tex]3H_2(g,1atm)+2Au^{3+}(aq,?M)\rightarrow 6H^{+}(aq,1.0M)+2Au(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=1.50-(0.0)=1.50V[/tex]
To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = 1.23 V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.50 V
n = number of electrons exchanged = 6
[tex][H^{+}]=1.0M[/tex]
[tex][Au^{3+}]=?M[/tex]
Putting values in above equation, we get:
[tex]1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})[/tex]
[tex][Au^{3+}]=-1.0906\times 10^{-6},1.096\times 10^{-6}[/tex]
Neglecting the negative value because concentration cannot be negative.
Hence, the concentration of [tex]Au^{3+}[/tex] is [tex]1.096\times 10^{-6}[/tex]
Ammonia, NH3, acts as an Arrhenius base because it Ammonia, N H 3, acts as an Arrhenius base because it blank the concentration of hydroxide ion, \rm OH^-, in aqueous solution. the concentration of hydroxide ion, OH−, in aqueous solution.
Answer:
Ammonia acts as an Arrhenius base because it increases the concentration of OH⁻ in aqueous solution.
Explanation:
The acid-base theory of Arrhenius explains that in aqueous solutions both acid and base dissociate, releasing ions in the solution. The acid release the ion H⁺ and some anion, and the base release the ion OH⁻ and some cation.
In water, the reaction of ammonia is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Because of that, ammonia is an Arrhenius base.
Ammonia acts as an Arrhenius base because it increases the concentration of hydroxide ions in solution through a reaction with water molecules where it accepts a proton, forming NH4+ and OH-. The Arrhenius definition is focused on hydroxide ion production; however, broader definitions like Brønsted-Lowry and Lewis provide a more general understanding of acid-base chemistry.
Ammonia, NH3, acts as an Arrhenius base because it increases the concentration of hydroxide ion, OH-, in aqueous solution by accepting a proton from water. This reaction can be represented by:
NH3(aq) + H2O(l) ⇒ NH4+(aq) + OH-(aq)
In this reaction, ammonia takes a hydrogen ion (H+) from a water molecule to produce an ammonium ion (NH4+) and a hydroxide ion (OH-). Unlike what might be initially assumed, ammonia does not release hydroxide ions into a solution directly; rather, the concentration of hydroxide ions increases due to this proton transfer.
Additionally, more fundamental definitions of acids and bases, beyond the Arrhenius model, include the Brønsted-Lowry theory, where bases are substances that accept hydrogen ions (H+), and the Lewis model, where a base is an electron pair donor.
Overall, while the Arrhenius model was originally seen as limited, it helps to explain how substances like ammonia function as bases without containing hydroxide ions in their formula.
Choose the molecule or compound that exhibits dipole-dipole forces as its strongest intermolecular force. a) BCl3 b) N2 c) H2 d) SO2 e) CBr4.
Answer:
SO₂
Explanation:
The dipole-dipole force is not only determined by the electron density around each atoms in the molecule (dependent of electronegativity difference), but also how the atoms in the molecules are arranged. In general, the more symmetric a molecule is, the less dipole force it exerts as each dipole moments cancels each other out.
Now let's examine each answer
b and c, N₂ and H₂ are composed of same type of atoms, therefore, no dipole moment occurs, and no dipole-dipole forces are exerted a and e, BCl₃ and CBr₄ are composed of atoms with different electronegativity, but are symmetric. With BCl₃ having trigonal planar structure and CBr₄ has tetragonal structure, each B-Cl and C- Br bond cancels out each other dipole moment, and thus, no dipole moments were generated.d, SO₂ structure is not linear, but is a little bent, this allows net dipole moment to occurs in this molecule as dipole moment from each S=O bond do not cancels each other outWhat is the polymerase chain reaction (PCR)?
a. A method to amplify a fragment of DNA.
b. A method to propagate a gene in bacteria.
c. A method to join two fragments of DNA together.
d. A method to determine the sequence of bases in a gene.
Answer: D.) A method to determine the sequence of bases in a gene.
Explanation:Polymerase chain reaction (PCR) is a method widely used in molecular biology to make several copies of a specific DNA segment. Using PCR, copies of DNA sequences are exponentially amplified to generate thousands to millions of more copies of that particular DNA segment
Two different ionic compounds each contain only copper and chlorine. Both compounds are powders, one white and one brown. An elemental analysis is performed on each powder. Which of the following questions about the compounds is most likely to be answered by the results of the analysis?
A. What is the density of each pure compound?
B. What is the formula unit of each compound?
C. What is the chemical reactivity of each compound?
D. Which of the two compounds is more soluble in water?
Answer:
B.
Explanation:
Elemental analysis involves the actual percentage compositions of each element in the sample of the compounds. We know that there are two different compounds here that both contain copper and oxygen. What we don't know is the actual number of copper or oxygen composed in the compounds.
An elemental composition of these compounds would thus tell us this and will give an idea of the chemical formula of each since we now know quite well the percentage composition of each of these elements
The question about the compounds which is most likely to be answered by the results of the analysis is "What is the formula unit of each compound?"
According to the law of multiple proportions, when two elements combine to form different compounds, the proportion of the elements in the different compounds are in simple ratio.
In accordance to this law, the proportion of copper and chlorine in the white and brown compounds vary.
One question that can be answered by an elemental analysis is the question; "What is the formula unit of each compound?"
The elemental analysis shows the amount of each element contained in each of the compounds from which its formula unit can be obtained.
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A possible mechanism for the overall reaction represented above is the following.
NO(g) + NO(g) → N2O2(g) slow
N2O2(g) + O2(g) → 2NO2(g) fast
Which of the following rate expressions agrees best with this possible mechanism?
A) Rate = k[NO]2
(D) Rate = k[NO]2[O2]
(B) Rate = k[NO]
(E) Rate = k[N2O2][O2][O2]
(C) Rate = k[NO]2[O2]
Answer:
A
Explanation:
The slow step of the reaction is the rate determining step. This means to get the rate law for a chemical activity with slow and fast steps, we have to consider the slow step if we are to write the rate law successfully.
Since the compound NO is reacting with itself, we have to raise the value of the concentration in the square bracket by 2. That is why we have the concentration squared.
Answer: it’s c
Explanation:
A bath tub is half full of warm water. How could the thermal energy of the water be increased? A) Add hot water to the tub. B) Heat the water in the tub without adding more water. C) Add more water at the same temperature to the tub. D) Transfer the water to a smaller container.
Answer:
A, B, & C
Explanation:
got it right on edu
The thermal energy of the bathtub can be increased by adding hot water, heating the water, and adding more water to the tub at the same temperature. Thus, options A, B, and C are correct.
What is thermal energy?Thermal energy is the increase in the collision and the kinetic energy of particles of matter due to a rise in the temperature of the substance. This movement produces energy called thermal energy.
The thermal energy of the substance can be increased by providing them with the same energy molecules. The addition of the hot water will increase the energy due to increased collision of the particles of the hot water.
Heating the water will increase the kinetic energy and the rate of a collision resulting in increased thermal energy. Also, water with the same temperature can be added to increase energy.
Therefore, in options A, B, and C adding hot water and heating water increases the thermal energy.
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The ΔHcomb value for anethole is -5539 kJ/mol. Assume 0.840 g of anethole is combusted in a calorimeter whose heat capacity (Calorimeter) is 6.60 kJ/°C at 20.6 °C. What is the final temperature of the calorimeter
Answer:
Final temperature of calorimeter is 25.36^{0}\textrm{C}
Explanation:
Molar mass of anethole = 148.2 g/mol
So, 0.840 g of anethole = [tex]\frac{0.840}{148.2}moles[/tex] of anethole = 0.00567 moles of anethole
1 mol of anethole releases 5539 kJ of heat upon combustion
So, 0.00567 moles of anethole release [tex](5539\times 0.00567)kJ[/tex] of heat or 31.41 kJ of heat
6.60 kJ of heat increases [tex]1^{0}\textrm{C}[/tex] temperature of calorimeter.
So, 31.41 kJ of heat increases [tex](\frac{1}{6.60}\times 31.41)^{0}\textrm{C}[/tex] or [tex]4.76^{0}\textrm{C}[/tex] temperature of calorimeter
So, the final temperature of calorimeter = [tex](20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}[/tex]
To determine the final temperature of the calorimeter, we need to calculate the heat released by the combustion of anethole using its enthalpy change. Then, we can use the equation q = mcΔT to find the change in temperature of the calorimeter. The final temperature is 16.65 °C.
Explanation:To determine the final temperature of the calorimeter, we can make use of the equation q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we need to consider the heat capacity of the calorimeter as well. We can start by calculating the heat released by the combustion of anethole using the given enthalpy change of -5539 kJ/mol. Then, we can use the equation to find the change in temperature of the calorimeter.
First, we need to calculate the moles of anethole. Using the molar mass of anethole, which is 178.26 g/mol, we can find:
moles of anethole = 0.840 g / 178.26 g/mol = 0.004716 molNext, we can calculate the heat released by combustion:
heat released = ΔHcomb x moles of anethole = -5539 kJ/mol x 0.004716 mol = -26.1 kJNow, we can consider the calorimeter's heat capacity:
heat absorbed by the calorimeter = heat released by combustioncalorimeter x ΔT = -26.1 kJΔT = -26.1 kJ / 6.60 kJ/°C = -3.95 °CSince the initial temperature of the calorimeter is 20.6 °C, the final temperature will be:
final temperature = initial temperature + ΔT = 20.6 °C - 3.95 °C = 16.65 °CLearn more about Final temperature of the calorimeter here:https://brainly.com/question/31042074
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Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to the following reaction. Molar Mass Fe2O3 = 159.7 g/mol
Answer:
The percent yield of the reaction is 82%
Explanation:
First step: make the chemist equation.
2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)
As the statement says that aluminun is in excess, the limiting reactant is the Fe2O3
Second step: Find out the moles in the reactant.
Molar weight Fe2O3: 159.7 g/m
Mass / Molar weight = moles
50 g /159.7 g/m = 0.313 moles
Third step: Analyse the reaction. 1 mol of Fe2O3 makes 2 moles of Fe.
1 mol Fe2O3 ____ 2Fe
0.313 mol Fe2O3 ____ 0.626 moles
Molar weight Fe = 55.85 g/m
Moles . molar weight = mass
55.85g/m . 0.626m = 34.9 grams
This will be the 100% yield of the reaction but we only made 28.65 g
34.9 g ____ 100%
28.65 g ____ 82.09 %
Answer:
83%
Explanation:
We first get the chemical reaction :
2 Al(s) + Fe2O3(s) ------> 2Fe(s) + Al2O3(s)
From the reaction we can see that one mole of the oxide yielded 2 moles of iron.
Firstly, we need to calculate the theoretical yield of the iron. This is done as follows. The number of moles of the oxide equals the mass of the oxide divided by the molar mass of the oxide = 50g ÷ 159.7 = 0.313moles
From the first relation, one mole oxide yielded 2 moles iron, hence 0.313 mole oxide will yield 2 × 0.313 mole iron = 0.616 moles
The mass of iron thus generated = 0.616 × 56 = 34.496g
% yield = Actual yield/theoretical yield × 100%
%yield = 28.65/34.396 × 100% = 83%
You sealed an Erlenmeyer flask that was determined to have a volume of 272.2 mL with a stopper. The density of air at 25°C is 0.001185 g/mL. Air has an effective molar mass of 28.96 g/mol. What amount of air was sealed in the flask?
Answer:
The amount of air is 11.1 mmol
Explanation:
Density shows the relation between mass and volume, so you can know the mass which has been sealed in the Erlenmayer.
Density = mass/ volume
Density . volume = mass
0.001185 g/ml . 272.2 ml = 0.322 g
We have the mass now, so the molar weight determinates the mols
Mass/molar weight = mol
0.322g/ 28.96 g/m = 0.0111 mol
To get a better number we can inform on milimols
Mols . 1000 = milimols
The amount of air sealed in the 272.2 mL Erlenmeyer flask is approximately 0.0111 mol, calculated using the given density and molar mass of air at 25°C.
Explanation:To determine the amount of air sealed in the flask, we need to use the density of the air (given as 0.001185 g/mL at 25°C), the volume of the flask and the molar mass of air.
First, using the density and the volume of the flask, we can calculate the mass of the air in the flask with the formula density = mass/volume. We multiply the given density of 0.001185 g/mL by the volume of 272.2 mL to find the mass of the air to be around 0.3224 g.
Next, to convert this mass to moles, we use the given molar mass of air (28.96 g/mol). Moles = mass / molar mass, hence the air in the flask is approximately 0.0111 mol. Thus, there is approximately 0.0111 mol of air sealed in the Erlenmeyer flask.
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One difference between a chemical reaction and a nuclear reaction is that in a nuclear reaction:
a) atoms retain their identity.
b) only the valence electrons are involved.
c) atoms often change from one element to another.
d) only small amounts of energy are absorbed or emitted.
Answer:
C
Explanation:
A nuclear reaction is one in which 2 atoms combine or an atom divides to give rise to entirely different atoms. This is basically the difference it has with a chemical reaction. In a chemical reaction, only new compounds are formed but not new elements. In the case of a nuclear reaction however, even if new elements are not formed, a different atom would be formed.
Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed.
If no reaction occurs, write "NO REACTION."
a. potassium carbonate and lead(II) nitrate
b. lithium sulfate and lead(II) acetate
c. copper(II) nitrate and magnesium sulfide
d. strontium nitrate and potassium iodi
Answer:
a. K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)
b. Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)
c. Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)
d. NO REACTION
Explanation:
For the reactions, the cation and the anion of the compounds will be replaced. The reaction will occur if at least one of the products is insoluble and will form a precipitated.
a. Potassium carbonate = K₂CO₃
Lead(II) nitrate = Pb(NO₃)₂
Products = KNO₃ and PbCO₃.
According to the solubility rules, all K⁺ ions are soluble, with no exceptions, so KNO₃ is soluble. All CO₃⁻² ions are insoluble, and Pb⁺² is not an exception, so PbCO₃ will be insoluble and will form a precipitated, so the reaction happen:
K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)
b. Lithium sulfate = Li₂SO₄
Lead(II) acetate = Pb(C₂H₃O₂)₂
Products = Li(C₂H₃O₂) and PbSO₄
All Li⁺ are solubles, without exceptions, so Li(C₂H₃O₂) is soluble, and all SO₄⁻² are soluble, but Pb⁺² is an exception, so PbSO₄ is insoluble and will form a precipitated, then the reaction happens:
Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)
c. Copper(II) nitrate = Cu(NO₃)₂
Magnesium sulfide = MgS
Products = CuS and Mg(NO₃)₂
All NO₃⁻ are soluble, with no exceptions, so Mg(NO₃)₂ is soluble, and all S⁺² are insoluble, and Cu⁺² is not an exception, so CuS is insoluble, and will form a precipitated, then the reaction happens:
Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)
d. Strontium nitrate = Sr(NO₃)₂
Potassium iodi = KI
Products = K(NO₃)₂ and SrI₂
All K⁺ are soluble, with no exceptions, so K(NO₃)₂ is soluble, and all I⁻ are soluble, and Sr⁺² are not an exception, then SrI₂ is soluble. Therefore, no precipitated is formed and the reaction doesn't happen.
Precipitation reactions occur for some pairs of aqueous solutions, while others do not.
Explanation:a. The combination of potassium carbonate and lead(II) nitrate will result in the formation of a precipitate. The balanced molecular equation for this reaction is:
Pb(NO3)2 + K2CO3 → PbCO3 + 2 KNO3
b. The combination of lithium sulfate and lead(II) acetate will not result in a precipitation reaction.
c. The combination of copper(II) nitrate and magnesium sulfide will result in the formation of a precipitate. The balanced molecular equation for this reaction is:
Cu(NO3)2 + MgS → CuS + Mg(NO3)2
d. The combination of strontium nitrate and potassium iodide will not result in a precipitation reaction.
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Distillation is a process of vaporization a substance and chilling the vapor to collect it back the liquid form. How much heat is removed from 74.2 grams of ethanol vapor at 83 °C (Tb = 78.37 °C) if the collected liquid ethanol has a temperature of 26ºC?
Answer:
72 kJ of heat is removed.
Explanation:
First, the ethanol vapor will reduce its temperature until the temperature of the boiling point, then it will occur a phase change from vapor to liquid, and then the temperature of the liquid will decrease. The total heat will be:
Q = Q1 + Q2 + Q3
Q1 = n*cv*ΔT1, Q2 = m*Hl, and Q3 = n*cl*ΔT2
Where n is the number of moles, cv is the specific heat of the vapor (65.44 J/K.mol, cl is the specific heat of the liquid (111.46 J/K.mol), Hl is the heat of liquefaction (-836.8 J/g), m is the mass, and ΔT is the temperature variation (final - initial).
Q = n*cv*ΔT1 + m*Hl + n*cl*ΔT2
The molar mass of ethanol is 46 g/mol, and the number of moles is the mass divided by the molar mass:
n = 74.2/46 = 1.613 moles
Q = 1.613*65.44*(78.37 - 83) + 74.2*(-836.8) + 1.613*111.46*(26 - 78.37)
Q = -72000 J
Q = -72 kJ (because it is negative, it is removed)
Copper, silver, and gold have all been know since ancient times because they appear in nature in____ and were thus discovered thousands of years ago. But the majority of elements__________ and, consequently, are hard to find in nature.
Answer:
elemental form, readily form compounds
Explanation:
Copper, silver, and gold have all been know since ancient times because they appear in nature in_elemental form___ and were thus discovered thousands of years ago. But the majority of elements__readily form compounds_and, consequently, are hard to find in nature.
Select all that apply. Which two of the following equations represent an exothermic reaction? NH3(g) + 12.0 kcal → ½N2(g) + 3/2H2(g) C(graphite) → C(diamond), ΔH = -0.45 kcal/mole C C + 2S → CS2, ΔH = 27,550 cal CH4 + 2O2 → CO2 + 2H2O + 212,800 cal 2H2 O → 2H2 + O2, ΔH = +58 kcal/mole H2O
Answer:
C(graphite) → C(diamond), ΔH = -0.45
CH4 + 2O2 → CO2 + 2H2O + 212,800 cal
Explanation:
NH3(g) + 12.0 kcal → ½N2(g) + 3/2H2(g) in this reaction we see that energy is added to the reactants so this is an endorthermic process. It takes in energy
C(graphite) → C(diamond), ΔH = -0.45 kcal/mole The change in enthalpy for this reaction is listed and it is negative, telling us that this is an exothermic reaction.
C + 2S → CS2, ΔH = 27,550 cal The change of enthalpy for this reaction is listed and it is positive which tells us that the reaction is endothermic reaction
CH4 + 2O2 → CO2 + 2H2O + 212,800 cal In this reaction, the energy is part of the equation on the products side. It is given out. this tells us that the reaction is exothermic
2H2 O → 2H2 + O2, ΔH = +58 kcal/mole H2O In this reaction, the energy is part of the equation on the products side. It is given out. this tells us that the reaction is exothermic