Answer:
0.00564 moles
Explanation:
Given that:
The rate constant, k = [tex]6.82\times 10^{-3}[/tex] s⁻¹
Initial concentration [A₀] = [tex]2.90\times 10^{-2}[/tex] mol
Time = 4.0 min = [tex]4.0\times 60[/tex] sec = 240 sec
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
So,
[tex][A_t]=2.90\times 10^{-2}\times e^{-6.82\times 10^{-3}\times 240}=2.9\times \frac{1}{10^2}\times \frac{1}{e^{1.6368}}[/tex]
[tex][A_t]=0.00564\ moles[/tex]
The concentration after four minutes is 3.3 ×10−3.
Let us recall that for a first order reaction;
ln[A] = ln[A]o - kt
Where;
[A] = concentration at time t
[A]o = initial concentration
k = rate constant
t = time
[A]o = 2.90×10−2 mol/1.7 L = 0.0171 M
k = 6.82×10−3 s−1
t = 4 min or 240 s
Substituting values;
ln[A] = ln[0.0171 M] - (6.82×10−3 s−1 × 240 s)
[A] = e^ln[0.0171 M] - (6.82×10−3 s−1 × 240 s)
[A] = 3.3 ×10−3.
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There are three ways in which to define acids and bases: the Arrhenius concept, the Brønsted-Lowry concept, and the Lewis concept. Arrhenius acids are substances that, when dissolved in water, increase the concentration of the H+ ion; Arrhenius bases are substances that, when dissolved in water, increase the concentration of the OH− ion. Brønsted-Lowry acids are substances that can donate a proton (H+) to another substance; Brønsted-Lowry bases are substances that can accept a proton (H+). A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor. Part A Using the Arrhenius concept of acids and bases, identify the Arrhenius acid and base in each of the following reactions: 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l) NH3(g)+HCl(g)→NH4Cl(s)
Answer:
(a) [tex]KOH[/tex] acts as base.
[tex]H_2SO_4[/tex] acts as acid.
(b) [tex]NH_3[/tex] acts as base.
[tex]HCl[/tex] acts as acid.
Explanation:
Arrhenius theory:-
The Arrhenius theory was introduced introduced by Swedish scientist named Svante Arrhenius in 1887.
According to the theory, acids are the substances which dissociate in the aqueous medium to produce electrically charged atoms ( may be molecule). Out of these species furnished, one must be a proton or the hydrogen ion, [tex]H^+[/tex].
Base are the substances which dissociate in the aqueous medium to produce electrically charged atoms ( may be molecule). Out of these species furnished, one must be a hydroxide ion, [tex]OH^-[/tex].
Thus, according to the reaction:-
(a)
[tex]2KOH_{(aq)}+H_2SO_4_{(aq)}\rightarrow K_2SO_4_{(aq)}+2H_2O_{(l)}[/tex]
[tex]KOH[/tex] dissociates as:-
[tex]KOH\rightarrow K^++OH^-[/tex] and hence, acts as base.
[tex]H_2SO_4[/tex] dissociates as:-
[tex]H_2SO_4\rightarrow 2H^++SO_4^{2-}[/tex] and hence, acts as acid.
(b)
[tex]NH_3_{(g)}+HCl_{(g)}\rightarrow NH_4Cl_{(s)}[/tex]
[tex]NH_3[/tex] dissociates as:-
[tex]NH_3+H_2O\rightarrow NH_4^++OH^-[/tex] and hence, acts as base.
[tex]HCl[/tex] dissociates as:-
[tex]HCl\rightarrow H^++Cl^{-}[/tex] and hence, acts as acid.
Carbon Monoxide at 25°C and steam at 150°C are fed to a continuous water-gas shift reactor. The product gas, which contains 50.0 mole% H2, 40.0% ????????2, and the balance H2????, emerges at 500°C at a rate of 2.50m3/h and goes to a condenser. The gas and liquid streams leaving the condenser are in equilibrium at 15°C and 1 bar. The liquid may be taken to be pure w"a) Calculate the % excess steam fed to the reactor and rate of condensation of the water (kg/h) leaving the condensor.b) Calculate the rate (kW) at which heat must be removed from the condensor.c) Calculate the rate of heat transfer (kW) to or from the reactor (state which it is).
The final answer for the solution is Q= -8.8171*[tex]10^{-4}[/tex] kW
Explanation:
[tex]CO +H_2O ⇄ CO_2+H_2[/tex]
basic:
100 moles of product gas
[tex]CO_2[/tex] = 40 moles
[tex]H_2[/tex] = 40 moles
[tex]H_20[/tex] = 20 moles
for 40 moles of[tex]CO_2[/tex] we need 40 moles of CO and 40 moles of [tex]H_2O[/tex]
therefore,
Inlet :
No of moles of CO = 40
No of moles of [tex]H_20[/tex] = 40
but [tex]H_20[/tex] is in the outlet stream = 20 modes
Excess = [tex]\frac{20}{40}\ \times 100[/tex]%
= 50%
50% excess stream is fed to the reactor
a) Rate of conversion:
ρ_water (density ) =[tex]\frac{1g}{cm^{3}}[/tex]
Given volume rate = 3.5 [tex]\frac{cm^{3} }{hr}[/tex]
3.5 * density = 3.5 * [tex]10^{-3}[/tex] [tex]\frac{kg}{hr}[/tex]
b) Rate (kW)
ΔH = ΔH_1 + ΔH_2 +ΔH_3
Rate of condensation Q° =n° ΔH
Q = n° (ΔH_1 + ΔH_2 +ΔH_3)
ΔH_1 is sensible heat
V =[tex]\int\limits^T_S {C_p(T) } \, dT[/tex]
C_p(T) = a + bT + CT^2 +dT^3
H_2O
a=32.218
b=o.192 * 10^-2
c= 1.055 * 10^-5
d= -3.593 * 10^-9
ΔH_2 = ∫(a+bT +a^2 +dT^3)dT (of limits 100 to 500 )
= -13497.6
ΔH_3 =∫(a +bT +T^2 +dT^3)dT (of limit 95 to 100)
= -2751.331
phase change
ΔH_2 = - ΔH_v
ΔH_v = 7.3 @ 100° C
liquid → vapor ⇒ ΔH_v
vapor → liquid ⇒ -ΔH_v
Q= n°(-16324.231)
=350 *[tex]10^{-5}[/tex] (-16324.231)
=[tex]\frac{350* 10^{-5} }{18}[/tex] (-16324.231) *[tex]\frac{kg}{hr}[/tex][tex]\frac{J}{mol}[/tex]
=[tex]\frac{350* 10^{-5} }{18}[/tex] (-16324.231) * \frac{J}{mol}[/tex]
=[tex]\frac{350* 10^{-5} }{18}[/tex] (-16324.231) *kW
Q= -8.8171*[tex]10^{-4}[/tex] kW
The solution involves various calculations related to a chemical reactor, including determining the excess steam fed into the reactor, condensation rate, heat removal rate, and heat transfer rate.
Explanation:This is a complex chemical engineering problem involving numerous calculations, including finding the excess steam fed into the reactor, condensation rate, heat removal rate, and heat transfer rate. Firstly, the percentage excess steam is found by comparing the stochiometric requirement of the reaction with actual steam flow. Next, we use the molar mass of water and the gas constant to find the condensation rate of water. For the heat removal rate, we need to consider the latent heat of condensation along with the energy balance of the condenser.
Lastly, for the heat transfer rate in the reactor, we need to take into account the enthalpy of reaction, heat capacities of reactants and products, and the reactor temperature profile to decide whether heat is added or removed from the reactor.
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The great French chemist Antoine Lavoisier discovered the Law of Conservation of Mass in part by doing a famous experiment in 1775. In this experiment Lavoisier found that mercury(II) oxide, when heated, decomposed into liquid mercury and an invisible and previously unknown substance: oxygen gas.
1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid mercury(II) oxide (HgO) into liquid mercury and gaseous dioxygen.
2. Suppose 50.0ml of dioxygen gas are produced by this reaction, at a temperature of 90°C and pressure of exactly 1atm. Calculate the mass of mercury(II) oxide that must have reacted. Be sure your answer has the correct number of significant digits.
Answer:
(a) [tex]2HgO_{(s)}\rightarrow 2Hg_{(l)}+O_2_{(g)}[/tex]
(b) 0.726 g
Explanation:
(a)
The balanced chemical equation is shown below as:-
[tex]2HgO_{(s)}\rightarrow 2Hg_{(l)}+O_2_{(g)}[/tex]
(b)
Given that:
Temperature = 90 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (90 + 273.15) K = 363.15 K
V = 50.0 mL = 0.05 L ( 1 mL = 0.001 L )
Pressure = 1 atm
Using ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
1 atm × 0.05 L = n ×0.0821 L atm/ K mol × 363.15 K
⇒n = 0.001677 mol
Thus, moles of [tex]O_2[/tex] = 0.001677 mol
According to the reaction shown above,
1 mole of [tex]O_2[/tex] is produced when 2 moles of mercury(II) oxide are reacted
0.001677 mole of [tex]O_2[/tex] is produced when [tex]2\times 0.001677[/tex] moles of mercury(II) oxide are reacted
Moles of mercury(II) oxide = 0.003354 moles
Molar mass of mercury(II) oxide = 216.59 g/mol
Mass of mercury(II) oxide = [tex]Moles\times Molar\ mass[/tex] = [tex]0.003354\times 216.59\ g=0.726\ g[/tex]
0.726 g is the mass of mercury(II) oxide that must have reacted.
The balanced chemical equation for the decomposition of mercury(II) oxide into mercury and dioxygen is: 2HgO(s) → 2Hg(l) + O2(g). The mass of mercury(II) oxide that reacted to produce 50.0 ml of dioxygen gas at 90°C and 1 atm pressure is approximately 0.97 g.
Explanation:1. The balanced chemical equation for the reaction is: 2HgO(s) → 2Hg(l) + O2(g).
2. First, we need to use the ideal gas law to calculate the number of moles of dioxygen gas produced. At standard temperature and pressure (STP), one mole of any gas occupies a volume of 22.4 L. Therefore, the number of moles of dioxygen is 50.0 ml / 22.4L/mol = 0.00223 moles. Next, we use the balanced chemical equation, which tells us that one mole of HgO produces a half mole of dioxygen gas. So, the amount of HgO reacted is 2 x 0.00223 moles = 0.00446 moles. Finally, we use the molar mass of HgO (216.6 g/mol) to calculate the mass of HgO reacted: 0.00446 moles x 216.6 g/mol = 0.97 g (to three significant digits).
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Write electron (subshell) configurations for the following atoms: a.15P b. 19K c. 28Ni d. 35Br e. 58Ce
Answer:
15P) 3s² 3p³
19K) 4s¹
28Ni) 3d⁸ 4s²
35Br) 3d¹⁰ 4s² 4p⁵
58Ce) 4f¹ 5d¹ 6s²
Explanation:
1) 15P
Phosphorus has 15 electrons
The order of filling of the energy levels is 1s, 2s, 2p, 3s, 3p, 4s, . . .
The 15 electrons of the phosphorus atom will fill up to the 3p orbital, which will contain three electrons
The electron configuration for 15P is: 1s2 2s2 2p6 3s2 3p3
3s² 3p³
2) 19K
Potassium has 19 electrons
The electron configuration for 19K is: 1s2 2s2 2p6 3s2 3p6 4s1
4s¹
3) 28Ni
Nickel has 28 electrons
The electron configuration for 28Ni is: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 8
3d⁸ 4s²
4) 35Br
Bromine has 35 electrons
The electron configuration for 35Br is: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
3d¹⁰ 4s² 4p⁵
5) 58Ce
Cerium has 58 electrons
The electron configuration for 58Ce is: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f1 5d1
4f¹ 5d¹ 6s²
Use the kinetic-molecular theory to explain the following phenomena:a.When 1 mol of a real gas is condensed to a liquid, the volume shrinks by a factor of about 1000.Molecules in a gas are far apart. They are much closer together in a liquid. Molecules in a gas are easily squeezed closer together as the gas is compressed
Answer:
The particles in a gas are comparatively very far from each other. Further an ideal gas has not interactions between the particles i.e repulsion. This explains why a gas that is compressed into a liquid collapses into a volume that is small by a factor of 1000. The particles get much closer to each other to form a liquid and they do not repel each other
If you have 10 grams of Lithium Oxide what will the volume be? Show your work
Answer:
Volume occupied by 10 grams of Lithium Oxide at STP is 7.52 L
Explanation:
STP (Standard Temperature and Pressure) : It is used while performing calculation on gases and provide standards of measurements while performing experiment.
According to IUPAC (International Union of Pure And applied Chemistry) , standard temperature is 273.15 K and Pressure is 100 kPa (0.9869 atm). At STP condition 1 mole of substance occupies 22.4 L volume .
Molar mass of Lithium Oxide = 29.8 g/mol
[tex]Li_{2}O[/tex] = 2(6.9) + 15.99 = 29.8
Mass of 1 mole [tex]Li_{2}O[/tex] = 29.8 g
1 mole [tex]Li_{2}O[/tex] occupies, Volume = 22.4 L
29.8 g [tex]Li_{2}O[/tex] occupies, V = 22.4 L
1 g [tex]Li_{2}O[/tex] occupies ,V = [tex]\frac{22.4}{29.8}[/tex]
1 g [tex]Li_{2}O[/tex] occupies ,V = 0.7516 L
10 g tex]Li_{2}O[/tex] occupies ,V = [tex]0.7516 \times10[/tex] L
V = 7.52 L
So, volume occupied by Lithium Oxide At STP is 7.52 L
From this list, choose all of the ionic compounds. Check all that apply. View Available Hint(s) Check all that apply. POCl3 KOCH2CH3 CH3OH SOCl2 CH3CH2CO2Na NaOH CH2
Answer: [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa[/tex], NaOH are all ionic compounds.
Explanation:
It is known that ionic compounds are the compounds in which one atom transfer its valence electrons to another atom. Hence, during this transfer partial opposite charges tend to develop on the combing atoms due to which strong force of attraction exists between the atoms.
An ionic bond is always formed between a metal and a non-metal.
For example, [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa[/tex], NaOH are all ionic compounds.
On the other hand, a compound formed due to sharing of electrons between the combining atoms is known as a covalent compound. Generally, a non-metal with same or different non-metal tends to form a covalent bond.
For example, [tex]POCl_{3}[/tex], [tex]SOCl_{2}[/tex] etc are all covalent compounds.
Thus, we can conclude that [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa, NaOH are all ionic compounds.
Select the true statements about the electron transport chain.
a. The major reactants in the electron transport chain are O 2 and either NADH or FADH 2 .
b. The electron transport chain produces two ATP.
c. In the electron transport chain, a series of reactions moves electrons through carriers.
d. The electron transport chain is an anaerobic process.
e. Coenzyme Q and cytochrome c are components of the electron transport chain.
Final answer:
The true statements about the electron transport chain are (a) it uses O₂ and NADH or FADH2 as reactants, (c) it involves a series of reactions moving electrons through carriers, and (e) Coenzyme Q and cytochrome c are components of the chain. Statements (b) and (d) are false.
Explanation:
Among the given statements about the electron transport chain (ETC), the following are true:
(a) The major reactants in the electron transport chain are O₂ and either NADH or FADH2.(c) In the electron transport chain, a series of reactions moves electrons through carriers.(e) Coenzyme Q and cytochrome c are components of the electron transport chain.The electron transport chain is a series of electron carriers that transfer electrons from NADH and FADH2 to oxygen, with the formation of water and ATP as end products. It involves multiprotein complexes and shuttle electron carriers in the inner mitochondrial membrane and is a crucial step in aerobic respiration. The electron transport chain is not an anaerobic process (d is false), and it produces significantly more than two ATP as stated in (b), which is also false.
A direct-current power supply of low voltage (less than 10 volts) has lost the markings that indicate which output terminal is positive and which is negative. A chemist suggests that the power supply terminals be connected to a pair of platinum electrodes that dip into 0.1-molar KI solution. Which of the following correctly identifies the polarities of the power supply terminals?
(A) A gas will be evolved only at the positive electrode.
(B) A gas will be evolved only at the negative electrode.
(C) A brown color will appear in the solution near the negative electrode.
(D) A metal will be deposited on the positive electrode.
(E) None of the methods above will identify the polarities of the power supply terminals.
Answer:
A
Explanation:
As electrolysis progresses, the iodide ion migrates towards the anode (positive electrode) and is discharged as iodine gas while the metal migrates towards the cathode (negative electrode). The evolution of gas at the positive electrode signifies the reaction:
2I-=I2 +2e
The brown color formed near an electrode in a KI solution during electrolysis indicates the positive terminal due to oxidation of iodide ions. However, the provided choices have an incongruity as the brown color appears near the positive electrode, not the negative one, as option (C) incorrectly states.
To identify the polarities of the power supply terminals using a 0.1-molar KI solution with platinum electrodes, we need to understand the processes occurring at each electrode during electrolysis. When direct current is applied, oxidation occurs at the anode (positive electrode) and reduction occurs at the cathode (negative electrode).
For a solution containing potassium iodide (KI), oxidation at the anode results in the liberation of iodine, which can cause a brown coloration in the solution near the positive electrode. At the cathode, reduction occurs, which could lead to the evolution of hydrogen gas. Hence, the brown color appearing near the anode indicates that it is the positive terminal.
If the terminals are connected and the solution becomes brown near one of the electrodes, we can identify that electrode as the anode (positive). Therefore, the correct answer is:
(C) A brown color will appear in the solution near the negative electrode. However, this statement seems to contradict the previous explanation—it should state that the brown color appears near the positive electrode due to the oxidation of iodide ions. Hence, the correct phenomenon described does not match any of the given answer choices perfectly, and ambiguities in (C) suggest that a more detailed investigation may be required to ensure the most accurate determination of the terminal polarities.
A certain metal fluoride crystallizes in such a way that the fluoride ions occupy simple cubic lattice sites, while the metal ions occupy the body centers of half the unit cells. The formula for the metal fluoride is
Answer:
[tex]MF_2[/tex]
Explanation:
In a simple cubic lattice lattice, the atoms are present at the eight corners of the cibe.
Since it is mentioned that the fluorine is present at the corners and also 1 corners are shared by 8 unit cells. So, share of atom in one unit cell is:- [tex]8\times \frac{1}{8}=1[/tex]
Also, the metal, M occupy half of the body centre. A cube has only one body centre. So, share of M in each unit cell:- [tex]\frac{1}{2}[/tex]
Thus, the formula is:-
[tex]M_{\frac{1}{2}}F[/tex] Or simplifying [tex]MF_2[/tex]
Answer:
The formula of the metal fluoride is = [tex]MF_2[/tex]
Explanation:
given that fluoride ions occupy simple cubic lattice and metal ions occupy the body center of half the cube
therefore,
The number of [tex]F^-[/tex] ions per unit cell = [tex]\frac{1}{8}*1[/tex]
[tex]= \frac{1}{8}[/tex]
There is only one body center per unit cell
The total number of metal ions per unit cell = [tex]\frac{1}{2}*1[/tex]
[tex]= \frac{1}{2}[/tex]
Therefore the formula for metal fluoride
[tex]= MF_2[/tex]
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Bombardment of cobalt-59 with a neutron produces a manganese-56 atom and another particle. What is this particle?A. an alpha particleB. a beta particle (electron)C. a positronD. a gamma rayE. a neutron
Answer:A
Explanation:
A nuclear reaction is balanced by ensuring that the Masses and charges of te reactants and products are exactly balanced on the left and right hand side of the reaction equation. If there are 60 mass units on the LHS and manganese has only 56 mass units then four mass units are left. If there is no charge on a neutron and there is a charge of 27 on the cobalt, then two charges are left. Four mass units and a charge of +2 corresponds to a helium which is actually an alpha particle.
In the nuclear reaction resulting from the bombardment of cobalt-59 with a neutron, a manganese-56 atom and a beta particle (electron) are produced.
Explanation:The bombardment of cobalt-59 with a neutron produces a manganese-56 atom and another particle. This process involves a certain type of nuclear reaction known as beta decay. In beta decay, a neutron is converted into a proton and a beta particle, also known as an electron.
Considering the atomic number and mass number balances in this nuclear reaction, the other particle produced can be determined. Cobalt (Co) has an atomic number of 27 and a mass number of 59. When it captures a neutron and becomes manganese (Mn), it has an atomic number of 25 and a mass number of 56. So, there is a difference of two units in the atomic number and three units in the mass number. Therefore, this leftover particle that is formed due to this difference is a beta particle or electron (B), making the correct answer Option B.
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Suppose of potassium nitrate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.
Answer:
0.0269 M
Explanation:
There is some info missing. I think this is the original question.
Suppose 0.816 g of potassium nitrate is dissolved in 300 mL of a 14.0 mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.
The molecular equation corresponding to this reaction is:
2 KNO₃(aq) + Na₂CrO₄(aq) ⇄ K₂CrO₄(aq) + 2 NaNO₃(aq)
The full ionic equation is:
2 K⁺(aq) + 2 NO₃⁻(aq) + 2 Na⁺(aq) + CrO₄²⁻(aq) ⇄ 2 K⁺(aq) + CrO₄²⁻(aq) + 2 Na⁺(aq) + 2 NO₃⁻(aq)
As we can see, the moles of K⁺ are equal to the initial moles of KNO₃. The molar mass of KNO₃ is 101.10 g/mol. The moles of KNO₃ (and K⁺) are:
0.816 g × (1 mol/ 101.10 g) = 8.07 × 10⁻³ mol
The molarity of K⁺ is:
8.07 × 10⁻³ mol / 0.300 L = 0.0269 M
Effective nuclear charge, Zeff, is defined as:
Zeff=Z−S
where Z is true nuclear charge and S is the amount of shielding.
In 1930, John C. Slater devised the following set of empirical rules to estimate S for a designated ns or np electron:
1. Write the electron configuration of the element, and group the subshells as follows: (1s), (2s, 2p), (3s, 3p), (3d), (4s, 4p), (4d), (4f ), (5s, 5p), and so on.
2. Electrons in groups to the right of the (ns, np) group contribute nothing to the shielding constant for the designated electron.
3. All the other electrons in the (ns, np) group shield the designated electron to the extent of 0.35 each.
4. All electrons in the n−1 shell shield to the extent of 0.85 each.
5. All electrons in the n−2 shell, or lower, shield completely—their contributions to the shielding constant are 1.00 each.
When the designated electron is in an nd or nf group, rules (i), (ii), and (iii) remain the same but rules (iv) and (v) are replaced by the following: Each electron in a group lying to the left of the nd or nf group contributes 1.00 to the shielding constant. These rules are a simplified generalization based on the average behavior of different types of electrons.
Part A) Calculate Zeff for a valence electron in an oxygen atom.
Answer:
The effective nuclear charge for a valence electron in oxygen atom: [tex]Z_{eff} = 4.55[/tex]
Explanation:
Effective nuclear charge [tex][Z_{eff}][/tex] is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.
It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s).
[tex]\Rightarrow Z_{eff} = Z - s[/tex]
For an oxygen atom-
Electron configuration: (1s²) (2s² 2p⁴)
The atomic number (actual nuclear charge): Z = 8
The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:
⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45
Therefore, the effective nuclear charge for a valence electron in oxygen atom is:
[tex]Z_{eff} = Z - s = 8 - 3.45 = 4.55[/tex]
Therefore, the effective nuclear charge for a valence electron in oxygen atom: [tex]Z_{eff} = 4.55[/tex]
To calculate the effective nuclear charge (Zeff) for a valence electron in an oxygen atom, which has an atomic number of 8, we use Slater's rules to determine the shielding constant (S) and subtract it from Z. The result is Zeff = 8 - (2 * 0.85 + 5 * 0.35) = 4.55.
The effective nuclear charge (Zeff) for an electron can be determined using Slater's rules which consider the actual nuclear charge (Z) and the shielding constant (S). To calculate Zeff for a valence electron in an oxygen atom with an atomic number of 8:
Firstly, identify the electron configuration of oxygen which is 1s2 2s2 2p4.Group the subshells as Slater's rules suggest.Since we are interested in a valence electron in the 2p orbital, the two 1s electrons shield with a factor of 0.85 each, and the six electrons in the 2s and 2p orbitals (excluding the electron in question) shield with a factor of 0.35 each.Therefore, the shielding constant S is calculated as 2(0.85) + 5(0.35).The effective nuclear charge Zeff is then Z - S, which equals 8 - (2 * 0.85 + 5 * 0.35) = 4.55.The calculated Zeff for a valence electron in an oxygen atom is 4.55.
How many grams of nickel metal are plated out when a constant current of 15.0
A is passed through aqueous NiCl2 for 60.0 minutes?
A) 10.9 gB) 16.4 gC) 32.8 gD) 36.3 g
Answer:
B) 16.4
Explanation:
Given that:-
Current, I = 15.0 A
Time, t = 60.0 minutes
Also, 1 minute = 60 seconds
So, t = [tex]60\times 60[/tex] s = 3600 s
F is Faraday constant = 96485 C
Atomic weight of Nickel = 58.69 g/mol
Also, Charge on Ni in [tex]NiCl_2[/tex] = 2
So, equivalent weight of Ni , E = [tex]\frac{58.69}{2}\ g/mol[/tex] = 29.34 g/mol
Thus, according to the Faraday's Law:-
[tex]W=\frac{EIt}{96485}[/tex]
Where, W is the mass of the metal deposited.
So,
[tex]W=\frac{29.34\times 15\times 3600}{96485}\ g=16.4\ g[/tex]
Weight of Ni metal plated out = 16.4 g
Final answer:
To determine the mass of nickel metal plated out, use Faraday's law of electrolysis and the given values of current and time. The answer is 16.4 grams.
Explanation:
To determine the number of grams of nickel metal plated out, we need to use Faraday's law of electrolysis. The formula is:
Mass (g) = (Current (A) * Time (s)) / (Molar mass (g/mol) * Faraday's constant (C/mol))
Given that the current is 15.0 A and the time is 60.0 minutes (or 3600 seconds), we can use the molar mass of nickel (58.69 g/mol) and Faraday's constant (96,500 C/mol) to calculate the mass:
Mass (g) = (15.0 A * 3600 s) / (58.69 g/mol * 96,500 C/mol) = 16.4 g
Therefore, the answer is Choice B) 16.4 g.
CH4 and CH3CH2CH3 have the lowest boiling points because they experience only \rm CH_4 and \rm CH_3CH_2CH_3 have the lowest boiling points because they experience only _____________, the effects of which increase with increasing ______., the effects of which increase with increasing \rm CH_4 and \rm CH_3CH_2CH_3 have the lowest boiling points because they experience only _______________, the effects of which increase with increasing _____________..
Answer:
Dispersion forces
Relative molecular mass
Explanation:
Alkanes experience only dispersion forces. Dispersion forces increase with increasevin the relative molecular mass of the compounds. Hence a higher relative molecular mass implies greater dispersion forces and a greater boiling point.
Methane (CH4) and Propane (CH3CH2CH3) have low boiling points because they only experience weak dispersion forces, which increase with increasing molecular size. This is due to their small size which leads to weaker forces and, thereby, lower boiling points.
Explanation:Methane (CH4) and Propane (CH3CH2CH3) have the lowest boiling points due to their experience of London dispersion forces, which increase with increasing molecular size. Dispersion forces are weak intermolecular forces that exist between all molecules, regardless of polarity, which are stronger in larger molecules due to greater electron cloud polarizability. Therefore, the reason why compounds like CH4 and CH3CH2CH3 have low boiling points is that they are small in size, leading to weaker dispersion forces and, thus, lower boiling points.
For example, the molar masses of CH4, SiH4, GeH4, and SnH4 are approximately 16 g/mol, 32 g/mol, 77 g/ mol, and 123 g/mol, respectively. Therefore, it can be inferred that CH4 with the lowest molar mass out of the group will have the lowest boiling point, and SnH4 with the highest molar mass would exhibit the highest boiling point.
Ultimately, the relationship between molecular size and type of forces impacts the boiling point of molecular compounds. Nonpolar compounds like CH4 and CH3CH2CH3 only experience dispersion forces, the effects of which increase with increasing molecular size.
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A chemist makes 240. mL of zinc nitrate (Zn(NO) working solution by adding distilled water to 70.0 mL of a 2.50stock solution of zinc nitrate in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.
Answer:
M = 0.729 M
Explanation:
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dillute solution of this reactant.
The stock solution of the nitrate has a concentration of 2.5 M, and he wants to prepare 240 mL of a more dillute concentration of the same solution. He adds 70 mL of the stock and complete with water until it reach 240 mL.
We want to know the concentration of this dilluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the dilluted solution so:
n1 = n2 (1)
and we also know that:
n = M*V (2)
If we replace this expression in (1) we have:
M1*V1 = M2*V2
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know concentration and volume used of the stock solution and the desired volume of the dilluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M2:
2.5 * 70 = 240M2
M2 = 2.5 * 70 / 240
M2 = 0.729 MThis is the concentration of the solution prepared.
Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the copper(II) acetate is dissolved in it. Be sure your answer has the correct number of significant digits.
Answer:
0.0714 M for the given variables
Explanation:
The question is missing some data, but one of the original questions regarding this problem provides the following data:
Mass of copper(II) acetate: [tex]m_{(AcO)_2Cu} = 0.972 g[/tex]
Volume of the sodium chromate solution: [tex]V_{Na_2CrO_4} = 150.0 mL[/tex]
Molarity of the sodium chromate solution: [tex]c_{Na_2CrO_4} = 0.0400 M[/tex]
Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:
[tex](CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)[/tex]
Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:
[tex]n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol[/tex]
Moles of the sodium chromate solution would be found by multiplying its volume by molarity:
[tex]n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol[/tex]
Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.
Write the net ionic equation for this reaction:
[tex]Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)[/tex]
Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:
[tex]n_{(AcO)_2Cu} = 0.0053515 mol[/tex]
According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:
[tex]n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol[/tex]
The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:
[tex]c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M[/tex]
The final molarity of the acetate anion in the solution, after copper(II) acetate is dissolved in a sodium chromate solution, is 0.1 M.
Explanation:The acetate anion, CH3 CO₂¯, found in copper(II) acetate, is the conjugate base of acetic acid. When copper(II) acetate is dissolved in a solution of sodium chromate, it yields a solution of inert cations and weak base anions, resulting in a basic solution. Given the initial acetate concentration, if the newly formed acetate ion gives a final acetate concentration of (1.0 × 10−²) + (0.01 × 10−²) = 1.01 × 10-² mol NaCH3 CO₂, this would mean there are both 9.9 × 10-3 mol/L of copper(II) acetate and 1.01 × 10-2 mol/L of sodium chromate.
To compute the molarity of acetate anion, you can divide the moles of the acetate by the volume of the solution, yielding a final molarity of the acetate anion in the solution. Hence the molarity of acetate anion in the solution would be 1.01 x 10^-2 mol /0.101 L = 0.1 M.
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Express the equilibrium constant for the following reaction.2 K(s) + 2 H2O(l) ↔ 2 KOH(aq) + H2(g)K = [H2][KOH]^2K = [KOH]^2[H2]/[K]^2[H2O]^2K = [H2][KOH]^-2K = [K]^2[H2O]^2/[KOH]^2[H2]K = [KOH]^1/2[H2]/[K]^1/2[H2O]^1/2
The equilibrium constant for the reaction 2 K(s) + 2 H2O(l) ↔ 2 KOH(aq) + H2(g) is Kc = [KOH]^2[H2], where Kc represents the ratio of the concentration of the products to the reactants.
Explanation:In the given chemical reaction, 2 K(s) + 2 H2O(l) ↔ 2 KOH(aq) + H2(g), the equilibrium constant, denoted by Kc, represents the ratio of the concentrations of the products to the reactants. Compounds in the solid or liquid state, like K(s) and H2O(l) in this case, are usually not included in the Kc expression. Hence, the equilibrium constant for this reaction would be expressed as:
Kc = [KOH]^2[H2]
This equation indicates that the equilibrium constant is the square of the concentration of KOH, times the concentration of H2. It is important to note that the coefficients become exponents in the equilibrium constant expression, hence [KOH] is squared.
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Final answer:
In chemistry, students are often asked to express equilibrium constants for reactions. This involves determining the respective equilibrium constant expression for each reaction given. Each equilibrium constant formula is unique to the specific reaction and helps understand the extent of the reaction at equilibrium.
Explanation:
Equilibrium Constant (K) Expressions:
For reaction a: K = [NO₂]² / [NO]²[O₂]
For reaction b: K = [NO₂]²[H₂O] / [N₂H₂][O₂]³
For reaction c: K = [H₂][O₂]² / [O₂][H₂O]
A galvanic cell consists of a Ni2+/ Ni half-cell and a standard hydrogen electrode.
If the Ni2+/ Ni half-cell standard cell functions as the anode, and the standard cell potential is 0.26 V, what is the standard reduction potential for the Ni2+/Ni half-cell?
A) - 0.26 V
B) - 0.13 V
C) + 0.13 V
D) + 0.26 V
Answer:
A) - 0.26 V
Explanation:
Here Ni undergoes oxidation by loss of electrons, thus act as anode. Standard hydrogen electrode undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
Given that:- [tex]E^0=0.26\ V[/tex]
Also, [tex]E^0_{[H^{+}/H_2]}=0\ V[/tex]
So,
[tex]E^0=E^0_{[H^{+}/H_2]}-E^0_{[Ni^{2+}/Ni]}[/tex]
[tex]0.26\ V=0\ V- E^0_{[Ni^{2+}/Ni]}[/tex]
[tex]E^0_{[Ni^{2+}/Ni]}=-0.26\ V[/tex]
The standard reduction potential for the Ni2+/Ni half-cell is - 0.26 V.
What is standard hydrogen electrode?The standard hydrogen electrode is a reference electrode that was arbitrarily assigned an electrode potential of 0.00V. We know that metals that are above hydrogen in the electrochemical series will always have a negative electrode potential.
As such, the standard reduction potential for the Ni2+/Ni half-cell is - 0.26 V.
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4 kmol Methane (CH4) is burned completely with the stoichiometric amount of air during a steady-flow combustion process. If both the reactants and the products are maintained at 25°C and 1 atm and the water in the products exists in the liquid form, determine the heat transfer from the combustion chamber during this process. What would your answer be if combustion were achieved with 100 percent excess air?
Answer:
Heat transfer = 3564 Jolues
The same value
Explanation:
The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):
Qtransfer = - mol x ΔHºc Qtransfer
So look up in appropiate reference table ΔHºc and solve the problem:
ΔHºc = - 891 kJ/mol
Qtransfer = - (4 x 10³ mol x -891 kJ/mol ) = 3564 J
if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.
Final answer:
The heat transfer from the combustion chamber when 4 kmol of methane is burned completely with stoichiometric air is 3561.6 kJ. With 100 percent excess air, the heat transfer remains the same as excess air does not change the heat released from the combustion reaction.
Explanation:
Heat Transfer During Combustion of Methane
To determine the heat transfer from the combustion chamber when 4 kmol of methane (CH4) is burned with the stoichiometric amount of air, we must use the thermochemical equation for the combustion of methane. The equation indicates that when 1 mol of methane is combusted, 890.4 kilojoules of heat is released. Therefore, for 4 kmol, the heat release would be 4 kmol × 890.4 kJ/kmol = 3561.6 kJ.
If the combustion is achieved with 100 percent excess air, the amount of heat transferred would not change because the stoichiometry of the reaction remains consistent and the same amount of methane is being burned. However, additional air does not contribute to the heat of the reaction unless it causes incomplete combustion or other side reactions, which in this case is disregarded.
Since the conditions specify that reactants and products are maintained at 25°C and 1 atm, and the water in the products is in the liquid form, we can assume that no extra energy is required for heating up or cooling down the substances and the entire heat of combustion is transferred out from the system.
In a saturated solution of Zn(OH)2at 25^oC the value of [OH–] is 2.0 x 10^–6M. What is the value of the solubility-product constant, Ksp, for Zn(OH)2at 25^oC?
(A) 4.0 x 10–18
(B) 8.0 x 10–18
(C) 1.6 x 10–17
(D) 4.0 x 10–12
(E) 2.0 x 10–6
Answer:B
Explanation:
The solubility product is the product of the concentration of all the ions present as found in the rate equation. The solubility product is the equilibrium expression that shows the extent to which a substance dissolves in water. The full solution is shown in the image attached.
Answer:
4.0E^-18
which makes the answer A
1. From the following standard electrochemical potential (and your knowledge of biomedically important metals), answer the questions below. Reaction ΔEº(V) Ti -> Ti+3 2.00 Al -> Al+3 1.70 Cr -> Cr+2 0.56 H2-> 2H+ 0.00 Au -> Au+ -1.68 a. Which metal would be the most anodic? b. Which metal would corrode the most? c. Which metal would be the most reactive? d. Why can Ti and Au be used for making implants, but not Al?
Answer:
The following answer are based only on emf values
a).Most anodic element ,Titanium - Ti
b).Metal That will corrode easily is : Titanium - Ti
c). Most reactive metal is Titanium-Ti
d).Al can't be used for making implants because it undergo corrosion easily and do not form strong passive oxide layer.
Ti forms very strong passive layer which prevents further corrosion.
Au is least reactive and does not corrode.
Note : You have to give answer according to E values
In actual(without considering E values) most anodic, reactive , corrosive element is AlExplanation:
Here the oxidation potential data is given (since elements are getting oxidised)
[tex]\Delta^{0} E[/tex] : Standard Reduction Potential : It is tendency of the element to gain electron and get reduced.
Oxidation Potential is opposite to reduction potential
Standard Oxidation Potential :It is tendency of the element to loose electron and get oxidised.
It is measured in Volts/V
More positive (greater)the value of the oxidation potential , stronger is the tendency of the element to get oxidised.More negative(lesser) the value of oxidation potential lesser the tendency of the element to get oxidised. Hence the element undergo Reduction in comparison to other with more positive value .The main application [tex]\Delta^{0} E[/tex] values is in electrochemical cell to predict the products of the reaction. or predict whether the reaction is spontaneous or not.
Anodic means tendency to undergo oxidationSince [tex]Ti\rightarrow Ti^{+}[/tex] has maximum value of [tex]\Delta^{0}E[/tex] = 2.00V , so it get oxidised as compared to other elements.
Corrode , The metal which oxidise easily will also undergo corrosion easilyFrom the , E values Ti show maximum tendency of oxidation hence it also get corroded easily.
Ti is the most reactive because it oxidised fast as compare to other elements Au (gold) is the least reactive element as its[tex]\Delta^{0}E[/tex] value is negative. So gold don't get corrode and used for making implants. Ti can also be used , since it is most reactive so it react with oxygen present in atmosphere and form a passive layer of oxide which prevent its corrosion. Al also form passive oxide but form very thin layer which is not able to protect it from corrosionFinal answer:
Titanium is the most anodic and most reactive metal given its high standard electrode potential of +2.00V, which also makes it the most prone to corrosion. Titanium and Gold are suitable for biomedical implants due to their corrosion resistance and biocompatibility, while Aluminum is not used due to potential toxicity.
Explanation:
To address the questions related to electrochemical potentials and the reactivity of metals, we must consider both the given standard electrode potentials and practical implications such as corrosion resistance in the case of biomedically important metals.
Most Anodic and Most Corrosive
The most anodic metal would be the one with the lowest reduction potential, which correlates with the highest oxidation potential. Since oxidation is the loss of electrons, the metal with the highest potential to lose electrons (most positive ΔEº) will be the most anodic and the most susceptible to corrosion. In this list, Titanium (Ti) has the highest standard electrode potential at +2.00 V, indicating it would be the most anodic and would corrode the most in an environment where it can react.
Most Reactive Metal
The reactivity of a metal in chemical terms is often reflected by how easily it can lose electrons (oxidize). By the given data, Titanium (Ti) would be the most reactive given its high standard electrode potential for oxidation.
Biomedical Implant Materials
Titanium (Ti) and Gold (Au) are used for biomedical implants because they are biocompatible and highly resistant to corrosion in the body. Titanium forms a stable, protective oxide layer when exposed to air or bodily fluids, preventing further oxidation. While Aluminum (Al) has a high standard reduction potential, which may suggest it is resistive to corrosion, it is not used for implants due to potential toxicity and less stable oxide formation than Ti, which could lead to corrosion in the body.
When the alkene below is brominated with NBS reaction can occur at more than one carbon.
List all of the carbons in the structure below to which bromine can become attached upon reaction of the alkene with NBS.
NOTE: When the structure is symmetrical, you need to list every numbered carbon to which bromine may be attached even if it leads to same product as reaction at another carbon.
Answer:
Four products are possible.(in attachment)
2,3,4,5 carbons attached to the bromine.
Explanation:
NBS is N-bromosuccinamide which is used to brominate the allylic and benzylic positions of the compound.
Mainly used for the allylic bromination Here, four products are possible among them two are formed by the direct allylic bromination and remaining two products are formed by the rearrangement of the radical formation.
2,3,4,5 carbons of the compound are attached to the bromine.
Final answer:
When the alkene is brominated with NBS, bromine can become attached to multiple carbons in the structure.
Explanation:
When the alkene is brominated with NBS, the reaction can occur at multiple carbons. In this case, the alkene is a hexagonal structure of benzene with two hydrogen atoms replaced by bromine atoms. The bromine atoms can become attached to the carbons adjacent to each other, 1 carbon apart, or 2 carbons apart. So, the bromine can attach to all the numbered carbons in the structure, leading to different isomers of the product.
What is the mole fraction of 25.0 g of ethanol (c2h5oh) is added to 75.0 g of water to form 96.4 mL of solution
Answer:
0.116
Explanation:
The molar mass of ethanol is 46.07 g/mol. The moles corresponding to 25.0 g are:
25.0 g × (1 mol/46.07 g) = 0.543 mol
The molar mass of water is 18.02 g/mol. The moles corresponding to 75.0 g are:
75.0 g × (1 mol/18.02 g) = 4.16 mol
The total number of moles is:
total moles = mol ethanol + mol water = 0.543 mol + 4. 16 mol = 4.70 mol
The mole fraction of ethanol is:
X(Ethanol) = mol ethanol / total moles = 0.543 mol / 4.70 mol = 0.116
a chemist reacts 150.0 grams of HCl with an excess of MnO2. if the following reaction occurs: MnO2+4HCl=MnCl2+2H2O+Cl2 how many grams of mncl2 are formed
Answer:
There are formed 98.05 g of MnCl₂
Explanation:
The reaction is this one:
MnO₂ + 4 HCl → MnCl₂ + 2 H₂O + Cl₂
First of all, determinate moles. Divide mass /molar mass
150 g / 36.45 g/m = 4.11 moles of HCl
Ratio between HCl and MnCl₂ is 4:1
4 moles of HCl produce 1 mol of Chloride
4.11 moles of HCl 'll produce (4.11 . 1)/ 4 =1.03 moles of chloride
Molar mass . Moles = Mass
Molar Mass MnCl₂ = 95.2 g/m
95.2 g/m . 1.03 moles = 98.05 grams
Answer:
There are 129.4 grams of MnCl2 formed
Explanation:
Step 1: Data given
Mass of HCl = 150.0 grams
MnO2 = excess
Molar mass of HCl = 36.46 g/mol
Step 2: The balanced equation
MnO2+4HCl → MnCl2+2H2O+Cl2
Step 3: Calculate moles of HCl
Moles HCl = Mass HCl / molar mass HCl
Moles HCl = 150.0 grams / 36.46 g/mol
Moles HCl = 4.114 moles
Step 4: Calculate Moles of MnCl2
For 1 mol of MnO2 we need 4 moles of HCl to produce 1 mol of MnCl2, 2 moles of H2O and 1 mol Cl2
For 4.114 moles of HCl we'll have 4.114/4 = 1.0285 moles of MnCl2
Step 5: Calculate mass of MnCl2
Mass MnCl2 =moles MnCl2 * molar mass MnCl2
Mass MnCl2 = 1.0285 * 125.84 g/mol
Mass MnCl2 = 129.4 grams
There are 129.4 grams of MnCl2 formed
A researcher suspects that the pressure gauge on a 54.3-L gas cylinder containing nitric oxide is broken. An empty gas cylinder weighs 90.0 lb. The weight of the partially full cylinder is 116.5 lb. This cylinder is located in a relatively chilly service hallway at 287 K. Follow the steps below to use the compressibility charts to estimate the pressure of the gas and the reading that the pressure gauge should have. How many moles of nitric oxide, NO, are in the cylinder?
Answer:
Number of moles nitric acid in the cylinder is 400.539g/mol.
Explanation:
From the given,
Weight of empty gas cylinder [tex]W_{1}= 90.0 lb= 40823.3 grams
Weight of full cylinder[tex]W_{2} = 116.5 lb= 52843.511 grams
The critical temperature = 287 K
The critical pressure 54.3 L
Molar mass of nitric acid = [tex]M_{NO}[/tex] = 30.01 g/mol
Number of moles nitric acid = [tex]n_{NO}[/tex] =?
The mass of nitric acid in the cylinder = [tex]W_{NO}=W_{2}-W_{1}[/tex]
[tex]=52843.511-40823.3 =12,020.2g[/tex]
Number of moles of nitric acid =
[tex]\frac{Given\,mass}{Molar\,mass}=\frac{12,020.2}{30}=400.539g/mol[/tex]
Therefore, number of moles nitric acid in the cylinder is 400.539g/mol.
In a hexagonal-close-pack (hcp) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes is 1:1:2.
What is the general formula for a compound if anions occupy the hcp lattice points and cations occupy all the octahedral holes?What is the general formula for a compound if anions occupy the hcp lattice points and cations occupy all the octahedral and tetrahedral holes?
Explanation:
In a hexagonal-close-pack (HCP) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes = 1 : 1 : 2
let the :
Number of lattice point = 1x.
Number of octahedral points = 1x
Number of tetrahedral points = 2x
If anions occupy the HCP lattice points and cations occupy half of the octahedral holes.
Number of anions occupying the HCP lattice points, A= 1x
Number of cations occupying the octahedral points, B = 1x
The formula of the compound will be = [tex]A_{1x}B_{1x}=AB[/tex]
If anions occupy the HCP lattice points and cations occupy all of the octahedral and the tetrahedral holes.
Number of anions occupying the HCP lattice points, A= 1x
Number of cations occupying the octahedral points, B = x
Number of cations occupying the tetrahedral points, B = 2x
total number of cations = x + 2x = 3x
The formula of the compound will be = [tex]A_{1x}B_{3x}=AB_3[/tex]
What is undergoing oxidation in the redox reaction represented by the following cell notation?
Fe(s) Fe3+(aq) Cl2(g) Cl(aq) Pt
Answer:
[tex]Fe\rightarrow Fe^{2+}+2e^-[/tex]
Explanation:
The given cell notation is:-
[tex]Fe_{(s)}|Fe^{3+}_{(aq)}||Cl_2_{(g)}|Cl^-_{(aq)}|Pt[/tex]
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
The oxidation half reaction of the cell is:-
[tex]Fe\rightarrow Fe^{2+}+2e^-[/tex]
Calculate the standard entropy of vaporization of ethanol at its boiling point 285 K. The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol
Answer:
standard entropy of vaporization of ethanol = 142.105 J/K-mol
Explanation:
given data
enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × [tex]10^{3}[/tex] J/mol
entropy of vaporization of ethanol boiling point = 285 K
to find out
standard entropy of vaporization of ethanol
solution
we get here standard entropy of vaporization of ethanol that is expess as
standard entropy of vaporization of ethanol ΔS = [tex]\frac{\Delta H}{T}[/tex] .............1
here ΔH is enthalpy of vaporization of ethanol and T is temperature
put value in equation 1
standard entropy of vaporization of ethanol ΔS = [tex]\frac{40.5*10^3}{285}[/tex]
standard entropy of vaporization of ethanol = 142.105 J/K-mol
Answer:
The standard entropy of vaporization is 142 J/mol*K
Explanation:
Step 1: Data given
The boiling point = 285 Kelvin
The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol
Step 2: Calculate the standard entropy of vaporization
ΔS = ΔH /T
⇒ with ΔS = the change in entropy
⇒ with ΔH = enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 *10³ J/mol
⇒ with T = the temperature =285 Kelvin
ΔS = (40.5 * 10³ J/mol)/ 285 Kelvin
ΔS = 142.1 ≈ 142 J/mol*K
The standard entropy of vaporization is 142 J/mol*K
(Note: The boiling point of ethanol is not 285K but 352 K)
Why are hydrogen bonds holding DNA bases together instead of covalent bonds?
Hydrogen bonds in DNA are used instead of covalent bonds because they offer the right balance of strength and flexibility, allowing DNA strands to easily 'unzip' for replication and transcription while maintaining the double helix structure. They enable the precise pairing of complementary bases, crucial for genetic information integrity.
Hydrogen bonds, rather than covalent bonds, are responsible for holding the bases in DNA together. The reason for this lies in their strength and the functionality they provide to the DNA molecule. While covalent bonds are much stronger and hold the atoms of the individual DNA molecules together, hydrogen bonds are perfectly suited for their role in the double helix structure.
Hydrogen bonds are strong enough to keep the complementary bases paired but sufficiently weak to allow the strands to 'unzip' for essential biological processes such as replication and transcription.
The interplay between adenine and thymine, as well as guanine and cytosine, through specific hydrogen bonding arrangements is crucial for the stability and integrity of DNA. The cumulative effect of millions of these bonds provides stability for the double helix structure while still allowing it to separate when needed. The precise alignment of hydrogen bonds ensures the proper matching of bases, which is vital for preserving genetic information and ensuring accurate DNA copying in cells.
In summary, hydrogen bonds are essential because they provide a balance between stability and flexibility, which is paramount for the dynamic role DNA plays within living organisms.