The molar concentrations of Fe2+ and Fe3+ in the original solution are 18.4 mM.
Explanation:To calculate the molar concentrations of Fe2+ and Fe3+ in the original solution, we can use stoichiometry based on the net ionic equation provided:
MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O
From the equation, we can see that one mole of MnO4- reacts with 5 moles of Fe2+. In the first titration, 23.0 mL of 0.0200 M KMnO4 solution is required to react with all of the Fe2+ ions, which means there is a total of 23.0 mL x 0.0200 M = 0.460 mmol of Fe2+ in the original solution.
In the second titration, 40.0 mL of the same KMnO4 solution is required to react with all of the Fe2+ ions converted from Fe3+. This means there is a total of 40.0 mL x 0.0200 M = 0.800 mmol of Fe2+ in the original solution.
Since all of the Fe2+ ions are converted to Fe3+ in the first titration, the total amount of Fe3+ in the original solution is also 0.460 mmol.
Therefore, the molar concentrations of Fe2+ and Fe3+ in the original solution are:
[Fe2+] = 0.460 mmol / 25.0 mL = 18.4 mM
[Fe3+] = 0.460 mmol / 25.0 mL = 18.4 mM
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The energy of any one-electron species in its nth state (n = principal quantum number) is given by E = –BZ2 /n2 where Z is the charge on the nucleus and B is 18 2.180 10 J. a) Find the ionization energy of the Be3+ ion in its first excited state in kilojoules per mole. b) Find the wavelength of light given off from the Be3+ ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels.
Explanation:
(a) The given data is as follows.
B = [tex]2.180 \times 10^{-18} J[/tex]
Z = 4 for Be
Now, for the first excited state [tex]n_{f}[/tex] = 2; and [tex]n_{i} = \infinity[/tex] if it is ionized.
Therefore, ionization energy will be calculated as follows.
I.E = [tex]\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})[/tex]
= [tex]8.72 \times 10^{-18} J[/tex]
Converting this energy into kJ/mol as follows.
[tex]8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol[/tex]
= 5249 kJ/mol
Therefore, the ionization energy of the [tex]Be^{3+}[/tex] ion in its first excited state in kilojoules per mole is 5249 kJ/mol.
(b) Change in ionization energy is as follows.
[tex]\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}[/tex]
[tex]\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}[/tex]
[tex]\lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}[/tex]
= [tex]303.7 \times 10^{-10} m[/tex]
or, = [tex]303.7^{o}A[/tex]
Therefore, wavelength of light given off from the [tex]Be^{3+}[/tex] ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels [tex]303.7^{o}A[/tex].
The ionization energy of the Be³+ ion in its first excited state is approximately 5250 kJ/mol. The wavelength of light emitted when an electron transitions from n=4 to n=2 in a Be³+ ion is approximately 75.9 nm.
Ionization Energy and Wavelength Calculation for Be³+
The problem involves calculating the ionization energy of a Be³+ ion in its first excited state and the wavelength of light emitted during an electron transition.
a) Ionization Energy
For a one-electron species like Be³+, the energy in the nth state is given by:
E = –BZ²/n², where Z is the charge on the nucleus and B = 2.180 × 10⁻¹⁸ J.
In the first excited state (n = 2) for Be³+ (Z = 4), we get:
En = –B(4)²/2² = –2.180 × 10⁻¹⁸ J × 16/4 = –8.72 × 10⁻¹⁸ J.
To find the ionization energy, we need the energy to liberate 1 mole of Be³+ ions:
Ionization Energy = |En| × Avogadro's Number = 8.72 × 10⁻¹⁸ J × 6.022 × 10²³ mol⁻¹ = 5.25 × 10⁶ J/mol = 5250 kJ/mol.
b) Wavelength of Light
The energy difference between the fourth (n=4) and second (n=2) energy levels is calculated using:
ΔE = E₄ - E₂ = -BZ² / 4² - (-BZ² / 2²) = BZ² (1/4² - 1/2²)
ΔE = 2.180 × 10⁻¹⁸ J × 16 × (1/16 - 1/4) = 2.180 × 10⁻¹⁸ J × 16 × (-¾) = -2.616 × 10⁻¹⁷ J
The emitted photon's wavelength is given by:
λ = hc /ΔE = (6.626 × 10⁻³⁴ Js × 3.00 × 10⁸ m/s) / 2.616 × 10⁻¹⁷ J ≈ 7.59 × 10⁻⁸ m = 75.9 nm
5. When doing an experiment, aside from doing calculations, what can you do to
determine which reactant is the limiting reactant and which is the excess
reactant? *
Answer:
check which reactant is totally consumed and which one remains in the mixture
Explanation:
Apart from doing calculations during an experiment, one can determine which reactant is limiting and which one is in excess by checking the resulting mixture for the presence of reactants.
A limiting reactant is one that determines the amount of product formed during a reaction. It is usually a reactant that is lower than stoichiometry amount.
On the other hand, an excess reactant is one that is present in more than the stoichiometrically required amount during a reaction.
Limiting reactants will be totally consumed in a reaction while excess reactant would still be seen present in mixture after the reaction has stopped.
Hence, apart from using stoichiometric calculation to determine which reactant is limiting or in excess during an experiment, one can just check the final mixture of the reaction for the presence of any of the reactants. The reactant that is detected is the excess reactant while the one without traces in the final mixture is the limiting reactant.
21. A good transition state analog
A) binds covalently to the enzyme
B) binds to the enzyme more tightly than the substrate,
C) binds very weakly to the enzyme
D) is toonstable to solate
E) must be almost identical to the substrate
Answer:
hhgjghjfgjfghj
Explanation:
I need help please answer
Answer:
A
Explanation:
Just thinking logically.
Upon adding solid potassium hydroxide pellets to water the following reaction takes place: KOH(s) → KOH(aq) + 43 kJ/mol Answer the following questions regarding the addition of 14.0 g of KOH to water: Does the beaker get warmer or colder? Is the reaction endothermic or exothermic? What is the enthalpy change for the dissolution of the 14.0 grams of KOH?
Answer:
a) Warmer
b) Exothermic
c) -10.71 kJ
Explanation:
The reaction:
KOH(s) → KOH(aq) + 43 kJ/mol
It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.
Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.
The enthalpy change for the dissolution of 14 g of KOH is:
[tex] n = \frac{m}{M} [/tex]
Where:
m: is the mass of KOH = 14 g
M: is the molar mass = 56.1056 g/mol
[tex] n = \frac{m}{M} = \frac{14 g}{56.1056 g/mol} = 0.249 mol [/tex]
The enthalpy change is:
[tex] \Delta H = -43 \frac{kJ}{mol}*0.249 mol = -10.71 kJ [/tex]
The minus sign of 43 is because the reaction is exothermic.
I hope it helps you!
How much 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl?
Answer:
54 L
Explanation:
Given data
Initial concentration (C₁): 9.0 MInitial volume (V₁): 450 mLFinal concentration (C₂): 0.075 MFinal volume (V₂): to be calculatedWe have a concentrated NaCl solution and we will add water to it to obtain a diluted NaCl solution. We can find the volume of the final solution using the dilution rule.
[tex]C_1 \times V_1 = C_2 \times V_2\\V_2 = \frac{C_1 \times V_1}{C_2} = \frac{9.0M \times 450mL}{0.075M} = 5.4 \times 10^{4} mL = 54 L[/tex]
The volume of the final solution should be 54L.
Calculation of the volume:Since
Initial concentration (C₁): 9.0 M
Initial volume (V₁): 450 mL
Final concentration (C₂): 0.075 M
We know that
C1 * V1 = C2 * V2
V2 = C1 * V1/ C2
= 9.0M * 450 mL/ 0.075 M
= 54 L
hence, The volume of the final solution should be 54L.
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The pH of a solution is 6.5.
What is the pOH?
Answer:
7.5
Explanation:
pH + pOH = 14
pOH = 14 - 6.5
pOH = 7.5
The pOH of a solution, given a pH of 6.5 at temperature 25°C, is 7.5. This is calculated by subtracting the pH from 14.
Explanation:The pH and pOH of a solution are related by the equation pH + pOH = 14 at 25°C. Given that the pH of the solution is 6.5, to find the pOH you simply subtract the pH from 14. So, pOH = 14 - pH. In this case, pOH = 14 - 6.5 = 7.5. Therefore, the pOH of a solution with a pH of 6.5 is 7.5.
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Which scenario best describes an increase in the entropy of a system?
Answer:
A) A solid salt dissolves in water.
Explanation:
A solid, like a salt, dissociates into ions as it dissolves in liquid. The particles (ions) become more spaced apart and with greater randomness. This is increasing entropy.
Arthropods do not have internal bones. Instead, an arthropod's body is supported by:
a) two or three segments.
b) an exoskeleton.
c) a set of antennae.
Answer:an exoskeleton
Explanation:jus did it
Answer:
B
Explanation:
Chlorophenols impart unpleasant taste and odor to drinking water at concentrations as low as 5 mg/m3. They are formed when the chlorine dis- infection process is applied to phenol-containing waters. What is the threshold for unpleasant taste and odor in units of (a) mg/L, (b) mg/L, (c) ppmm, and (d) ppbm
Answer:
See explaination
Explanation:
Chlorine’s disinfection properties have helped improve the lives of billions of people around the world. Chlorine also is an essential chemical building block, used to make many products that contribute to public health and safety, advanced technology, nutrition, security and transportation.
Please kindly check attachment for the step by step solution of the given problem.
The threshold for unpleasant taste and order in the respective units are;
A) 0.005 mg/L
A) 0.005 mg/LB) 5 μg/L
A) 0.005 mg/LB) 5 μg/LC) 0.005 ppm
A) 0.005 mg/LB) 5 μg/LC) 0.005 ppmD) 5 ppb
We are given the concentration of chlorophenols as; M = 5 mg/m³
A) Converting mg/m³ to mg/L means that;
1 mg/m³ = 0.001 mg/L. Thus;
5 mg/m³ = (5 × 0.001)/1
>> 0.005 mg/L
B) Converting mg/m³ to μg/L means that;
1 mg/m³ = 1 μg/L
Thus; 5 mg/m³ = 5 × 1/1 μg/L
>> 5 μg/L
C) Converting mg/m³ to ppm gives;
1 mg/m³ = 0.001 ppm
Thus;
5 mg/m³ = (5 × 0.001)/1 ppm
>> 0.005 ppm
D) Converting mg/m³ to ppb gives;
1 mg/m³ = 1 ppm
Thus;
5 mg/m³ = 5 × 1/1 ppb
>> 5 ppb
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Use the pump to add 250 molecules of the heavy species gas. How does the temperature change?
Type the correct answer in each box. Use numerals instead of words.
When you begin to add the molecules, the temperature inside the container is K. After all 250 molecules have been added, the temperature inside the container is K.
Answer:
When you begin to add the molecules, the temperature inside the container is 300 K. After all 250 molecules have been added, the temperature inside the container is 300 K.
When ignited, solid ammonium dichromate decomposes in a fiery display. This is the reaction for a "volcano" demonstration. The decomposition produces nitrogen gas, water vapor, and chromium(III) oxide. The temperature is constant at 25°C.
Substance H0f (kJ/mol) S0 (kJ/mol . K)
a. Cr2O3 (g) -1140 0.0812
b. H2O (I) -242 0.1187
c. N2 (g) 0 0.1915
d. (NH4)2 Cr2O7 - 22.5 0.1137
Answer:
Your question is half unfinished, regarding the chromium III oxide the correct option that expresses the inorganic formula of said compound is "A"
Explanation:
In the reaction an initial salt reacts giving as product water vapor, nitrogen gas and an oxide that is chromium oxide.
Chromium oxide is an oxide that adopts the structure of corundum, compact hexagonal. It consists of an anion oxide matrix with 2/3 of the octahedral holes occupied by chromium. Like corundum, Cr2O3 is a tough, brittle material.
It is used as a pigment, green in color.
Which anion would bond with K+ in a 1:1 ratio to form a neutral ionic compound?
Select the correct answer below:
A. O2-
B. F-
C. N3-
D. S2-
The anion that would bond with K+ in a 1:1 ratio to form a neutral ionic compound is F- (fluoride ion)
Explanation:The anion that would bond with K+ in a 1:1 ratio to form a neutral ionic compound is B. F- (fluoride ion).
Potassium (K+) is a cation with a +1 charge, and to form a neutral ionic compound, it needs to bond with an anion with a -1 charge. The fluoride ion (F-) has a -1 charge, and therefore, it can form a 1:1 ratio with K+ to create a neutral compound.
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The fuel value of hamburger is approximately 3.3 kcal/g. If a man eats 0.75 pounds of hamburger for lunch and none of the energy is stored in his body, estimate the amount of water that would have to be lost in perspiration to keep his body temperature constant. The heat of vaporization of water may be taken as 2.41 kJ/g.
Answer:
[tex]m_{w} = 1755.323\,g[/tex]
Explanation:
The energy stored in the hamburger is:
[tex]Q = (0.75\,pd)\cdot \left(\frac{453\,g}{1\,pd} \right)\cdot \left(3.3\,\frac{kcal}{g} \right)\cdot \left(4.186\,\frac{kJ}{kcal} \right)[/tex]
[tex]Q = 4693.239\,kJ[/tex]
The amount of water perspired is derived from the following formula:
[tex]Q = m_{w}\cdot (c_{p}\cdot \Delta T + L_{v})[/tex]
[tex]m_{w} = \frac{Q}{c_{p}\cdot \Delta T + L_{v}}[/tex]
[tex]m_{w} = \frac{4693.239\,kJ}{\left(4.186\times 10^{-3}\,\frac{kJ}{g\cdot ^{\circ}C} \right)\cdot (100^{\circ}C-37^{\circ}C)+2.41\,\frac{kJ}{g} }[/tex]
[tex]m_{w} = 1755.323\,g[/tex]
The rust that appears on steel surfaces is iron(III) oxide. If the rust found spread over the surfaces of a steel bicycle frame contains a total of 9.62×1022 oxygen atoms, how many grams of rust are present on the bicycle frame?
Answer:
The grams of rust present in the bicycle frame is [tex]x = 8.50 g[/tex]
Explanation:
From the question we are told that
The number of oxygen atom contained is [tex]n = 9.62 *10^{22} \ atoms[/tex]
The molar mass of the compound is [tex]M_{Fe_3O_3} = 159.69 g[/tex]
At standard temperature and pressure the number of oxygen atom in one mole of iron(III) oxide is mathematically evaluated as
[tex]N_o = 3 * Ne[/tex]
Where Ne is the avogadro's constant with a value [tex]N_e = 6.023 *10^{23} \ atoms[/tex]
So
[tex]N_o= 3 * 6.023*10^{23} \ atoms[/tex]
[tex]N_o= 1.8069*10^{24} \ atoms[/tex]
So
[tex]1.8069*10^{24} \ atoms[/tex] is contained in [tex]M_{Fe_3O_3} = 159.69 g[/tex]
[tex]9.62 *10^{22} \ atoms[/tex] is contained in x
So
[tex]x = \frac{159.69 * 9.62 *10^{22}}{1.8069 *10^{24}}[/tex]
[tex]x = 8.50 g[/tex]
Final answer:
To calculate the grams of rust on the bicycle frame, we need to use stoichiometry. By converting the number of oxygen atoms to moles and then to grams using the molar mass of iron(III) oxide, we find that there are approximately 5.37x10^22 grams of rust on the bicycle frame.
Explanation:
To determine the grams of rust present on the bicycle frame, we need to calculate the molar mass of iron(III) oxide and then use stoichiometry to convert the number of oxygen atoms to grams of rust. The molar mass of Fe2O3 is 159.69 g/mol. Given that 1 mole of Fe2O3 contains 3 moles of oxygen atoms, we can calculate the moles of Fe2O3 using the given number of oxygen atoms and then convert it to grams using the molar mass.
9.62x10^22 oxygen atoms x (1 mol Fe2O3 / 3 mol O) x (159.69 g Fe2O3 / 1 mol Fe2O3) = 5.37x10^22 g Fe2O3
Therefore, there are approximately 5.37x10^22 grams of rust present on the bicycle frame.
The complete combustion of methane is: CH4 + 2O2 ! 2H2O + CO2 a. Calculate the standard Gibbs free energy change for the reaction at 298 K (i.e. ). b. Calculate the energetic (ΔH) and entropic contributions (TΔS) to the favorable standard Gibbs free energy change at 298 K and determine which is the dominant contribution,? c. Estimate the equilibrium constant at 298 K.
Answer:
a
The standard Gibbs free energy change for the reaction at 298 K is
[tex]\Delta G^o_{re} = -800.99 kJ/moles[/tex]
b
The energetic (ΔH) is [tex]\Delta H^o _{re} = -802.112 \ kJ/mole[/tex]
The entropic contributions is [tex]T \Delta S = -1.126 \ kJ/mole[/tex]
Energetic is the dominant contribution
c
The equilibrium constant at 298 K is [tex]K = 2.53[/tex]
Explanation:
From the question we are told that
The chemical reaction is
[tex]CH_4 + 2 O_2 ----> 2 H_2 O + CO_2[/tex]
Generally ,
The free energy of formation of [tex]CH_4[/tex] is a constant with a value
[tex]\Delta G^o_f __{CH_4}} = -50.794 \ kJ / moles[/tex]
The free energy of formation of [tex]O_2[/tex] is a constant with a value
[tex]\Delta G^o_f __{O_2}} = 0 \ kJ / moles[/tex]
The free energy of formation of [tex]H_2O[/tex] is a constant with a value
[tex]\Delta G^o_f __{H_2O}} = -228.59 \ kJ / moles[/tex]
The free energy of formation of [tex]CO_2[/tex] is a constant with a value
[tex]\Delta G^o_f __{H_2O}} = -394.6 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]CH_4[/tex] at standard condition i is a constant with a value
[tex]\Delta H^o_f __{CH_4}} = -74.848 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]CO_2[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{CO_2}} = -393.3 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]O_2[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{O_2}} = 0 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]H_2O[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{H_2O}} = -241.83 \ kJ / moles[/tex]
The standard Gibbs free energy change for the reaction at 298 K is mathematically represented as
[tex]\Delta G^o_{re} = (\Delta G^o_f __{H_2O}} + (2 * \Delta G^o_f __{H_2O}} )) - ((\Delta G^o_f __{CH_4}} + (2 * \Delta G^o_f __{O_2}}))[/tex]
Substituting values
[tex]\Delta G^o_{re} =\Delta G= ( (-394.6 ) + (2 * (-228.59)) ) - ((-50.794) +(2* 0))[/tex]
[tex]\Delta G^o_{re} = -800.99 kJ/moles[/tex]
The Enthalpy of formation of the reaction is
[tex]\Delta H^o _{re} =( \Delta H^o_f __{CH_4}} + (2 * (\Delta H^o_f __{H_2O}} ))) - ( \Delta H^o_f __{CH_4}} + (2 * \Delta H^o_f __{O_2}}))[/tex]
Substituting values
[tex]\Delta H^o _{re} = \Delta H = ((-393.3) + 2 * ( -241.83)) - ( -74.848 + (2 * 0))[/tex]
[tex]\Delta H^o _{re} = -802.112 \ kJ/mole[/tex]
The entropic contributions is mathematically represented as
[tex]T \Delta S = \Delta H -\Delta G[/tex]
Substituting values
[tex]T \Delta S =-802 .112-(-800.986)[/tex]
[tex]T \Delta S = -1.126 \ kJ/mole[/tex]
Comparing the values of [tex]T \Delta S \ and \ \Delta G[/tex] we see that energetic is the dominant contribution
The standard Gibbs free energy change for the reaction at 298 K can also be represented mathematically as
[tex]\Delta G = -RT lnK[/tex]
Where R is the gas constant with as value of [tex]R = 8.314 *10^{-3} kJ/mole[/tex]
K is the equilibrium constant
T is the temperature with a given value of [tex]T = 298K[/tex]
Making K the subject we have
[tex]K = e ^{- \frac{\Delta G }{RT} }[/tex]
Substituting values
[tex]K = e ^{- \frac{-800.99 }{(8.314 *10^{-3} ) * (298)} }[/tex]
[tex]K = 2.53[/tex]
Final answer:
To calculate the standard Gibbs free energy change for the combustion of methane, use the formula ΔG = ΔH - TΔS, where ΔG is the standard Gibbs free energy change, ΔH is the standard enthalpy change, T is the temperature in Kelvin, and ΔS is the standard entropy change.
Explanation:
The standard Gibbs free energy change for a reaction can be calculated using the formula:
ΔG = ΔH - TΔS
where ΔG is the standard Gibbs free energy change, ΔH is the standard enthalpy change, T is the temperature in Kelvin, and ΔS is the standard entropy change.
In this case, we are given the thermochemical equation for the combustion of methane as: CH4 + 2O2 → 2H2O + CO2
To calculate the standard Gibbs free energy change, we need the values of ΔH and ΔS for this reaction at 298 K. Once we have these values, we can plug them into the formula to get the answer.
How much energy is released when 50.00 mol of glycerin melts? (Heat of fusion glycerin-91.7 KJ/mol)
Answer:
4585 KJ
Explanation:
Data obtained from the question. This includes following:
Number of mole (n) of glycerin = 50 moles
Heat of fusion glycerin (Hf) = 91.7 KJ/mol
Heat (Q) released =?
The heat released can be obtained as follow:
Q = n•Hf
Q = 50 x 91.7
Q = 4585 KJ
Therefore, 4585 KJ of heat is released.
A 50.0 mL sample containing Cd 2 + and Mn 2 + was treated with 50.0 mL of 0.0400 M EDTA . Titration of the excess unreacted EDTA required 19.5 mL of 0.0270 M Ca 2 + . The Cd 2 + was displaced from EDTA by the addition of an excess of CN − . Titration of the newly freed EDTA required 17.1 mL of 0.0270 M Ca 2 + . What are the concentrations of Cd 2 + and Mn 2 + in the original solution?
Answer:
Check the explanation
Explanation:
VOLUME OF newly freed EDTA
=(VOLUME OF Ca2* STRENGTH OF Ca2)/ STRENGTH OF EDTA
=(24.9*0.0220)/ 0.0700
=7.825mL
STRENGTH OF Cd2
=( VOLUME OF newly freed EDTA * STRENGTH OF EDTA) / VOLUME OF SAMPLE
=(7.825*0.0700)/50
=0.0109 M
VOLUME OF excess unreacted EDTA
=(VOLUME OF Ca2* STRENGTH OF Ca2)/ STRENGTH OF EDTA
=(19.5*0.0270)/ 0.0400
=13.16mL
VOLUME OF EDTA REQUIRED FOR SAMPLE CONTAINING Cd2 AND Mn2= (17.1-13.16) mL
=3.94 mL
VOLUME OF EDTA REQUIRED FOR Mn2
= (VOLUME OF EDTA REQUIRED FOR SAMPLE CONTAINING Cd2 AND Mn2 - VOLUME OF newly freed EDTA )
=40.02-7.825 mL
=32.20 mL
STRENGTH OF Mn2
=( VOLUME OF EDTA REQUIRED FOR Mn2* STRENGTH OF EDTA) / VOLUME OF SAMPLE
=(32.20*0.0700)/50
=0.045 M
Despite the destruction from Hurricane Katrina in September 2005, the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered an atmospheric pressure of 88.2 kPa on October 19, 2005, that was 2.0 kPa lower than Hurricane Katrina. What was the difference in pressure between the two hurricanes?
Complete Question
Despite the destruction from Hurricane Katrina in Sept 2005,the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered at atmospheric pressure of 88.2 kPa on Oct 19, 2005,about 2 kPa lower than Hurricane Katrinia. What was the difference between the two hurricanes in:
millimeters of Hg? ___ mm Hg
atmospheres? ___ atm
millibars? ___ mb
Answer:
The difference in pressure in mm of Hg is [tex]Pg = 15.05 \ mm\ Hg[/tex]
The difference in pressure in atm is [tex]P_{atm} = 0.0198 \ atm[/tex]
The difference in pressure in millibar is [tex]P_{bar} = 20 \ mbar[/tex]
Explanation:
From the question we are told that
The pressure of Hurricane Wilma is [tex]P = 88 kPa[/tex]
The difference is pressure is [tex]\Delta P =2\ k Pa[/tex]
Generally
[tex]1\ atm = 1.01 *10^5 \ Pa[/tex]
[tex]1 \ atm = 760\ mmHg[/tex]
The pressure difference in mm Hg is mathematically evaluated as
[tex]Pg = \Delta P * \frac{1000 Pa}{1kPa} * \frac{1\ atm}{1.01 *10^5} * \frac{760 mm Hg}{1 \ atm}[/tex]
Substituting the value
[tex]Pg = 2 kPa * \frac{1000 Pa}{1kPa} * \frac{1\ atm}{1.01 *10^5} * \frac{760 mm Hg}{1 \ atm}[/tex]
[tex]Pg = 15.05 \ mm\ Hg[/tex]
The pressure difference in atm is
[tex]P_{atm } = 2kPa * \frac{1000}{1\ kPa} *\frac{1 \ atm}{1.01 *10^5 Pa}[/tex]
[tex]= 0.0198 \ atm[/tex]
The pressure in millibars is
[tex]= 2kPa * \frac{1000Pa}{1 kPa} * \frac{1\ bar }{1*10^5 Pa} * \frac{1000 \ milli bar}{1 bar}[/tex]
[tex]= 20 \ m bar[/tex]
Final answer:
The pressure difference between Hurricane Wilma and Hurricane Katrina was 2.0 kPa, with Wilma's pressure being 88.2 kPa and Katrina's being 90.2 kPa. The pressure drop during a hurricane on the northeast U.S. coast in torr is 50.8 torr. A nonvolatile liquid is used in barometers and manometers because volatility would affect the accuracy of pressure measurements.
Explanation:
The difference in atmospheric pressure between Hurricane Wilma and Hurricane Katrina can be calculated using the information provided. If Wilma's pressure was 88.2 kPa and that was 2.0 kPa lower than Katrina's, then Katrina's pressure was 88.2 kPa + 2.0 kPa = 90.2 kPa. Therefore, the difference in pressure between the two hurricanes was 2.0 kPa.
To calculate the drop in pressure in torr during a hurricane on the northeastern United States coast: the pressure usually at 30.0 in. Hg can drop to 28.0 in. Hg. Since 1 in. Hg is equivalent to 25.4 mm Hg, and 1 mm Hg is the same as 1 torr, the drop in pressure is (30.0 in. Hg - 28.0 in. Hg) x 25.4 mm Hg/in. Hg = 50.8 torr.
It is necessary to use a nonvolatile liquid in a barometer or manometer because a volatile liquid would evaporate and form a vapor that would affect the pressure measurement, leading to inaccurate readings.
substances change temperature at different rates because of their
density
specific heat
melting point
boiling point
Answer:
ok
Explanation:
I have the same question
Draw the structure(s) of the product(s) of the Claisen condensation reaction between ethyl propanoate and ethyl formate. Draw only the condensation product, including the self-condensation product if applicable, do not draw the structure of the leaving group. Do not consider stereochemistry.
Answer:
See explanation below
Explanation:
A claisen reaction, is often used between two esters (One of them, usually having alpha hydrogens atoms) to form Beta Keto esters. Depending of the reagents, this reaction is taking place with base where one ester acts as nucleophile and the other as electrophile.
Now, we have here ethyl propanoate and ethyl formate. The ethyl propanoate has two alpha hydrogens, while ethyl formate do not have that. So, the base will react with the ethyl propanoate first, and then, it will react as nucleophile with the formate.
Now, the base to be used in this case will have to be a relatively strong base such sodium ethanoate, because if we use a strong base such NaOH, this will cause the saponification of the ester and not the condensation. So, in order to promove the claisen condensation, we need to use a base not too strong.
In the picture attached you have the mechanism and structure of the final product.
Which process is a chemical change?
TIL
00
burning a match
boiling water
melting ice
O breaking glass
Answer:
Burning a match is a chemical change while the rest are physical changes.
Explanation:
Burning a match represents a chemical change, where a substance transforms into new substances with distinct chemical properties. In contrast, boiling water, melting ice, and breaking glass are examples of physical changes, altering the substance's form but not its identity.
Explanation:Among the given options, burning a match is a chemical change. A chemical change is a process that involves a substance changing into a new substance with different chemical properties. In this case, when a match is burned, it undergoes a chemical reaction leading to the formation of new substances such as heat, light, water vapor, and carbon dioxide, which were not there before the match was lit. On the other hand, boiling water, melting ice, and breaking glass are all examples of physical changes, which involve changes in the form of a substance but do not change the identity of the substance itself.
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When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of ammonium chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for ammonium chloride in . Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.
Answer:
The figures and parameters to answer this question are not clearly stated, so I will answer it in the closest best way I can.
Explanation:
Molality of alanine = mole / weight of solvent( kg)
= (170×1000)/(89×600)
= 3.18 molal
We know ∆ Tf = Kf × m
Kf = ∆ Tf / molality
= 7.9/3.18 = 2.48 °c.kg.mol-1
Now using the value of cryoscopic constant we calculate can't Hoff factor for Ammonium chloride.
i = ∆ Tf /( Kf × molality of NH4Cl)
= (24.7 × 53.5 × 600) /(2.48 × 170 × 1000)
= 1.8
175 mal of Cl2 gas is held in a flexible vessel at STP. If the vessel is transported to the bottom of the impact basin Hellas planitia on the surface of Mars where is 1.16 kPa and the temperature is -5.0ºC. What is the new volume of Cl2 gas in liters
Answer: it’s 15.0 L
Explanation: Trust me
A 175 ml sample of neon had its pressure changed from 75 kPa to 150 kPa. What is its new volume?
Answer:
The new volume is 87.5 ml.
Explanation:
We have,
Volume of sample of neon is 175 ml
The pressure changed from 75 kPa to 150 kPa.
We need to find new volume.
It is based on the concept of Boyle's law. Let V₂ is new volume. So,
[tex]P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{175\times 75}{150}\\\\V_2=87.5\ ml[/tex]
So, new volume is 87.5 ml.
Using Boyle's Law, the new volume of the neon gas when the pressure is changed from 75 kPa to 150 kPa is calculated to be 87.5 mL.
To find the new volume of a gas when its pressure is changed, we use Boyle's Law.
Boyle's Law states that the pressure and volume of a gas are inversely proportional when temperature and the number of gas molecules remain constant.
The formula is expressed as:
P₁V₁ = P₂V₂
In this problem:
Initial Pressure, P₁ = 75 kPaInitial Volume, V₁ = 175 mLFinal Pressure, P₂ = 150 kPaWe need to find the final volume, V₂. Rearranging Boyle's Law, we get:
V₂ = (P₁V₁) / P₂
Plugging in the values:
V₂ = (75 kPa * 175 mL) / 150 kPa
V₂ = 13125 mL / 150 kPa
V₂ = 87.5 mL
Therefore, the new volume of the neon gas sample is 87.5 mL.
For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l) + oxygen (g) carbon monoxide (g) + water (g) What is the maximum amount of carbon monoxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams
Answer:
The maximum amount of CO2 that can be formed is 9.15 grams CO2
O2 is the limiting reactant
There will remain 4.82 grams of benzene
Explanation:
Step 1: Data given
Mass of benzene = 7.53 grams
Mass of oxygen gas = 8.33 grams
Molar mass of benzene = 78.11 g/mol
Molar mass oxygen gas = 32.00 g/mol
Step 2: The balanced equation
2C6H6 + 15O2 → 12CO2 + 6H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles C6H6 = 7.53 grams / 78.11 g/mol
Moles C6H6 = 0.0964 moles
Moles O2 = 8.33 grams / 32.00 g/mol
Moles O2 = 0.2603 moles
Step 4: Calculate the limiting reactant
For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O
O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles
There will remain 0.0964 - 0.0347 = 0.0617 moles benzene
This is 0.0617 moles * 78.11 g/mol = 4.82 grams benzene
Step 5: Calculate moles CO2
For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O
For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2
Step 6: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.208 moles * 44.01 g/mol
Mass CO2 = 9.15 grams
When HClO2 is dissolved in water, it partially dissociates according to the equation
HClO2 → H+ + ClO2-. A solution is prepared that contains 5.850 g of HClO2 in 1.000 kg of water. Its freezing point is -0.3473 °C. Calculate the fraction of HClO2 that has dissociated.
Answer:
x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates
Explanation:
Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF
0.3473 = m * 1.86
Solving, m = 0.187 m
Moles of HClO2 = mass / molar mass = 5.85 / 68.5 = 0.0854 mol
Molality = moles / mass of water in kg = 0.0854 / 1 = 0.0854 m
Initial molality
Assuming that a % x of the solute dissociates, we have the ICE table:
HClO2 H+ + ClO2-
initial concentration: 0.0854 0 0
final concentration: 0.0854(1-x/100) 0.0854x/100 0.0854x / 100
We see that sum of molality of equilibrium mixture = freezing point molality
0.0854( 1 - x/100 + x/100 + x/100) = 0.187
2.1897 = 1 + x / 100
x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates
The rate constant of the elementary reaction C2H5CN(g) → CH2CHCN(g) + H2(g) is k = 7.21×10-3 s-1 at 634 °C, and the reaction has an activation energy of 247 kJ/mol. (a) Compute the rate constant of the reaction at a temperature of 741 °C. _________ s-1 (b) At a temperature of 634 °C, 96.1 s is required for half of the C2H5CN originally present to be consume. How long will it take to consume half of the reactant if an identical experiment is performed at 741 °C? (Enter numbers as numbers, no units. For example, 300 minutes would be 300. For letters, enter A, B, or C. Enter numbers in scientific notation using e# format. For example 1.43×10-4 would be 1.43e-4.)
Answer:
the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]
it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C
Explanation:
Given that :
[tex]k_1 = 7.21*10^{-3} s^{-1} \\ \\ E_a = 247 kJ/mol \ \ \ = 247*10^3 \ J/mol \\ \\ T_1 = 634 ^ {^0} C= (273 + 634) K = 907 \ K \\ \\ T_2 = 741^{^0 } C = (273+ 741) K = 1014 \ K \\ \\ R =8.314 \ \ J/mol/K[/tex]
a)
According to Arrhenius Equation ;
[tex]In\frac{k_2}{k_1} = -\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = 3.456412[/tex]
[tex]\frac{k_2}{7.21*10^{-3}} =e^{ 3.456412[/tex]
[tex]k_2 = 31.70302 * 7.21*10^{-3}[/tex]
[tex]k_2 = 0.22858 \ s^{-1}[/tex]
Therefore , the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]
b) Given that :
[tex]t_{(1/2)} = 96.1 \ s[/tex]
[tex]k_1 = \frac{0.693}{ t_{(1/2)}}[/tex]
[tex]k_1 = \frac{0.693}{ 96.1 \ s}[/tex]
[tex]k_1 = 7.211*10^{-3} \ s^{-1}[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = 3.456412[/tex]
[tex]\frac{k_2}{7.211*10^{-3}} =e^{ 3.456412[/tex]
[tex]k_2 =e^{ 3.456412 * 7.21*10^{-3}[/tex]
[tex]k_2 = 0.22862 \ s^{-1}[/tex]
[tex]t_{(1/2)_2}} = \frac{0.693}{k_2}[/tex]
[tex]t_{(1/2)_2}} = \frac{0.693}{0.22862}[/tex]
[tex]t_{(1/2)_2}} =3.0313 \ s[/tex]
it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C
In your lab you are studying the kinetics of the degradation of a pain killer in the human liver. You are monitoring the concentration of the pain killer over a period of time. The initial concentration of the pain killer in your experiment was 1.59 M. After 0.80881 hours the concentration was found to be 0.795 M. In another 0.80881 hours the concentration was found to be 0.3975 M (t=27.868 hours overall).
If another experiment were set up where the initial concentration of the pain killer was 0.479 M, how long would it take for the pain killer concentration to reach 0.0161 M?
Answer:
Explanation:
From the data it is clear that after each .80881 hours , the concentration becomes half . so its decomposition appears to be first order .
k = 1 / t ln A₀ / A
= (1/.80881) ln 2
= .857
for the given case
A = .0161 , A₀ = .479
t = ?
t = 1/k ln( .479 / .0161)
= (1 / .857 )ln (.479 / .0161 )
= (1 / .857 ) x ln 29.75
= 3.959 hours
Which is an example of a beneficial mutation?
Answer:
b.one that results in lighter flower petal colors without changing the plant’s ability to reproduce
Explanation:
on edge
Beneficial mutations improve the survival and reproductive chances of an organism. Examples include mosquitos' resistance to insecticides, the color change in peppered moths during the Industrial Revolution, and the Antennapedia mutation in Drosophila.
Explanation:A beneficial mutation can be understood as the type of mutation that improves an organism's chances of survival and reproduction. For instance, consider the case of mosquitoes. The mutation that has provided them with resistance to some insecticides is an example of a beneficial mutation. This resistance has allowed them to survive even when faced with these chemicals, thus enhancing their chances of propagation.
Similarly, during the Industrial Revolution, peppered moths underwent a beneficial mutation where their coloration turned from light to dark. As industrial pollution darkened the environment, the dark coloration assisted in camouflage, aiding their survival.
The mutant allele can also interfere with the normal gene's function or its distribution in the body, which can result in beneficial mutations. For example, the Antennapedia mutation in Drosophila that allowed for the development of legs in place of antennae. However, whether a mutation is beneficial or not primarily depends on its effects on the organism's ability to mature sexually and reproduce successfully.
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