Answer:
1. Na3[Fe(CN)6]
Oxidation number of iron is +3
Sodium hexacyanoferrate (III)
2. [Cr (en)2Cl2]+
Oxidation number of chromium is +3
Dichlorobis (ethylenediamine)chromium (III) ion
3. [Co (en)3]Cl3
Oxidation number of cobalt is +3
Tris (ethylenediamine)cobalt (III) chloride
Determine whether the following reaction is a Lewis acid/base reaction and if so, what is the Lewis acid: Fe3+(aq) + 6 CN−(aq) → Fe(CN)63−(aq) Group of answer choices This is not a Lewis acid/base reaction It is a Lewis acid/base reaction, and Fe3+ is the Lewis acid It is a Lewis acid/base reaction, and CN- is the Lewis acid
Answer: It is a Lewis acid/base reaction, and [tex]Fe^{3+}[/tex] is the Lewis acid.
Explanation:
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
[tex]Fe^{3+}[/tex] can readily accept electrons and thus act as a lewis acid which is short of electrons.
[tex]CN^-[/tex] can readily lose electrons and thus act as a lewis base which has excess of electrons.
It is a Lewis acid/base reaction, and [tex]Fe^{3+}[/tex] is the Lewis acid.
Which of the following is a false statement about the isotopes carbon-12 (12C) and carbon-13 (13C)? Please choose the correct answer from the following choices, and then select the submit answer button. 12C and 13C have the same atomic weight. 12C and 13C have different numbers of neutrons. 13C has one more electron than 12C. 13C has more subatomic particles in its nucleus than 12C. 13C has the same atomic number as 12C.
The given isotopes of carbon, i.e., 12C and 13C, have the same number of protons but the number of neutrons differs. So, the statement 12C and 13C have different numbers of neutrons, is correct.
Isotopes are members of the same element's family but have variable numbers of neutrons despite having the same number of protons.
The element carbon can have different isotopes, based on the number of neutrons.
The isotopes of carbon include, C10, C11, C12, C13, C14, C15, and C16.
The given isotope, 12C and 13C, possess similar chemical properties and the same atomic number.
But the mass of the 13C isotope of carbon is more, because it possesses, one extra neutron, as compared to the 12C isotope of carbon.
Thus, the statement about the isotopes, “12C and 13C have different numbers of neutrons”, is correct.
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The statement that 13C has one more electron than 12C is incorrect. Both carbon-12 (12C) and carbon-13 (13C) isotopes have the same number of electrons due to both having the same atomic number. They only differ in the number of neutrons they possess.
Explanation:The statement '13C has one more electron than 12C' is false. Both isotopes, carbon-12 (12C) and carbon-13 (13C) have the same atomic number '6', which represents the number of protons and, under stable conditions, also views it as the same number of electrons.
So, while 12C and 13C indeed differ in the number of neutrons (12C has 6 neutrons and 13C has 7), they possess the same number of protons and electrons. Furthermore, their atomic weights are different as a result of the difference in the number of neutrons. The atomic weight of 12C is closer to 12 and that of 13C is closer to 13.
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Nitroglycerine decomposes violently according to the chemical equation below. What mass of carbon dioxide gas is produced from the decomposition of 10.4 g C3H5(NO3)3?
Explanation:
Below is an attachment containing the solution.
Even though we have known of many nuclear energy technologies for about one century some of these technologies have not yet been fully developed. Obtaining energy from thorium remains a challenge for many reasons. In a reactor, thorium is bombarded with neutrons and becomes U-233 and then U-234, which can undergo fission and produce energy. Complete the following U-234 fission reaction. Hint: Do not ignore the amount of energy given off by the reaction! 234.0 tlu . i37 154Xe + 3?Sr + 3 1.0087n + 180 MeV
Answer: The nuclear fission reaction of U-234.043 is written below.
Explanation:
In a nuclear reaction, the total mass and total atomic number remains the same.
For the given fission reaction:
[tex]^{234.043}_{92}\textrm{U}\rightarrow ^{137.159}_{54}\textrm{Xe}+^{A}_{38}\textrm{Sr}+3^{1.0087}_0\textrm{n}+180MeV[/tex]
To calculate A:
Total mass on reactant side = total mass on product side
234.043 = 137.159 + A + 3(1.0087)
A = 93.858
Now, the chemical equation becomes:
[tex]^{234.043}_{92}\textrm{U}\rightarrow ^{137.159}_{54}\textrm{Xe}+^{93.858}_{38}\textrm{Sr}+3^{1.0087}_0\textrm{n}+180MeV[/tex]
Hence, the nuclear fission reaction of U-234.043 is written above.
a metal sample has a mass of 7.56 g. the sample is placed into a graduated cylinder previously filled with 20.00 mL of water. If the final volume in the cylinder is 21.68 mL, what is the density of the substance?
Answer:
4.5g/ml
Explanation:
metal sample has a mass = 7.56 g
cylinder previously filled with water = 20.00 mL
final volume in the cylinder = 21.68 mL
[tex]density = \frac{mass}{V_f - V_i}[/tex]
[tex]density = \frac{7.56}{(21.68 - 20.00)}[/tex]
density = 4.5g/ml
Be sure to answer all parts. Zinc is an amphoteric metal, meaning it reacts with both acids and bases. The standard reduction potential is −1.36 V for the following reaction: (1)Zn(OH)42−(aq) + 2e− → Zn(s) + 4OH−(aq) Calculate the formation constant Kf for the reaction: (2)Zn2+(aq) + 4OH−(aq) ⇌ Zn(OH)42−(aq)
Answer:
[tex]1.86*10^{20[/tex]
Explanation:
The equation for the reaction is:
[tex]Zn^{2+}_{(aq)} + 4OH^-_{(aq)}[/tex] ⇄ [tex]Zn(OH)^{2-}_{4(aq)}[/tex]
Oxidation can be defined as the addition of oxygen, removal of hydrogen and/or loss of electron during an electron transfer. Oxidation process occurs at the anode.
On the other hand; reduction is the removal of oxygen, addition of hydrogen and/ or the process of electron gain during an electron transfer. This process occurs at the cathode.
The oxidation-reduction process with its standard reduction potential is as follows:
[tex]Zn(OH)^{2-}_{4(aq)} + 2e^- ----->Zn_{(s)} +4OH^-_{(aq)}[/tex] [tex]E^0_{anode} = -1.36 V[/tex]
At the zinc electrode (cathode); the reduction process of the reaction with its standard reduction potential is :
[tex]Zn^{2+}_{(aq)} +2e^- -----> Zn_{(s)}[/tex] [tex]E^0_{cathode} = -0.76 V[/tex]
The standard cell potential [tex]E^0_{cell}[/tex] is given as:
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
[tex]E^0_{cell}[/tex] = -0.76 V - (- 1.36 V)
[tex]E^0_{cell}[/tex] = -0.76 V + 1.36 V
[tex]E^0_{cell}[/tex] = +0.60 V
Now to determine the formation constant [tex]k_f[/tex] of the [tex]E^0_{cell}[/tex] ; we use the expression:
[tex]E^0_{cell}[/tex] = [tex]\frac{RT}{nF}Ink_f[/tex]
where;
[tex]E^0_{cell}[/tex] = +0.60 V
R = universal gas constant = 8.314 J/mol.K
T = Temperature @ 25° C = (25+273)K = 298 K
n = numbers of moles of electron transfer = 2
F = Faraday's constant = 96500 J/V.mol
[tex]+ 0.60 V = \frac{(8.314)(298)}{n(96500)} Ink_f[/tex]
[tex]+0.60 V = \frac{(0.0257)}{n}Ink_f[/tex]
[tex]+0.60V = \frac{0.0592}{n}log k_f[/tex]
[tex]logk_f = \frac{+0.60V*n}{0.0592}[/tex]
[tex]logk_f = \frac{+0.60V*2}{0.0592}[/tex]
[tex]logk_f = 20.27[/tex]
[tex]k_f= 10^{20.27}[/tex]
[tex]k_f = 1.86*10^{20}[/tex]
Therefore, the formation constant [tex]k_f[/tex] for the reaction is = [tex]1.86*10^{20[/tex]
Acetylenic compounds may be used as dienophiles in the Diels-Alder reaction. Write the structure for an adduct that you expect from the reaction of 1,3-butadiene with hexafluoro-2-butyne:
Answer:
Explanation:
An electron-donating heteroatom substituent at position-2 of a furan promotes regiospecific opening of the 7-oxa bridge of the Diels-Alder cycloadduct with hexafluoro-2-butyne, producing a 4-heterosubstituted 2,3-di(trifluoromethyl)phenol building block in a single step. The phenol and heteroatom substituent are easily transformed to the corresponding iodide or triflate that readily undergoes Heck, Suzuki, and Stille reactions to install a variety of substituents in high yields. This methodology provides a facile and general synthesis of 1,4-disubsituted 2, 3-di(trifluoromethyl)benzenes.
Final answer:
The Diels-Alder reaction of 1,3-butadiene with hexafluoro-2-butyne would form a six-membered ring with a hexafluoroalkyl side group, yielding a bicyclic compound.
Explanation:
The Diels-Alder reaction between 1,3-butadiene and hexafluoro-2-butyne would yield a six-membered ring product, specifically a bicyclic compound due to the formation of a cyclohexene derivative where the double bond is part of the ring and the hexafluoroalkyl group is attached to two adjacent carbons in the ring. It involves the 1,3-butadiene acting as the diene and the hexafluoro-2-butyne acting as the dienophile in an intermolecular reaction. The Diels-Alder reaction creates two new sigma bonds and a pi bond across the alkene and the alkyne to form the new cyclohexene ring.
Determine the rate constant for each of the following fi rst-order reactions, in each case expressed for the rate of loss of A: (a) A S B, given that the concentration of A decreases to one-half its initial value in 1000. s; (b) A S B, given that the concentration of A decreases from 0.67 molL1 to 0.53 molL1 in 25 s; (c) 2 A S B C, given that [A]0 0.153 molL1 and that after 115 s the concentration of B rises to 0.034 molL1
Explanation:
The integrated first law is given by :
[tex][A]=[A]_o\times e^{-k\times t}[/tex]
Where:
[tex][A]_o[/tex] = initial concentration of reactant
[A] = concentration of reactant after t time
k = rate constant
a)
[tex][A_o]=x[/tex]
[tex][A]=\frac{x}{2}[/tex]
t = 1000 s
[tex]\frac{x}{2}=x\times e^{-k\times 1000 s}[/tex]
Solving for k:
[tex]k=0.06934 s^{-1}[/tex]
The rate constant for this reaction is [tex]0.06934 s^{-1}[/tex].
b)
[tex][A_o]=0.67 mol/L[/tex]
[tex][A]=0.53 mol/L[/tex]
t = 25 s
[tex]0.53 mol/L=0.67 mol/L\times e^{-k\times 25s}[/tex]
Solving for k:
[tex]k=0.009376 s^{-1}[/tex]
The rate constant for this reaction is [tex]0.009376 s^{-1}[/tex].
c) 2 A → B +C
[tex][A_o]=0.153 mol/L[/tex]
[tex][A]=?[/tex]
[tex][B]=0.034 mol/L[/tex]
According to reaction, 1 mole of B is obtained from 2 moles of A.
Then 0.034 mole of B will be obtained from:
[tex]\frac{2}{1}\times 0.034 mol= 0.068 mol[/tex] of A
So, the concentration left after 115 seconds:
[tex][A]=0.153 mol/L-0.068 mol/L=0.085 mol/L[/tex]
t = 115 s
[tex]0.085mol/L=0.53 mol/L\times e^{-k\times 115 s}[/tex]
Solving for k:
[tex]k=0.01592 s^{-1}[/tex]
The rate constant for this reaction is [tex]0.01592 s^{-1}[/tex].
PCL was mixed with gelatin to make a blend for elctrospun fibrous scaffold encapsulating growth factor that was admixed in the polymer solution and then filled in the syringe for electrospinning. Three scaffolds were made with (A: only PCL), (B: PCL:gelatin=3:1), (C: PCL:gelatin = 1:1). The scaffolds were immersed in PBS at 37 °C! and its degradation rate was noted: Scaffold A, B and C degraded 1%, 25% and 50% respectively, in 21 days. Assuming the growth factor is trapped and cannot freely diffuse out of the scaffold. Assuming, growth factor release is dependent on scaffold degradation A. Which scaffold can be used for delivering growth factor (encapsulated in the scaffold) needed to be delivered in first 7 days of incubation in an in-vivo experiment and why? (5pts) B. The scaffold A and C were prepared and freeze dried. Their weights were Dry weight of A: 10 mg Dry weight of C: 10 mg Both (A and C) the scaffolds were kept in PBS for 7 days at 37 °C. The scaffolds were freeze dried and their weight was recorded. Weight of A after degradation: 9.967 mg Weight of C after degradation: 8.33 mg Calculate the percentage weight remaining for A and C scaffold and comment on why one scaffold degraded faster than the other.
Answer:
Explanation:
First PCL to form a scaffold, combined with gelatin.
They are made by first three forms A) made by just PCL.B) made by gelatin ratio PCL is 3:1 and last is C) made by gelatin PCL.
The decoration rate is 1%, 25% and 50% respectively.
A) Growth factor is stuck in 21 days, and can not spread. In this case in vivo experiment for 7 days used highly degradable scaffold use the ability to break down due to decomposition in vivo degradation rate depends on the scaffold's acid byproduct impact.
B) the amount of scaffolded degradation.
First, with scaffold A.
10 mg scaffold weight A= 1 per cent degradation.
Following degradation wt is 9.967=? Degradation per cent.
So, degradation (9.967* 1)/10= 0.9967 per cent.
Likewise for C) scaffold c 10 mg wt. Loss of 50 per cent.
After wt 8.33=? Degradation per cent.
Degradation (8.33* 50)/10=41.65 per cent.
Scaffold c degrades significantly, since the loss of wt is even greater.
Calculate the relative atomic mass of this sample of Germanium, giving your answer to two decimal places. ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... .........................................................................................................................
Answer:
Germanium is an element in the same group with Carbon and Silicon. The atomic number is 32. The relative atomic mass is usually measured with the Sample of an isotope. In this case Germanium has a relative atomic mass of 72.63
Answer:
The relative atomic mass of the sample of Germanium is 72.89 u
Explanation:
For a sample of germanium we have the following compositions:
m70Ge = 70 u
%70Ge = 22.6%
m72Ge = 72 u
%72Ge = 25.45%
m74Ge = 74 u
%74Ge = 36.73%
m76Ge = 76 u
%76Ge = 15.22%
To calculate the atomic mass of Germanium we must add all the atomic masses of its isotopes multiplied by its percentage of abundance and all this divided by 100, in this way:
mGe = [(m70Gex%70Ge)+(m72Gex%72Ge)+(m74Gex%74Ge)+(m76Gex%76Ge)]/100
replacing values:
mGe = [(70x22.6)+(72x25.45)+(74x36.73)+(76x15.22)]/100 = 72.89 u
t physiological pH, the carboxylic acid group of an amino acid will be ________, while the amino group will be ________, yielding the zwitterion form.
Answer:
At physiological pH,the carboxylic acid group of an amino acid will be deprotonated while the amino group will be protonated,yielding the zwitter ion form.
Explanation:
Deprotonated means removal of protons in an acid base reaction and protonated means addition of protons in an acid base reaction.
Both protonated and de protonated reaction takes place in catalytic acid bade reaction by changing either it's mass or it's charge.
During formation of zwitter ion, the carboxylic acid will be deprotonated by donating the H+ ion while the amino acid is protonated by taking the H+ ion.
R-CH-COOH R-CH-COO-
I ⇒ I
NH₂ NH₃ (Zwitter ion)
At physiological pH, the carboxylic acid group of an amino acid will be negatively charged, while the amino group will be positively charged, yielding the zwitterion form.
Explanation:Amino acids can exist as zwitterions at physiological pH. The carboxylic acid group of an amino acid will be negatively charged as a carboxylate ion (-COO-) at physiological pH. On the other hand, the amino group of an amino acid will be positively charged as an ammonium ion (-NH3+) at physiological pH. This combination of positive and negative charges gives rise to the zwitterion form of an amino acid.
If a reaction has a very low activation energy, the products would be expected to form quickly at high temperatures. Select the reason why such a reaction might form products very slowly, or not at alla. negative delta G
b. lack of a catalyst
c. positive delta G
Answer: Option (c) is the correct answer.
Explanation:
It is known that when Gibb's free energy, that is, [tex]\Delta G[/tex] has a negative value then the reaction will be spontaneous and the formation of products is favored more rapidly.
Activation energy is defined as the minimum amount of energy required to initiate a chemical reaction.
So, when reactants of a chemical reaction are unable to reach towards its activation energy then a catalyst is added to lower the activation energy barrier so the reaction can take place rapidly.
Since, the given reaction has low activation energy. Therefore, there is no need to add a catalyst.
And, when value of [tex]\Delta G[/tex] is positive then the reaction is spontaneous in nature and formation of products is less favored.
Thus, we can conclude that for the given situation positive delta G is the reason that a reaction might form products very slowly, or not at all.
Lab group B determines that 0.003 moles of bicarbonate was present in their portion of an antacid tablet. How many grams of bicarbonate are in their tablet? Report your answer to 3 decimal places.
Answer:
There are 0.183 grams of bicarbonate in their tablet
Explanation:
Step 1: Data given
0.003 moles of bicarbonate was present in their portion of an antacid tablet
Bicarbonate = HCO3-
Molar mass of bicarbonate = 61.02 g/mol
Step 2: Calculate mass bicarbonate
Mass bicarbonate = moles bicarbonate * molar mass bicarbonate
Mass bicarbonate = 0.003 moles * 61.02 g/mol
Mass bicarbonate = 0.183 grams
There are 0.183 grams of bicarbonate in their tablet
) PABA refers to para-aminobenzoic acid which is used in some sunscreen formulations. If a 0.055 M solution of PABA has a pH of 2.96, determine the Ka of PABA. What kind of problem do you think thisis
Answer: [tex]K_a[/tex] of PABA is 0.000022
Explanation:
[tex]NH_2C_6H_5COOH\rightarrow H^++NH_2C_6H_5COO^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.055 M and [tex]\alpha[/tex] = ?
[tex]K_a=?[/tex]
[tex]pH=-log[H^+][/tex]
[tex]2.96=-log[H^+][/tex]
[tex][H^+]=1.09\times 10^{-3}[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex]1.09\times 10^{-3}=0.055\times \alpha[/tex]
[tex]\alpha=0.02[/tex]
Putting in the values we get:
[tex]K_a=\frac{(0.055\times 0.02)^2}{(0.055-0.055\times 0.02)}[/tex]
[tex]K_a=0.000022[/tex]
Thus [tex]K_a[/tex] of PABA is 0.000022
What tests can you do to identify gases, based on their chemical properties?
Calculate the friction of the person on ski. We use the situation from example 5.1 "Skiing Exercise" from the text book, but we change the slope to 15 degrees. Tips: - The gravity force of the person is 608 N. - You first have to calculate the perpendicular component of the gravity force. This is also the normal force.
Answer: Friction (Fk) = 46.2N.
Explanation: In the attachment, there is the drawing of the forces acting on the skiing person. As the person is in movement, the formula to calculate the friction is Fk = μk.N, where μk is the coefficient of kinetic friction and N is normal force. Normal force is the force necessary to keep a person "above" the ground. To find the normal force as suggested, we use the representation of the forces in the attachment. With it, it's shown that to calculate N:
cos 15° = [tex]\frac{N}{Fg}[/tex]
N = Fg·cos 15°
N = 608·(-0.76)
N = - 461.9N
Now, with N and knowing the coefficient, which can be found in Physics Text Books and it is 0.1, we have:
Fk = 0.1·(-461.9)
Fk = - 46.2N
The friction is negative because it is pointing to the opposite side of the reference.
An aqueous solution of sodium sulfate is allowed to react with an aqueous solution of barium nitrate. What is the coefficient of the solid in the balanced equation
Answer: The coefficient of the solid in the balanced equation is 1.
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
The balanced chemical equation is:
[tex]Na_2SO_4(aq)+Ba(NO_3)_2(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)[/tex]
Thus a coefficient of 1 is placed in front of the solid.
The coefficient of the solid in the balanced equation is 1.
The double substitution reaction is a reaction in which ion exchange takes place. Salts that are soluble in water are indicated by the symbol (aq), and salts that are insoluble in water and remain solid are represented by (s) according to the chemical formula.The balanced chemical equation is:
Ba(NO3)2 + Na2SO4 → BaSO4 + NaNO3
Thus, the coefficient of the solid in the balanced equation is 1.
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One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?
A 55-gram serving of cereal supplies 270 mg of sodium, which is 11% of the recommended daily allowance. The recommended daily allowance contains 2454.55 mg of sodium and 106.82 mol of sodium. It also contains 6.43 x 10²⁵ atoms of sodium.
Explanation:To find the moles of sodium in the recommended daily allowance, we need to know the amount of sodium in the daily allowance. The problem states that a 55-gram serving of the cereal supplies 270 mg of sodium, which is 11% of the recommended daily allowance.
So, the total amount of sodium in the daily allowance is
270 mg / 0.11 = 2454.55 mg.
To convert this to moles, we use the molar mass of sodium, which is 22.99 g/mol.
Therefore, the number of moles of sodium in the daily allowance is 2454.55 mg / 22.99 g/mol = 106.82 mol.
To find the number of atoms of sodium in the daily allowance, we use Avogadro's number (6.022 x 10²³ atoms/mol). Therefore, the number of atoms of sodium in the daily allowance is 106.82 mol x (6.022 x 10²³ atoms/mol) = 6.43 x 10²⁵ atoms.
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To find the moles and atoms of sodium in the recommended daily allowance, divide the mass of sodium in mg by its molecular weight, then multiply the resulting moles by Avogadro's number to get atoms. There are roughly 0.107 moles and 6.44 x 1022 atoms of sodium in the allowance.
One 55-gram serving of cereal provides 270 mg of sodium, which is 11% of the recommended daily allowance. To calculate the total recommended daily allowance, we divide 270 mg by 0.11, getting approximately 2455 mg of sodium. The molecular weight of sodium is about 22.99 g/mol, so we can find the number of moles of sodium in the daily allowance by dividing the mass (in grams) by the molecular weight. This results in approximately 0.107 moles of sodium.
To find the number of atoms of sodium, we use Avogadro's number, which is approximately 6.022 x 1023 atoms/mole. Multiplying the number of moles by Avogadro's number gives us the number of atoms. So, there are roughly 6.44 x 1022 sodium atoms in the recommended daily allowance of sodium.
The formation constant* of [M(CN) 4 ]2− is 7.70 × 10 16 , where M is a generic metal. A 0.150 mole quantity of M(NO3)2 is added to a liter of 0.820 M NaCN solution. What is the concentration of M2+ ions at equilibrium?
The question involves calculation of equilibrium concentrations in a complex formation reaction in a solution, where the reaction is between a metal ion M²⁺ and cyanide ions CN⁻ to form a complex [M(CN)4]²⁻. The stated formation constant for the reaction allows formulating an equation for calculating the equilibrium concentration of M²⁺.
Explanation:To determine the concentration of M²⁺ ions at equilibrium, we need to consider the process of complex formation here, which is M²⁺ + 4CN⁻ → [M(CN)4]²⁻. The formation constant for this reaction is given as 7.7 x 10¹⁶. Given a 0.150 mole quantity of M(NO3)2 is added to a liter of 0.820 M NaCN solution, initially we will have [M²⁺] = 0.150 M and [CN⁻] = 0.820 M.
As the reaction proceeds to equilibrium, [M²⁺] drops by x amount reacting with 4x amount of [CN⁻]. At equilibrium, [M²⁺] = 0.150 - x, [CN-] = 0.820 - 4x and [M(CN)4]²⁻ = x. Applying the formation constant: 7.7 x 10¹⁶ = [M(CN)4]²⁻ / ([M²⁺][CN-]⁴) = x / (0.150 - x)(0.820 - 4x)⁴
This is a difficult equation to solve directly, but if we make the assumption that x, which relates to the equilibrium concentration of [M²⁺], is much less than initial concentrations of M²⁺ and CN⁻, this simplifies the equation and allows determination of the equilibrium concentration of M²⁺.
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To find the concentration of M2+ ions at equilibrium, use the formation constant and initial concentrations of M(NO3)2 and NaCN.
Explanation:The concentration of M2+ ions at equilibrium can be determined using the formation constant and the initial concentrations of M(NO3)2 and NaCN. First, calculate the initial concentration of M2+ ions by multiplying the concentration of M(NO3)2 (0.150 M) by its stoichiometric coefficient (2). Then, use the formation constant (7.70 × 10^16) to calculate the concentration of [M(CN)4]2- ions at equilibrium. Since the stoichiometric coefficient of M2+ ions is also 1, the concentration of M2+ ions at equilibrium will be equal to the concentration of [M(CN)4]2- ions.
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13. In animals, nitrogenous wastes are produced mostly from the catabolism of
A. phospholipids and glycolipids
B. proteins and nucleic acids.
C. starch and cellulose
D. triglycerides and steroids
E. fatty acids and glycerol
Answer:
The answer is B
Explanation:
2 UO2+ + 4 H+U4+ + UO22+ + 2 H2O is second order in UO2+ and third order overall. Complete the rate law for this reaction in the box below. Use the form k[A]m[B]n... , where '1' is understood for m, n ... (don't enter 1) and concentrations taken to the zero power do not appear
Answer:
[tex]r=k[UO_2^+]^2[H^+][/tex]
Explanation:
The given reaction is :-
[tex]2UO_2^++4H^+\rightarrow U^{4+}+UO_2^{2+}+2H_2O[/tex]
According to the law of mass action:-
The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.
Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.
The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.
m = 2 = is order with respect to [tex]UO_2^+[/tex]
n = order with respect to H+
overall order = m+n = 3
n = 3 - m = 3 - 2 = 1
Rate law is:-
[tex]r=k[UO_2^+]^2[H^+][/tex]
A system at equilibrium has two aqueous products and two aqueous reactants. An additional amount of one of the products is added to the system. After the addition, which of the following will change?Select the correct answer below:a. the amount of the added productb. the amount of the other productc. the amounts of the reactantsd. all of the above
If an extra amount of a product is added to a system at equilibrium, the system will shift towards the reactants to counteract this change. This shift can cause changes in the amounts of all substances involved, including the added product, the other product, and the reactants.
Explanation:When an extra amount of a product is added to a system at equilibrium, the system shifts to counteract this change and re-establish equilibrium, as described by Le Châtelier's principle. Specifically, the system will shift in the direction that reduces the excess amount of the product, i.e., towards the reactants. With this shift, the quantities of all substances in the system can change.
Therefore, the appropriate answer is d. all of the above. Adding more of one product will indeed increase the amount of that product initially (a), but the system will shift to remove the added product by moving towards the reactants. In this process, the amount of the other product may also change as it is consumed or generated (b), and the amounts of reactants will also adjust as they are generated or consumed (c).
Learn more about Le Châtelier's principle here:https://brainly.com/question/12099569
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A chemist adds of a sodium carbonate solution to a reaction flask. Calculate the mass in grams of sodium carbonate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
The question is incomplete, here is the complete question:
A chemist adds 180.0 mL of a 1.42 M sodium carbonate solution to a reaction flask. Calculate the mass in grams of sodium carbonate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answer: The mass of sodium carbonate that must be added are 40.9 grams
Explanation:
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Molar mass of sodium carbonate = 106 g/mol
Molarity of solution = 1.42 M
Volume of solution = 180.0 mL
Putting values in above equation, we get:
[tex]1.42mol/L=\frac{\text{Mass of sodium carbonate}\times 1000}{160g/mol\times 180}\\\\\text{Mass of sodium carbonate}=\frac{160\times 1.42\times 180}{1000}=40.9g[/tex]
Hence, the mass of sodium carbonate that must be added are 40.9 grams
A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa(50 ksi) is exposed to a stress of 1205 MPa(174800psi) . (a) If the largest surface crack is 0.8 mm (0.03150 in.) long, determine the critical stress ?c. (b) Will this specimen experience fracture? Assume that the parameter Y has a value of 0.99.
Explanation:
(a) Formula for critical stress is as follows.
[tex]\sigma_{c} = \frac{k_{IC}}{\tau \sqrt{\pi \times a}}[/tex]
Here, [tex]K_{IC}[/tex] = 54.8
[tex]\tau[/tex] = 0.99
a = 0.8 mm = [tex]0.8 \times 10^{-3}[/tex] m
Putting the given values into the above formula as follows.
[tex]\sigma_{c} = \frac{k_{IC}}{\tau \sqrt{\pi \times a}}[/tex]
= [tex]\frac{54.8}{0.99 \times \sqrt{3.14 \times 0.8 \times 10^{-3}}}[/tex]
= 1107 MPa
Hence, value of critical stress is 1107 MPa.
(b) Applied stress value is given as 1205 MPa and since it is more than the critical stress (1107 MPa) as a result, a fracture will occur.
H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles of H2(g) react at standard conditions. S°surroundings = J/K
Answer: The value of [tex]\Delta S^o[/tex] for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K
Explanation:
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change is of a reaction is:
[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]
For the given chemical reaction:
[tex]H_2(g)+F_2(g)\rightarrow 2HF(g)[/tex]
The equation for the entropy change of the above reaction is:
[tex]\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})][/tex]
We are given:
[tex]\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K[/tex]
Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K
We are given:
Moles of hydrogen gas reacted = 2.20 moles
By Stoichiometry of the reaction:
When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K
So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = [tex]\frac{-14.1}{1}\times 2.20=-31.02J/K[/tex]
Hence, the value of [tex]\Delta S^o[/tex] for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules. Several other gases make up the remaining 1% of air molecules. What is the partial pressure of the other gases in air (excluding oxygen and nitrogen) at an atmospheric pressure of 0.90 atm
Answer:
The partial pressure of the other gases is 0.009 atm
Explanation:
Step 1: Data given
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.
The atmospheric pressure = 0.90 atm
Step 2: Calculate mol fraction
If wehave 100 moles of air, 78 moles will be nitrogen,
21 moles will be oxygen, and 1 mol will be other gases.
Mol fraction = 1/100 = 0.01
Step 3: Calculate the partial pressure of the other gases
Pgas = Xgas * Ptotal
⇒ Pgas = the partial pressure = ?
⇒ Xgas = the mol fraction of the gas = 0.01
⇒Ptotal = the total pressure of the pressure = 0.90 atm
Pgas = 0.01 * 0.90 atm
Pgas = 0.009 atm
The partial pressure of the other gases is 0.009 atm
In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g H2O initially at 23.2°C. The temperature of the water increased to 31.8°C. What was the change in enthalpy for the dissolution of the compound? Give your answer in units of joules per gram of compound. Assume that the specific heat of the solution is the same as that of pure water, 4.18 J ⁄ (g ⋅ °C).
The change in enthalpy for the dissolution of the compound is 344.11 J/g.
Explanation:In a coffee-cup calorimeter experiment, the change in enthalpy for the dissolution of a compound can be determined using the equation q = mCΔT, where q is the amount of heat absorbed or released, m is the mass of the compound, C is the specific heat of the solution, and ΔT is the change in temperature. Given that the initial temperature of the water is 23.2°C and the final temperature is 31.8°C, the change in temperature is ΔT = 31.8°C - 23.2°C = 8.6°C.
Since 10.00 g of the compound was added to 75.0 g of water, the total mass of the solution is 75.0 g + 10.00 g = 85.0 g.
Using the equation q = mCΔT, where C is the specific heat of water (4.18 J/(g·°C)), the amount of heat involved in the dissolution can be calculated as q = (85.0 g)(4.18 J/(g·°C))(8.6°C) = 3,441.14 J.
The change in enthalpy for the dissolution of the compound is thus 3,441.14 J/10.00 g = 344.11 J/g of the compound.
A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.
The solubility product constant (Ksp) is calculated as 2.392.
To solve this problem, we can use the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
where:
Ecell is the measured potential difference between the rod and the SHE (0.5812 V)
E°cell is the standard cell potential
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (25°C = 298 K)
n is the number of electrons transferred in the balanced redox reaction
F is Faraday's constant (96485 C/mol)
Q is the reaction quotient
In this case, the balanced redox reaction is:
Ag2C2O4(s) ⇌ 2Ag+(aq) + C2O4^2-(aq)
The solubility product constant (Ksp) for silver oxalate can be expressed as:
Ksp = [Ag+]^2 * [C2O4^2-]
Since the rod is positive, it means that Ag+ ions are being reduced to Ag(s) on the rod, so we can write:
E°cell = E°red,cathode - E°red,anode
The standard reduction potential for Ag+ to Ag(s) is 0.7996 V, and the standard reduction potential for C2O4^2- to CO2(g) is 0.1936 V.
Therefore, E°cell = 0.7996 V - 0.1936 V = 0.606 V
Now we can substitute all the values into the Nernst equation:
0.5812 V = 0.606 V - (8.314 J/(mol·K) * 298 K / (2 * 96485 C/mol)) * ln(Q)
Simplifying:
0.5812 V = 0.606 V - (0.0257 V) * ln(Q)
0.0257 V * ln(Q) = 0.606 V - 0.5812 V
0.0257 V * ln(Q) = 0.0248 V
ln(Q) = 0.0248 V / 0.0257 V
ln(Q) = 0.964
Q = e^(0.964)
Q ≈ 2.622
Since Q = [Ag+]^2 * [C2O4^2-], we can write:
[Ag+]^2 * [C2O4^2-] ≈ 2.622
The solubility product constant (Ksp) is the product of the concentrations of the ions at equilibrium, so we can assume that the concentrations are equal to each other:
[Ag+] ≈ [C2O4^2-]
Therefore, we can substitute [Ag+] for [C2O4^2-]:
[Ag+]^2 * [Ag+] ≈ 2.622
[Ag+]^3 ≈ 2.622
Taking the cube root of both sides:
[Ag+] ≈ (2.622)^(1/3)
[Ag+] ≈ 1.378
Therefore, the approximate solubility of silver oxalate is 1.378 M.
The solubility product constant (Ksp) can now be calculated as:
Ksp = [Ag+]^2 * [C2O4^2-] ≈ (1.378)^2 * (1.378) ≈ 2.392
The solubility product constant for silver oxalate at [tex]25\°C[/tex] is approximately [tex]\( 10^{-22.125} \)[/tex].
The solubility product constant [tex](\( K_{sp} \))[/tex] for silver oxalate can be calculated using the Nernst equation and the measured potential difference between the silver rod and the standard hydrogen electrode (SHE). The reaction occurring at the silver rod can be represented as:
[tex]\[ \text{Ag}_2\text{C}_2\text{O}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{C}_2\text{O}_4^{2-}(aq) \][/tex]
The Nernst equation for this reaction at [tex]25\°C[/tex] is given by:
[tex]\[ E = E^{\circ} - \frac{0.0592}{n} \log \frac{1}{[\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}]} \][/tex]
where [tex]\( E \)[/tex] is the measured potential (0.5812 V), [tex]\( E^{\circ} \)[/tex] is the standard reduction potential for the [tex]\( \text{Ag}^+ \)[/tex] / Ag couple, [tex]\( n \)[/tex] is the number of electrons transferred in the half-reaction (which is 2 for this reaction), and the concentrations of [tex]\( \text{Ag}^+ \)[/tex] and [tex]\( \text{C}_2\text{O}_4^{2-} \)[/tex] are equal to the solubility of silver oxalate since they come from its dissolution.
First, we need to find the standard reduction potential [tex]\( E^{\circ} \)[/tex] for the [tex]\( \text{Ag}^+ \)[/tex] / Ag couple, which is 0.7996 V.
Rearranging the Nernst equation to solve for the solubility \( S \), we get:
[tex]\[ S = [\text{Ag}^+] = [\text{C}_2\text{O}_4^{2-}] = 10^{\frac{(E^{\circ} - E)n}{0.0592}} \][/tex]
Plugging in the values:
[tex]\[ S = 10^{\frac{(0.7996 - 0.5812) \times 2}{0.0592}} \][/tex]
[tex]\[ S = 10^{\frac{0.2184 \times 2}{0.0592}} \][/tex]
[tex]\[ S = 10^{\frac{0.4368}{0.0592}} \][/tex]
[tex]\[ S = 10^{7.375} \][/tex]
[tex]\[ S \approx 10^{-7.375} \][/tex]
The solubility product constant [tex]\( K_{sp} \)[/tex] is given by:
[tex]\[ K_{sp} = [\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}] \][/tex]
[tex]\[ K_{sp} = (S)^2(S) \][/tex]
[tex]\[ K_{sp} = (10^{-7.375})^2(10^{-7.375}) \][/tex]
[tex]\[ K_{sp} = 10^{-7.375 \times 3} \][/tex]
[tex]\[ K_{sp} = 10^{-22.125} \][/tex]
One way to represent this equilibrium is: N2O4(g)2NO2(g) Indicate whether each of the following statements is true, T, or false, F. AT EQUILIBRIUM we can say that: 1. The concentration of NO2 is equal to the concentration of N2O4. 2. The rate of the dissociation of N2O4 is equal to the rate of formation of N2O4. 3. The rate constant for the forward reaction is equal to the rate constant of the reverse reaction. 4. The concentration of NO2 divided by the concentration of N2O4 is equal to a constant.
Answer:
1. The concentration of [tex]NO_2[/tex] is equal to the concentration of [tex]N_2O_4[/tex] : False
2. The rate of the dissociation of [tex]N_2O_4[/tex] is equal to the rate of formation of [tex]N_2O_4[/tex]: True
3. The rate constant for the forward reaction is equal to the rate constant of the reverse reaction: False
4. The concentration of [tex]NO_2[/tex] divided by the concentration of [tex]N_2O_4[/tex] is equal to a constant : False
Explanation:
[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]
The concentrations of reactant and product is constant and not equal.
The rate of forward and backward reactions are equal. Thus rate of the dissociation of [tex]N_2O_4[/tex] is equal to the rate of formation of [tex]N_2O_4[/tex]
For a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium constant for the backward reaction and not equal.
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
[tex]K_c=\frac{[NO_2]^2}{[N_2O_4]}[/tex]
Final answer:
Statements 1 and 3 are False, and 2 and 4 are True. At equilibrium, the rate of formation and dissociation of N₂O₄ are equal but their concentrations are not necessarily the same. Rate constants for forward and reverse reactions are different yet define the equilibrium constant. The concentration ratio of NO₂ to N₂O₄ is constant at a given temperature.
Explanation:
When dinitrogen tetroxide (N₂O₄) is in equilibrium with nitrogen dioxide (NO₂), the following statements can be made:
1. The concentration of NO₂ is equal to the concentration of N₂O₄. False - The stoichiometry of the reaction is 1:2, so for every molecule of N₂O₄ that dissociates, two molecules of NO₂ are formed.2. The rate of the dissociation of N₂O₄ is equal to the rate of formation of N₂O₄. True - At equilibrium, the rate of the forward reaction (dissociation of N₂O₄) is equal to the rate of the reverse reaction (formation of N₂O₄).3. The rate constant for the forward reaction is equal to the rate constant of the reverse reaction. False - The rate constants for the forward and reverse reactions are generally different, but their ratio is constant at a given temperature, defining the equilibrium constant.4. The concentration of NO₂ divided by the concentration of N₂O₄ is equal to a constant. True - At equilibrium, this ratio is equal to the equilibrium constant (Keq) for the reaction at a given temperature.Grey Goose ® vodka has an alcohol content of 40.0 % (v/v). Assuming that vodka is composed of only ethanol and water answer the following questions. Note: The molar masses of water and ethanol are 18.0 g and 46.0 g, respectively. The densities of water, ethanol, and this vodka mixture are 1.00 g/mL, 0.789 g/mL, and 0.935 g/mL, respectively
a. Calculate the molarity of ethanol in this vodka, assuming that water is the solvent.
b. Calculate the percent by mass of ethanol % (m/m) in this vodka.
c. Calculate the molality of ethanol in this vodka assuming that water is the solvent.
d. Calculate the mole fractions of ethanol and water in this vodka.
e. Calculate the vapor pressure, in torr, of this vodka at 25.0 oC if the vapor pressures of pure water and ethanol are 23.8 torr and 45.0 torr, respectively?
Explanation:
Grey Goose vodka has an alcohol content of 40.0 % (v/v).
Volume of vodka = V = 100 mL
This means that 40.0 mL of alcohol is present 100 mL of vodka.
Volume of ethanol=V' = 40.0 mL
Mass of ethanol = m
Density of the ethanol = d = 0.789 g/mL
[tex]m=d\times V' = 0.789 g/ml\times 40.0 mL=31.56 g[/tex]
Volume of water = V''= 100 ml - 40.0 mL = 60.0 mL
Mass of water = m'
Density of the water = d' = 1.00 g/mL
[tex]m'=d'\times V'' = 1.00 g/ml\times 60.0 mL=60.0 g[/tex]
a.)
Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Volume of vodka = V = 100 mL = 0.100 L ( 1mL=0.001 L)
Molarity of the ethanol:
[tex]=\frac{0.6861 mol}{0.100 L}=6.861 M[/tex]
6.861 M the molarity of ethanol in this vodka.
b) Mass of ethanol = 31.56 g
Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Volume of vodka = V = 100 mL
Mass of vodka = m
Density of the water = D = 0.935 g/mL
[tex]M=D\times V=0.935 g/ml\times 100 ml=93.5 g[/tex]
The percent by mass of ethanol % (m/m):
[tex]\frac{31.56 g}{93.5 g}\times 100=33.75\%[/tex]
33.75% is the percent by mass of ethanol % (m/m) in this vodka.
c)
Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Mass of solvent that is water = 60.0 g = 0.060 kg ( 1g = 0.001 kg)
Molality of ethanol in vodka :
[tex]m=\frac{0.6861 mol}{0.060 kg}=11.435 m[/tex]
11.435 m is the molality of ethanol in this vodka.
d)
Moles of ethanol = [tex]n_1=\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Moles of water = [tex]n_2=\frac{60.0 g}{18 g/mol}=3.333 mol[/tex]
Mole fraction of ethanol = [tex]\chi_1[/tex]
[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.6861 mol}{0.6861 mol+3.333 mol}[/tex]
= 0.1707
Mole fraction of water = [tex]\chi_2[/tex]
[tex]\chi_2=\frac{n_2}{n_1+n_2}=\frac{3.3333 mol}{0.6861 mol+3.333 mol}[/tex]
= 0.8290
e)
The vapor pressure of vodka = P
Mole fraction of ethanol = [tex]\chi_1=0.1707[/tex]
Mole fraction of water = [tex]\chi_2=0.8290[/tex]
The vapor pressures of ethanol = [tex]p_1=45.0 Torr[/tex]
The vapor pressures of pure water = [tex]p_2=23.8Torr[/tex]
[tex]P=\chi_1\times p_1+\chi_2\times p_2[/tex]
[tex]P=0.1707\times 45.0torr+0.8290\times 23.8 Torr=27.41 torr[/tex]
The vapor pressure of vodka is 27.41 Torr.