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.
a) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains only carbon and hydrogen, what is its empirical formula?
b) Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005-g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is the empirical formula for menthol? If menthol has a molar mass of 156 g>mol, what is its molecular formula?
Answer:
For a: The empirical formula for the given compound is [tex]CH[/tex]
For b: The empirical and molecular formula for the given organic compound are [tex]C_{10}H_{20}O[/tex]
Explanation:
For a:The chemical equation for the combustion of hydrocarbon follows:
[tex]C_xH_y+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', and 'y' are the subscripts of Carbon and hydrogen respectively.
We are given:
Conversion factor used: 1 g = 1000 mg
Mass of [tex]CO_2=5.86mg=5.86\times 10^{-3}g[/tex]
Mass of [tex]H_2O=1.37mg=1.37\times 10^{-3}g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in [tex]5.86\times 10^{-3}g[/tex] of carbon dioxide, [tex]\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in [tex]1.37\times 10^{-3}g[/tex] of water, [tex]\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g[/tex] of hydrogen will be contained.
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is [tex]0.133\times 10^{-3}[/tex] moles.
For Carbon = [tex]\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1[/tex]
For Hydrogen = [tex]\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H = 1 : 1
Hence, the empirical formula for the given compound is [tex]CH[/tex]
For b:The chemical equation for the combustion of menthol follows:
[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2[/tex] = 0.2829 g
Mass of [tex]H_2O[/tex] = 0.1159 g
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.2829 g of carbon dioxide, [tex]\frac{12}{44}\times 0.2829=0.077g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 0.1159 g of water, [tex]\frac{2}{18}\times 0.1159=0.013g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00065 moles.
For Carbon = [tex]\frac{0.0064}{0.00065}=9.84\approx 10[/tex]
For Hydrogen = [tex]\frac{0.013}{0.00065}=20[/tex]
For Oxygen = [tex]\frac{0.00065}{0.00065}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O = 10 : 20 : 1
The empirical formula for the given compound is [tex]C_{10}H_{20}O[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]
We are given:
Mass of molecular formula = 156 g/mol
Mass of empirical formula = 156 g/mol
Putting values in above equation, we get:
[tex]n=\frac{156g/mol}{156g/mol}=1[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O[/tex]
Hence, the empirical and molecular formula for the given organic compound are [tex]C_{10}H_{20}O[/tex]
Final answer:
The empirical formula of toluene, derived from combustion yielding 5.86 mg of CO2 and 1.37 mg of H2O, is C7H8. For menthol, from 0.1005 g combusted to give 0.2829 g of CO2 and 0.1159 g of H2O and a molar mass of 156 g/mol, its empirical and molecular formulas are both C10H20O.
Explanation:
Empirical and Molecular Formulas of Organic Compounds
Combustion analysis is commonly used to determine the empirical formulas of organic compounds. The process involves burning a sample of the compound to produce CO2 and H2O, which can then be analyzed to determine the amounts of carbon and hydrogen in the original compound.
The empirical formula of toluene, given that combustion yields 5.86 mg of CO2 and 1.37 mg of H2O with only carbon and hydrogen present, can be found by:
Converting the mass of CO2 to moles of carbon.
Converting the mass of H2O to moles of hydrogen.
Determining the simplest whole number ratio of carbon to hydrogen atoms to find the empirical formula.
Following these steps, we can determine that the empirical formula of toluene is C7H8.
For menthol, with a 0.1005-g sample yielding 0.2829 g of CO2 and 0.1159 g of H2O, and a known molar mass of 156 g/mol, we:
Convert the masses of CO2 and H2O to moles of carbon and hydrogen, respectively.
Assume the rest of the mass is due to oxygen and calculate its moles.
Determine the empirical formula.
Calculate the molecular formula using the empirical formula and the given molar mass.
By doing so, we find that menthol's empirical formula is C10H20O and its molecular formula is also C10H20O.
What is Δn for the following equation in relating Kc to Kp?2 SO2(g) + O2(g) ↔ 2 SO3(g)23-2-11
Answer:
-1
Explanation:
The relation between Kp and Kc is given below:
[tex]K_p= K_c\times (RT)^{\Delta n}[/tex]
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
[tex]2SO_2_{(g)}+O_2_{(g)}\rightleftharpoons2SO_3_{(g)} [/tex]
Δn = (2)-(2+1) = -1
Thus, Kp is:
[tex]K_p= K_c\times (RT)^{-1}[/tex]
A voltaic cell utilizes the following reaction: 4Fe2+(aq)+O2(g)+4H+(aq)→4Fe3+(aq)+2H2O(l).
The EMF under standard contions is .46 V.
What is the emf of this cell when [Fe2+]= 2.0M , [Fe3+]= 1.9
Answer:
The Emf of the given cell at [Fe2+] = 2.0 M and [Fe3+] = 1.9 M is 0.48 V
Explanation:
The half cell reaction can be written as :
Anode-Half (oxidation) :
[tex]Fe^{2+}\rightarrow Fe^{3+} + 1e^{-}[/tex] ......E = 0.77 V
(multiply this equation by 4 to balance the electrons)
Cathode-half (reduction)
[tex]4H^{+} +O_{2} + 4e^{-} \rightarrow 2H_{2}O[/tex]....E= 1.23 V
[tex]E^{0}_{cell} = E_{cathode} - E_{anode}[/tex]
[tex]E^{0}_{cell} = 1.23 - 0.77[/tex]
[tex]E^{0}_{cell} = 0.46 V[/tex]
According to Nernst Equation
[tex]E_{cell} = E^{0} - \frac{RT}{nF}lnQ[/tex]
[tex]E_{cell} = E^{0} - \frac{0.059}{n}logQ[/tex]
n = number of electron transferred in the cell reaction = 4
The balanced equation is :
[tex]4Fe^{2+} + 4H^{+} +O_{2} \rightarrow 4Fe^{3+} + 2H_{2}O[/tex]
[tex]E_{cell} = 0.46 - \frac{0.059}{4}logQ[/tex]
[tex]log Q = \frac{[Fe^{3+}]^{4}}{[Fe^{2+}]^{4}}[/tex]
[tex]log Q = \left ( \frac{[Fe^{3+}]}{[Fe^{2+}]} \right )^{4}[/tex]
[tex]log Q = \left ( \frac{1.9}{2.0} \right )^{4}[/tex]
Insert the value of log Q in Nernst Equation:
[tex]E_{cell} = 0.46 - \frac{0.059}{4} log\left ( \frac{1.9}{2.0} \right )^{4}[/tex]
(using :[tex]log^{a}b = a\log b[/tex]
)
[tex]E_{cell} = 0.46 - log\frac{1.9}{2.0}[/tex]
[tex]E_{cell} = 0.46 - log(0.95)[/tex]
[tex]E_{cell} = 0.46 -(-0.0227)[/tex]
[tex]E_{cell} = 0.482 V[/tex]
The cell EMF is obtained from Nernst equation as 0.451 V.
The reaction equation is given by;
4Fe2+ (aq) + O2(g) + 4H+ (aq)-------->4Fe3+(aq)+2H2O(l)
We have the following information;
EMF under standard conditions = 0.46 V
[Fe2+]= 2.0M
[Fe3+]= 1.9 M
From Nernst equation;
Ecell = E°cell - 0.0592/n logQ
Where;
E°cell = 0.46 V
n = 4
Q = [Fe3+]^4/ [Fe2+]^4
Q = [1.9]^4/[2.0]^4
Q = 0.8
Substituting values;
Ecell = 0.46 - 0.0592/4 log (0.8)
Ecell = 0.451 V
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2) What is the difference between spilling solid KOH on your skin, and spilling a solution of KOH in triethyleneglycol? 3) Why should you always monitor a sand bath with an external thermometer, as well as the temperature recorded by the hotplate?
Answer:
Solid KOH will corrode the skin but a solution of KOH in triethyleneglycol will not.
The temperature is monitored with an external thermometer also because the sand looses some heat to energy exchange with the surrounding.
Explanation:
KOH is a deliquescent solid which is very corrosive upon contact with skin. Its solution in an organic liquid is not corrosive.
Secondly, when exposed to the surrounding, heat is lost according to the laws of thermodynamics.
Question 9
A chemistry student needs 85.0mL of carbon tetrachloride for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of carbon tetrachloride is 1.59 gcm^-3 . Calculate the mass of carbon tetrachloride the student should weigh out.
Be sure your answer has the correct number of significant digits.
To find the mass of carbon tetrachloride needed, multiply the volumes by the density. The student needs to weigh out 135 g of carbon tetrachloride, rounded to the correct number of significant digits.
Explanation:To solve this problem, we need to multiply the volume of carbon tetrachloride needed by the student (85.0 mL) by its density (1.59 g/cm^3). However, note that the volume is given in milliliters (mL) while the density is given in grams per cubic centimeter (g/cm^3). You need to know that 1 mL is equivalent to 1 cm^3. So, we don’t have to convert the units. Hence, the calculation becomes:
Mass = Density x Volume = 1.59 g/cm^3 x 85.0 mL = 135.15 g.
Since the given values only have three significant digits, the answer should also be rounded to three significant digits so the student needs to weigh out 135 g of carbon tetrachloride.
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A given substance is a white, granular solid at 25 °C that does not conduct electricity. It melts at 750 °C wand the melt conducts electricity. It dissolves easily in water and produces a conductive solution. Identify the type of crystal and justify your answer.
Answer:
Ionic crystal
Explanation:
An ionic crystal has a high melting point. In an ionic crystal, the ions are tightly held in electrostatic attraction by their oppositely charged neighbors forming a rigid three dimensional lattice. However, when this solid melts, the rigid crystal structure collapses and the individual ions become free and mobile. Hence the melt conducts electricity.
Organic matter with a _______ ratio results in a low net release of nutrients during decomposition. This is because microbial growth is more limited by _______ than it is by _______. low C:N; energy; N supply high C:N; N supply; energy high C:H2O; water supply; energy low C:H2O; energy; water supply low N:H2O; N supply; water supply
Answer:
high C:N; N supply; energy
Explanation:
Nitrogen supply is required for microbial growth to synthesize nutrients such as amino acids and proteins. For a high C:N ratio, the amount of nitrogen supply is considerably small compared with the amount of carbon. As a result, a low amount of nutrients is released during decomposition.
In the given question, the organic matter with a __high C:N___ ratio results in a low net release of nutrients during decomposition. This is because microbial growth is more limited by __N supply____ than it is by __energy_____.
Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. These species are called ligands. In the past we have assumed all the d orbitals in some species are degenerate; however, they often are not. Sometimes the ligands bound to a central metal cation can split the d orbitals. That is, some of the d orbitals will be at a lower energy state than others. Ligands that have the ability to cause this splitting are called strong field ligands, CN- is an example of these. If this splitting in the d orbitals is great enough electrons will fill low lying orbitals, pairing with other electrons in a given orbital, before filling higher energy orbitals. In question 7 we had Fe2+, furthermore we found that there were a certain number (non-zero) of unpaired electrons. Consider now Fe(CN)64-: here we also have Fe2+, but in this case all the electrons are paired, yielding a diamagnetic species. How can you explain this?
A. There are 3 low lying d orbitals, which will be filled with 6 electrons before filling the 2, assumed to be degenerate, higher energy orbitals.
B. There are 4 low lying d orbitals, which will be filled with 8 electrons before filling the 1 higher energy orbital.
C. There is 1 low lying d orbital, which will be filled with two electrons before filling the 4, assumed to be degenerate, higher energy orbitals.
D. All the d orbitals are degenerate.
E. There are 2 low lying d orbitals, which will be filled with 4 electrons before filling the 3, assumed to be degenerate, higher energy orbitals.
Answer:
A
Explanation:
Iron has the ground state electronic configuration [Ar]3d64s2
Fe2+ has the electronic configuration [Ar]3d6.
In an octahedral crystal field, there are two sets of degenerate orbitals; the lower lying three t2g orbitals, and the higher level two degenerate eg orbitals. Strong field ligands cause high octahedral crystal field splitting, there by separating the two sets of degenerate orbitals by a tremendous amount of energy. This energy is much greater than the pairing energy required to pair the six electrons in three degenerate orbitals. Since CN- is a strong field ligand, it leads to pairing of six electrons in three degenerate orbitals
The Fe(CN)64- ion is diamagnetic due to the significant energy difference caused by strong field ligands like CN-, resulting in pairing of electrons in the 3 low energy d orbitals before filling the higher energy ones.
Explanation:The answer is A. There are 3 low lying d orbitals, which will be filled with 6 electrons before filling the 2, assumed to be degenerate, higher energy orbitals. When a metal ion is coordinated to ligands, such as in Fe(CN)64-, the degeneracy of the 3d orbitals is broken (they have different energies), due to the electrostatic interactions between the ligands and the orbitals. In the case of strong-field ligands like CN-, the energy difference is significant enough to cause pairing of electrons in the 3 low energy d orbitals before the 2 high energy orbitals are populated, resulting in a diamagnetic species.
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Write the values for the four quantum numbers for a 4f6 electron.2.Write the values for the four quantum numbers for a 3s1 electron.3.Write the values for the four quantum numbers for the last electron to fill 28Ni.4.Write the values for the four quantum numbers for the outermost electron in 22Ti.
The electron configuration shows the arrangement of electrons in atoms.
The electron configuration gives a description of the location of the electron in an atom.
Note that;
n = Principal quantum number
l = orbital quantum number
m = magnetic quantum number
s = spin quantum number
For the 4f electron;n = 4, l = 3, m = 3, -2, -1, 0, 1, 2, 3, s= ±1/2
For the 3s electron;n = 3, l = 0, m = 0, s = ±1/2
For the last electron in Ni of electron configuration; [Ar] 3d⁸ 4s²n = 4, l = 0, m = 0, s = ±1/2
For the outermost electron in Ti of electron configuration; [Ar] 3d² 4s²n = 4, l = 0, m = 0, s = ±1/2
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Final answer:
The quantum numbers for specific electrons in the 4f6, 3s1, 28Ni, and 22Ti configurations have been provided according to their respective principal, azimuthal, magnetic, and spin quantum numbers.
Explanation:
The four quantum numbers are principal (n), azimuthal (l), magnetic (ml), and spin (ms). Here are their specified values for the given electrons:
Final answer:
The quantum numbers for specific electrons in the 4f6, 3s1, 28Ni, and 22Ti configurations have been provided according to their respective principal, azimuthal, magnetic, and spin quantum numbers.
Explanation:
The four quantum numbers are principal (n), azimuthal (l), magnetic (ml), and spin (ms). Here are their specified values for the given electrons:
In the presence of excess oxygen, methane gas burns in a constant-pressure system to yield carbon dioxide and water:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O DH = -890.0 kJ
Calculate the value of q (kJ) in this exothermic reaction when 1.70 g of methane is combusted at constant pressure.
Answer:
94.56KJ
Explanation:
1.7 grams of CH4 contains 1.7/16 moles of CH4.
If 1 mole of CH4 give 890KJ
1.7/16 moles of CH4 gives 1.7/16*890 = 94.56KJ
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
(a) HNO3+H2O⟶H3O++NO3−
(b) CN− +H2O ⟶ HCN+OH−
(c) H2SO4+Cl− ⟶HCl+HSO4−
(d) HSO4−+OH− ⟶SO42−+H2O
(e) O2− + H2 O ⟶ 2OH−
Explanation:
As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.
The chemical species which accept proton are called Brønsted-Lowry base.
(a) [tex]HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-[/tex]
[tex]HNO_3[/tex] is Bronsted lowry acid and [tex]NO_3^-[/tex] is its conjugate base.
[tex]H_2O[/tex] is Bronsted lowry base and [tex]H_3O^+[/tex] is its conjugate acid.
(b)
[tex]CN^- + H_2O \rightarrow HCN + OH^-[/tex]
[tex]CN^-[/tex] is Bronsted lowry base and HCN is its conjugate acid.
[tex]H_2O[/tex] is Bronsted lowry acid and [tex]OH^-[/tex] is its conjugate base.
(c)
[tex]H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-[/tex]
[tex]H_2SO_4[/tex] is Bronsted lowry acid and [tex]HSO_4^-[/tex] is its conjugate base.
Cl^- is Bronsted lowry base and HCl is its conjugate acid.
(d)
[tex]HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O[/tex]
[tex]HSO_4^-[/tex] is Bronsted lowry acid and [tex]SO_4^{2-}[/tex] is its conjugate base.
OH^- is Bronsted lowry base and [tex]H_2O[/tex] is its conjugate acid.
(e)
[tex]O_{2-}+H_2O \rightarrow 2OH^-[/tex]
[tex]O_{2-}[/tex] is Bronsted lowry base and OH- is its conjugate acid.
[tex]H_2O[/tex] is Bronsted lowry acid and OH- is its conjugate base.
The identification and labelling of the Brønsted-Lowry acid and base, and their respective conjugate are done below.
What is Brønsted-Lowry acid and base?Brønsted-Lowry acids are chemical species that donate proton (H+) while the chemical species which accept proton (H+) are called Brønsted-Lowry base
Based on this question, the conjugate acid and base of the equations given are as follows:
For equation a:
HNO3 is the Bronsted lowry acid and NO3- is its conjugate base.H2O is Bronsted lowry base while H3O+ is its conjugate acid.For equation b:
Cyanide ion is the Bronsted lowry base while hydrogen cyanide is its conjugate acid. H2O is the Bronsted lowry acid while OH- is its conjugate base.For equation c:
H2SO4 is the Bronsted lowry acid while HSO4- is its conjugate base.Cl- is the Bronsted lowry base and HCl is its conjugate acid.For equation d:
HSO4- is the Bronsted lowry acid while SO4²- its conjugate base.OH- is the Bronsted lowry base and H2O is its conjugate acidFor equation e:
O²- is the Bronsted lowry base and OH- is its conjugate acid.H2O is the Bronsted lowry acid and OH- is its conjugate baseLearn more about Brønsted-Lowry base and acid at: https://brainly.com/question/21736327
Stan is a member of the Association for Consumer Research, but when he attends the conferences, he doesn't identify with the group. Even though his research is received favorably, he often feels detached and perceives himself as different from the other members. Stan can best be described as having a(n) _____ self-schema with respect to this group.idealizedseparatedactualizedconnectedambiguous
Answer:
Separated
Explanation:
Stan does not feel he identifies closely with the group. He feels the other members are close knit with each other but he is not part of the group.
For Stan, it would be an illusion to say he has a close relationship with the members of the group
This infers that Stan has a separated self-schema with respect to the group.
Certain glass objects are not meant to be heated and could shatter if exposed to a heat source. What two examples of heat-sensitive glassware are given in the video?
Answer: the heat-sensitive glassware that were given are : Volumetric and Graduated cylinder.
Explanation:glass material that reacts to ambient temperatures radiated off of other surfaces like hands or water is known as heat sensitive glassware. They are not meant to be heated and could shatter if exposed to a heat source. Examples from the video includes Volumetric and Graduated cylinder. Hope this helps. Thanks.
The glassware in the video that are shown to be heat sensitive are volumetric and graduated cylinders.
The exposure of heat to the system has been resulted in the atoms of the system to excite. The material that radiate hands or water with the exposure of the heat are termed as heat-sensitive materials.
What are heat-sensitive glassware?The heat sensitive materials are not heated at the high temperatures, as they may shatter. This has been the result, as there has been an increase in the volume of the glassware with the explosion of heat.
The heat results in the glass molecules to raise the temperature. The excited electrons emit the radiations, that has been the representation of the material to be heat-sensitive.
The glassware in the video that are shown to be heat sensitive are volumetric and graduated cylinders.
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c2 is diamagnetic with a bond order of 2 but o2 is paramagnetic with a bond of 2. Using molecular orbital theory, account for these facts.
Answer:
Orbital mixing or sigma pi crossover
Explanation:
Looking at the diagram below, the ordering of molecular orbitals differ between carbon and oxygen molecules. The reason for this is that, with increase in atomic number, the energies of sigma 2p bonding molecular orbital and pi 2p molecular orbital come close together but are energetically far apart in the lighter elements. Following this ordering of orbitals, the two degenerate pi- 2p antibonding orbitals are singly filled accounting for the paramagnetism of oxygen while the two degenerate pi-2p bonding orbitals in carbon molecule is doubly occupied hence the molecule is diamagnetic.
O₂ is paramagnetic with a bond order of 2 due to two unpaired electrons in antibonding molecular orbitals. C₂ is diamagnetic and also has a bond order of 2, but with all electrons paired due to two additional electrons filling the antibonding orbitals when compared to B₂.
The student is asking about the paramagnetic nature of O₂ and the diamagnetic nature of C₂ despite both having a bond order of 2. According to molecular orbital theory, the electronic configuration of O₂ includes two unpaired electrons in antibonding
(*12py, *12pz) molecular orbitals. These unpaired electrons cause O₂ to be paramagnetic. Calculating the bond order, we find that there are eight electrons in bonding orbitals and four electrons in antibonding orbitals, resulting in a bond order of 2 (8 - 4 = 2), which confirms the double covalent bond in O₂. In contrast, the C₂ molecule, with two additional electrons compared to B₂, fills up its antibonding orbitals completely, leading to all electrons being paired and thus making C₂ diamagnetic.
The Michaelis-Menten equation is an expression of the relationship between the initial velocity,V0, of an enzymatic reaction and substrate concentration, [S]. There are three conditions that are useful for simplifying the Michaelis-Menten equation to an expression from which the effect of [S] on the rate can be more readily determined. Match the condition (e.g. [S] = Km) with the statement(s) that describe it:
1. Doubling [S] will almost double the rate.
2. Half of the active sites are occupied by substrate.
3. About 90% of the active sites are occupied by substrate.
4. Doubling [S] will have little effect on the rate.
5. Less than 10% of the active sites are occupied by substrate.
6. This condition will result in the highest rate.
There are three conditions for simplifying the Michaelis-Menten equation: when [S] is significantly less than Km, when [S] equals Km, and when [S] is much larger than Km. These are associated with less than 10% and half of the active sites being occupied by the substrate, and with doubling [S] either almost doubling the rate or having little effect.
Explanation:The Michaelis-Menten equation expresses the relationship between the initial enzymatic reaction velocity, V₀, and the substrate concentration, [S]. It allows for a simplification of the relationship between [S] and rate, given certain conditions:
[S]<[S] = Km: This relates to statement 2: half of the active sites are occupied by the substrate.[S]>>Km: This condition, where [S] greatly exceeds Km, matches with statement 3: approximately 90% of the active sites are occupied by the substrate. Doubling [S] will have little impact on the rate (statement 4), and the highest rate will be reached (statement 6).Understanding these conditions can significantly simplify the usage of the Michaelis-Menten equation in exploring enzyme kinetics.
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Water alone does not easily remove grease from dishes or hands because grease is nonpolar and water is polar. The addition of soap to water, however, allows the grease to dissolve. Study the structure of sodium stearate (a soap) and describe how its structure allows it to interact with both nonpolar grease and polar water Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Reset Help The soap molecule has a nonpolar and a polar cationic end when it is dissolved in water. When it dissolves in water, the sodium stearate congregates to form small spheres (called micelles) with the on the inside and the on the surface. The anionic end can attract and interacts with the polar water molecules, while the interact with the nonpolar grease. This allows the soapy water to remove the grease by trapping the grease inside the micelle.
Answer:
Micelle Formation
Explanation:
The soap molecule has a nonpolar and a polar cationic end when it is dissolved in water. When it dissolves in water, the sodium stearate congregates to form small spheres (called micelles) with the on the inside and the on the surface. The anionic end can attract and interacts with the polar water molecules, while the interact with the nonpolar grease. This allows the soapy water to remove the grease by trapping the grease inside the micelle.
Because greased is electrically neutral and water is polarity, water alone will not simply remove oil off dishes or hands. The grease, on the other hand, dissolves when soap is added to water.
This traps the greasy within the micelle, allowing the soapy liquid to remove it.
So,
In case 1: Hydro-carbon
In case 2: Anionic
In case 3: Hydro-carbon
In case 4: Anionic
In case 5: Anionic
In case 6: Hydro-carbon
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The equilibrium constant is given for one of the reactions below.
Determine the value of the missing equilibrium constant.
2 HD(g) ⇌ H2(g) + D2(g) Kc = 0.28
2 H2(g) + 2 D2(g) ⇌ 4 HD(g)
To find the missing equilibrium constant, we can use stoichiometry and the first given equilibrium constant. The missing equilibrium constant for the second reaction is the square of the given Kc value.
Explanation:To determine the missing equilibrium constant, you can use the concept of equilibrium constant expressions.
In the first reaction, 2 HD(g) ⇌ H2(g) + D2(g), the equilibrium constant is given as Kc = 0.28.
To find the value of the missing equilibrium constant for the second reaction, 2 H2(g) + 2 D2(g) ⇌ 4 HD(g), we can use the stoichiometry.
Since the equilibrium constant is a ratio of product concentrations to reactant concentrations, and the stoichiometric coefficients for the second reaction are all multiplied by 2 compared to the first reaction, the missing equilibrium constant would be the square of the given Kc value.
Therefore, the missing equilibrium constant is 0.28 squared, which is approximately 0.0784.
The equilibrium constant for the reaction 2 H₂(g) + 2 D₂(g) ⇌ 4 HD(g) is D) 13
To determine the equilibrium constant for the reaction 2 H₂(g) + 2 D₂(g) ⇌ 4 HD(g) given the equilibrium constant for the reverse reaction, 2 HD(g) ⇌ H₂(g) + D₂(g), which is Kc = 0.28, we can use the concept that the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.
The given reaction is:
2 HD(g) ⇌ H₂(g) + D₂(g), Kc = 0.28
The reverse reaction is:
H₂(g) + D₂(g) ⇌ 2 HD(g), K'c = 1 / 0.28
K'c = 3.57
The target reaction is twice the reverse reaction:
2 H₂(g) + 2 H₂(g) ⇌ 4 HD(g)
When the reaction coefficients are doubled, the equilibrium constant is squared:
Kc = K'c²
Kc = 3.57² = 12.75
Therefore, the equilibrium constant for the reaction 2 H₂(g) + 2 D₂(g) ⇌ 4 HD(g) is 12.798
Rounding of to two significant figures, the Equilibrium constant is D) 13
Complete question is - The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant.
2 HD (g) = H2 (g) + D2 (g) Kc = 0.28
2 H2 (g) + 2 D2 (g) = 4 HD (g) Kc = ?
A) 7.8 x 10⁻²
B) 3.6
C) 0.53
D) 13
E) 1.9
You are measuring the concentration of Fe 2 in a sample is determined by measuring the absorbance of its complex with ferroxine. The sample, measured in a 1.00 cm cuvette, has an absorbance of 0.333 . The reagent blank in the same cuvette has an absorbance of 0.014 . What would be the absorbance reading for each of these two solutions if measured in a 5.00 cm cuvette?
Answer:
Sample: 1.67
Blank: 0.070
Explanation:
The absorbance of a solution (A) is explained by the Beer-Lambert law.
A = ε . l . c
where,
ε is the absorptivity of the species
l is the optical path length
c is the molar concentration of the species
As we can see, the absorbance is directly proportional to the path length, that is, the length of the cuvette. If l is increased 5 times (1.00 cm to 5.00 cm), the absorbance will also be increased 5 times.
The absorbance of the sample will be 5 × 0.333 = 1.67
The absorbance of the blank will be 5 × 0.014 = 0.070
According to Beer's law, if the cuvette size increases fivefold from 1.00 cm to 5.00 cm, the absorbance will also increase fivefold. Therefore, the absorbance of the sample would become 1.665, and the reagent blank would become 0.070.
Explanation:The absorbance of a solution is related to the path length (in this case, cuvette size) according to Beer's law, which states that A = εlc, where A is absorbance, ε is molar absorptivity, l is path length, and c is concentration. Thus, absorbance is directly proportional to the path length.
If the path length increases from 1.00 cm to 5.00 cm, the absorbance also increases by the same factor. Therefore, the absorbance of the sample and the reagent blank in a 5.00 cm cuvette would be:
Sample: 0.333 * (5.00 cm / 1.00 cm) = 1.665Reagent blank: 0.014 * (5.00 cm / 1.00 cm) = 0.070Learn more about Absorbance measurement here:https://brainly.com/question/32564177
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The drug molecules bind the protein in a 1:1 ratio to form a drug-protein complex. The protein concentration in aqueous solution at 25 ∘C is 1.60×10−6 M . Drug A is introduced into the protein solution at an initial concentration of 2.00×10−6M. Drug B is introduced into a separate, identical protein solution at an initial concentration of 2.00×10−6M. At equilibrium, the drug A-protein solution has an A-protein complex concentration of 1.00×10−6M, and the drug B solution has a B-protein complex concentration of 1.40×10−6M.a. Calculate the Kc value for the A-protein binding reaction.b. Calculate the Kc value for the B-protein binding reaction.c. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?
Answer:
a. kc = 1,67x10⁶
b. kc = 1,17x10⁷
c. Drug B is the better choice.
Explanation:
The bind of drug-protein is described as:
Protein + Drug ⇄ Drug-protein
Where kc is:
kc = [Drug-protein] / [Protein] [Drug] (1)
a. For A, the equilibrium concentration of each specie is:
[Protein]: 1,60x10⁻⁶M - x
[Drug]: 2,00x10⁻⁶M - x
[Drug-protein]: x = 1,00x10⁻⁶M
-Where x is reaction coordinate-
Thus:
[Protein]: 1,60x10⁻⁶M - 1,00x10⁻⁶M = 0,60x10⁻⁶M
[Drug]: 2,00x10⁻⁶M - 1,00x10⁻⁶M = 1,00x10⁻⁶M
Replacing in (1):
kc = [1,00x10⁻⁶] / [1,00x10⁻⁶] [0,60x10⁻⁶]
kc = 1,67x10⁶
b. For B:
[Protein]: 1,60x10⁻⁶M - x
[Drug]: 2,00x10⁻⁶M - x
[Drug-protein]: x = 1,40x10⁻⁶M
-Where x is reaction coordinate-
Thus:
[Protein]: 1,60x10⁻⁶M - 1,40x10⁻⁶M = 0,20x10⁻⁶M
[Drug]: 2,00x10⁻⁶M - 1,40x10⁻⁶M = 0,60x10⁻⁶M
Replacing in (1):
kc = [1,40x10⁻⁶] / [0,20x10⁻⁶] [0,60x10⁻⁶]
kc = 1,17x10⁷
c. The drug with the bigger kc will be the more effective because will be the drug that binds more strongly with the protein. Thus, drug B is the better choice.
I hope it helps!
The Kc values for the A-protein and B-protein binding reactions are 3.125 and 4.375, respectively. Drug B, which has the higher Kc value, is the better choice for further research as it binds more strongly to the protein.
Explanation:a. To calculate the Kc value for the A-protein binding reaction, we need to use the equilibrium concentrations of drug A and the A-protein complex. The Kc value is given by [A-protein complex] / ([A] * [protein]). Plugging in the values, we get Kc = (1.00×10^6) / ((2.00×10^6) * (1.60×10^−6)) = 3.125.
b. Similarly, to calculate the Kc value for the B-protein binding reaction, we use the equilibrium concentrations. Kc = (1.40×10^6) / ((2.00×10^6) * (1.60×10^−6)) = 4.375.
c. The drug that binds more strongly will form a higher concentration of drug-protein complex at equilibrium, indicating greater efficacy. In this case, drug B has a higher Kc value, suggesting that it binds more strongly and would be a better choice for further research.
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Lithium aluminum hydride reduces carboxylic acids to primary alcohols via what intermediate?A) a ketone. B) a methyl ester. C) an aldehyde. D) a secondary alcohol. E) an acid chloride.
Final answer:
Lithium aluminum hydride reduces carboxylic acids to primary alcohols via an aldehyde intermediate, although the aldehyde is not isolated and is reduced further to a primary alcohol in one step. So the correct option is C.
Explanation:
Lithium aluminum hydride reduces carboxylic acids to primary alcohols by first forming an aldehyde as the intermediate. The reduction process does not stop at the aldehyde stage but proceeds to reduce the aldehyde further to a primary alcohol. When a carboxylic acid is reduced, the process typically bypasses the isolation of the intermediate aldehyde due to the strong reducing power of lithium aluminum hydride (LiAlH4), which is capable of fully reducing the carboxylic acid to the alcohol in one step.
Heat of fusion (?Hfus) is used for calculations involving a phase change between solid and liquid, with no temperature change. For H2O, ?Hfus=6.02 kJ/mol.Specific heat capacity (C) is used for calculations that involve a temperature change, but no phase change. For liquid water, C=4.184 J/(g??C).Heat of vaporization (?Hvap) is used for calculations involving a phase change between liquid and gas, with no temperature change. For H2O, ?Hvap=40.7 kJ/mol.Part AHow much heat is required to raise the temperature of 92.0g of water from its melting point to its boiling point?Express your answer numerically in kilojoules.
Answer:
q = 38,5 kJ
Explanation:
In its melting point, at 0°C, water is liquid. The boiling point of water is 100°C. It is possible to estimate the heat you required to raise the temperature of water from 0°C to 100°C using:
q = C×m×ΔT
Where C is specific heat of water (4,184J/g°C), m is mass of water (92,0g) and ΔT is change in temperature (100°C-0°C = 100°C)
Replacing:
q = 4,184J/g°C×92,0g×100°C
q = 38493 J, in kilojoules:
q = 38,5 kJ
I hope it helps!
The heat energy required to raise the temperature of 92.0g of water from its melting point to its boiling point is calculated using the formula q = mcΔT. Using given values and converting to kilojoules, the answer is 38.51kJ.
Explanation:To find the heat required to raise the temperature of water from its melting point to its boiling point, we can make use of the formula for calculating heat (q):
q = mcΔT
Where:
q is the heat energym is mass in gramsc is the specific heat capacityΔT is the change in temperature in CelsiusIn your case, m = 92g, c = 4.184 J/(g??C), and ΔT = 100°C (boiling point of water) - 0°C (melting point of water) = 100°C.
Substitute these values into the formula:
q = (92g)(4.184 J/g°C)(100°C)
Calculate to get q = 38506.88 J, in kilojoules convert by dividing by 1000 to get 38.51 kJ.
Hence, the heat required to raise the temperature of 92.0g of water from its melting point to its boiling point is 38.51 kJ.
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The Bohr model of the atom described the position of an electron as circling the nucleus in set orbits with a specific radius. Current atomic theory describes the location of the electron more accurately. What are the three main parts of an electron's "address" used in the current atomic theory that describe an electron's location?
Answer: principal quantum number (n), azimuthal quantum number (l) and magnetic quantum number (m)
Explanation:
According to quantum mechanics, there are three sets of quantum numbers which accurately describes the electrons position; principal quantum number (n), azimuthal quantum number (l) and magnetic quantum number (m). These three quantum numbers are obtained from Schrödinger wave equation.
Be sure to answer all parts.
Write equations for the initiation and propagation steps for the formation of dichloromethane by free-radical chlorination of chloromethane. Draw the structure of the products of each equation.
Be sure
to include lone pair(s) of electrons where necessary.
Please clearly show both propogation steps and the organic + inorganic radical.
Answer:
CH3Cl + Cl2 -> CH2ClCH2Cl + HCl
Explanation:
Chlorination is an addition reaction that involves addition of Chlorine to a chemical compound. Methane undergo addition reaction with chlorine to produce chloromethane, which undergo further further chlorination to produce dichloromethane.
For the chlorination of methane, one of the hydrogen atoms of methane is replaced by a chlorine atom of the chlorine gas (molecule)
CH4 + Cl2 -> CH3Cl + HCl
methane chlorine chloromethane Hydrochloric acid
Using their structure:
H H
I I
H - C - H + Cl - Cl -> H - C - Cl + HCl
I I
H H
methane chloromethane
In further chlorination of chloromethane, one of the hydrogen atom of chloromethane is replaced by the chlorine atom in the chlorine gas (molecule)
Chlorination of Chloromethane to dichloromethane is represented below:
CH3Cl + Cl2 -> CH2ClCH2Cl + HCl
Chloromethane Chlorine dichloromethane Hydrochloric acid
Using their structure:
Cl Cl
I I
H - C - H + Cl - Cl -> H - C - Cl + HCl
I I
H H
Chloromethane chlorine dichloromethane
Organic compound Inorganic compound
From the chemical equation above, an organic compound (Chloromethane) reacts with an inorganic compound (Chlorine) .
Watch the animation and select the interactions that can be explained by hydrogen bonding. Check all that apply.
a. CH4 molecules interact more closely in the liquid than in the gas phase.
b. HF is a weak acid neutralized by NaOH.
c. Ice, H2O, has a solid structure with alternating H−O interactions.
d. H2Te has a higher boiling point than H2S.
e. HF has a higher boiling point than HCl.
Answer:
c. Ice, H₂O, has a solid structure with alternating H−O interactions.
e. HF has a higher boiling point than HCl.
Explanation:
For molecules to interact through hydrogen bonding, it is required that there is an H atom bonded to a more electronegative atom, such as N, O or F.
Select the interactions that can be explained by hydrogen bonding. Check all that apply.
a. CH₄ molecules interact more closely in the liquid than in the gas phase. NO. The electronegativity of C is not high enough to form hydrogen bondings.
b. HF is a weak acid neutralized by NaOH. NO. This reaction occurs in water and it is better explained by ion-ion forces.
c. Ice, H₂O, has a solid structure with alternating H−O interactions. YES. This structure is a consequence of the hydrogen bonding.
d. H₂Te has a higher boiling point than H₂S. NO. The electronegativities of Te and S are not high enough to form hydrogen bondings.
e. HF has a higher boiling point than HCl. YES. The stronger hydrogen bonding interactions in HF explain the higher boiling point.
Final answer:
Hydrogen bonding explains interactions such as HF being a weak acid, the solid structure of ice (H2O), and the higher boiling point of HF compared to HCl.
Explanation:
In the context of the provided information, the interactions that can be explained by hydrogen bonding are:
b. HF is a weak acid neutralized by NaOH.c. Ice, H2O, has a solid structure with alternating H-O interactions.e. HF has a higher boiling point than HCl.Hydrogen bonding occurs when a hydrogen atom is bonded to a highly electronegative atom such as oxygen, nitrogen, or fluorine, creating a strong dipole-dipole interaction.
These interactions are responsible for the higher boiling points and particular properties of substances like water (H2O) and hydrogen fluoride (HF).
In chemistry, some reactions are linked to other reactions. For example, in the decomposition of potassium chlorate, oxygen and potassium chloride are produced. Then, the oxygen can react with magnesium and produce magnesium oxide. Using the reactions below, if you start with exactly 6 moles of potassium chlorate, how much magnesium oxide can you produce? 2 KClO3 → 3 O2 + 2 KCl 2 Mg + O2 → 2 MgO
A. 18 moles
B. 5 moles
C. 6 moles
D. 2 moles
Answer:
C
Explanation:
If 2 moles of potassium chlorate produced 2 moles of MgO, then 2*6/2 moles of MgO are produced when 6 moles of potassium perchlorate is used.
Liquid ammonia, , was used as a refrigerant fluid before the discovery of the chlorofluorocarbons and is still widely used today. Its normal boiling point is –33.4 °C, and its vaporization enthalpy is 23.5 kJ/mol. The gas and liquid have specific heat capacities of 2.2 and 4.7 , respectively. Calculate the heat energy transfer required to raise the temperature of 13.0 kg liquid ammonia from –50.0 °C to 0.0 °C. Heat energy = kJ
Answer:
The heat energy transfer required to raise the temperature of 13.0 kg liquid ammonia from -50.0 °C to 0.0 °C is 36,896.44 kJ.
Explanation:
The process involved in this problem are :
[tex](1):NH_3(l)(-50^oC)\rightarrow NH_3(l)(-33.4^oC)\\\\(2):NH_3(l)(-33.4^oC)\rightarrow NH_3(g)(-33.4^oC)[/tex]
[tex](3):NH_3(g)(-33.4^oC)\rightarrow NH_3(g)(-0.0^oC)[/tex]
Now we have to calculate the amount of heat released or absorbed in both processes.
For process 1 :
[tex]Q_1=m\times c_{1}\times (T_{final}-T_{initial})[/tex]
where,
[tex]Q_1[/tex] = amount of heat absorbed = ?
m = mass of ammonia = 13000 g
[tex]c_1[/tex] = specific heat of liquid ammonia = [tex]4.7J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]-50.0^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]-33.4^oC[/tex]
Now put all the given values in [tex]Q_3[/tex], we get:
[tex]Q_1=13000 g\times 4.7 J/g^oC\times ((-33.4)-(-50.0))^oC[/tex]
[tex]Q_1=1,014,260 J=1.1014.260 kJ[/tex]
For process 2 :
[tex]Q_2=n\times \Delta H_{fusion}[/tex]
where,
[tex]Q_2[/tex] = amount of heat absorbed = ?
m = mass of solid ammonia = 13.0 Kg = 13000 g
n = Moles of ammonia = [tex]\frac{13000 g}{17 g/mol}=764.71 mol[/tex]
[tex]\Delta H_{vap}[/tex] = enthalpy change for vaporization=23.5 kJ/mol
Now put all the given values in [tex]Q_1[/tex], we get:
[tex]Q_2=764.71 mol\times 23.5 kJ/mol=17,970.6 kJ[/tex]
For process 3 :
[tex]Q_3=m\times c_{p,l}\times (T_{final}-T_{initial})[/tex]
where,
[tex]Q_3[/tex] = amount of heat absorbed = ?
m = mass of ammonia = 13000 g
[tex]c_2[/tex] = specific heat of gaseous ammonia = [tex]2.2J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]-33.4^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]0.0^oC[/tex]
Now put all the given values in [tex]Q_3[/tex], we get:
[tex]Q_3=13000 g\times 2.2J/g^oC\times (0.0-(-33.4))^oC[/tex]
[tex]Q_3=955,240 J=955.240 kJ[/tex]
The heat energy transfer required to raise the temperature of 13.0 kg liquid ammonia from -50.0 °C to 0.0 °C = Q
[tex]Q=Q_1+Q_2+Q_3=17,970.6 kJ+17,970.6 kJ+955.240 kJ[/tex]
Q = 36,896.44 kJ
Azomethane decomposes into nitrogen and ethane at elevated temperature: H3C–N=N–CH3 → N2 + C2H6 A chemist studying this reaction at 300°C begins an experiment with an azomethane concentration of 4.52 mM and obtains the following data: Time(s) 100 150 200 250 300 [Azomethane] (mM) 3.50 3.08 2.71 2.39 2.10 Calculate the rate constant. Give an answer with concentration units of molarity and time units of seconds.
Answer:
k = 0,0026 s⁻¹
Explanation:
To calculate the rate constant it is necessary to find out the order of reaction. The R² nearest 1 will be the order of reaction.
For zeroth order the integrated rate law is:
[A] = [A]₀ -kt
The graph of [A] vs t gives a correlation coefficient R² of 0,9944.
The first order is:
ln [A] = ln [A]₀ -kt
The graph of ln [A] vs t gives a R² of 1
The second order is:
1/[A] = 1/[A]₀ -kt
The graph of 1/[A] vs t gives a R² of 0,9942
As R² = 1 for first order, the descomposition of azomethane follows this kinetics order. The lineal correlation is:
y = b - mx
y = 1,5077 - 0,0026x
ln [A] = ln [A]₀ -kt
That means:
-k = - 0,0026 s⁻¹
k = 0,0026 s⁻¹
I hope it helps!
A 0.588 g sample of a nonvolatile, yellow crystalline solid dissolves in 11.5 g of benzene, producing a solution that freezes at 5.02 oC. Find the molar mass of the yellow solid. The following may be useful: The freezing point of benzene is 5.51 oC and the freezing point depression constant, kf, is 4.90 oC/m.
Answer:
The molar mass of the solid is 511.3 g/mol
Explanation:
Step 1: Data given
Mass of the sample = 0.588 grams
Mass of benzene = 11.5 grams
The solution freezes at 5.02 °C
The freezing point of benzene is 5.51 °C
The freezing point depression constant, kf = 4.90 °C/m
Step 2: Determine the temperature change
Δt = 5.51 - 5.02 = 0.49°C
Step 3: Determine number of moles
Δt = i*Kf*m
⇒ with i = the number of dissolved particles the solute produces = 1
⇒ with Kf = the molal freezing point depression constant Kf = 4.90 °C*Kg/mol
⇒ with m = the molality of the solute
0.49 °C = (1) (4.90 °C kg/mol) (x / 0.01150 kg)
x = 0.00115 mol
Step 4: Calculate molar mass
0.588 grams / 0.00115 mol = 511.3 g/mol
The molar mass of the solid is 511.3 g/mol
A sample of methane at a pressure of 1.00 atm and a temperature of 93.1 K is heated at constant pressure to a temperature of 158 K. Which of the following are true? Choose all that apply
a. The liquid initially present will solidify.
b. The final state of the substance is a gas.
c. The sample is initially a solid.
d. The sample is initially a liquid. One or more phase changes will occur.
Answer:
b. The final state of the substance is a gas.
d. The sample is initially a liquid. One or more phase changes will occur.
Explanation:
Methane has the following properties:
Normal melting point: 90.7 KNormal boiling point: 111.65 K*"Normal" refers to normal pressure (1 atm).
According to this, we can affirm:
Below 90.7 K, methane is solid.Between 90.7 K and 111.65 K, methane is liquid.Above 111.65 K, methane is gas.A sample of methane at a pressure of 1.00 atm and a temperature of 93.1 K is heated at constant pressure to a temperature of 158 K. Which of the following are true? Choose all that apply.
a. The liquid initially present will solidify. FALSE. The liquid will vaporize.
b. The final state of the substance is a gas. TRUE.
c. The sample is initially a solid. FALSE. The sample is initially a liquid.
d. The sample is initially a liquid. One or more phase changes will occur. TRUE.
The substance would start as a gas, then transition to a solid without becoming a liquid due to temperature conditions.
The substance would begin as a gas and as the pressure increases, it would compress and eventually solidify without liquefying as the temperature is below the triple point temperature.
Describe the phase changes from -80°C to 500°C at 2 atm.
The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid.
he equilibrium constant is equal to 5.00 at 1300 K for the reaction:2 SO2(g) + O2(g) ↔ 2 SO3(g).If initial concentrations are [SO2] = 6.00 M, [O2] = 0.45 M, and [SO3] = 9.00 M, the system isnot at equilibrium and will shift to the right to achieve an equilibrium state.not at equilibrium and will shift to the left to achieve an equilibrium state.at equilibrium.not at equilibrium and will remain in an unequilibrated state.
Answer:
The system is at equilibrium
Explanation:
Let's consider the following reversible reaction.
2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)
To determine whether a reaction is at equilibrium or not, we have to calculate the reaction quotient (Q).
[tex]Q=\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}.[O_{2}]} =\frac{(9.00)^{2} }{(6.00)^{2}\times 0.45} =5.0[/tex]
Since Q = K (equilibrium constant), regardless of the significant figures, the system is at equilibrium.
The correct answer is at equilibrium.
The initial concentrations given correspond to the equilibrium concentrations for the reaction at 1300 K, and no shift in the reaction direction is necessary to achieve equilibrium because [tex]\( Q = K \).[/tex]
The equilibrium position of the reaction under the given conditions, we need to compare the initial concentrations with the equilibrium concentrations and the equilibrium constant [tex]\( K \).[/tex]
Given.
- Equilibrium constant K = 5.00
- Initial concentrations.
- [tex]\([SO_2]_{\text{initial}} = 6.00 \)[/tex] M.
- [tex]\([O_2]_{\text{initial}} = 0.45 \)[/tex] M.
- [tex]\([SO_3]_{\text{initial}} = 9.00 \)[/tex] M.
The reaction is. [tex]\( 2 SO_2(g) + O_2(g) \leftrightarrow 2 SO_3(g) \)[/tex]
1.Calculate Reaction Quotient Q
[tex]\[ Q = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} \][/tex]
Substitute the initial concentrations into the expression for [tex]\( Q \).[/tex]
[tex]\[ Q = \frac{(9.00)^2}{(6.00)^2 \times 0.45} \][/tex]
[tex]\[ Q = \frac{81.00}{16.20} \][/tex]
[tex]\[ Q \approx 5.00 \][/tex]
2. Compare Q with K.
- Q = 5.00
- K = 5.00
Since Q the reaction quotient is equal to K the equilibrium constant the system is already at equilibrium.