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.
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.
Which of the following properties is shared by gases and liquids?
They have a definite shape.
They consist of charged particles.
Their particles vibrate in fixed positions.
Their particles are arranged randomly.
Answer:
Their particles are arranged randomly
Explanation:
The random arrangement of molecules cause the translational motion of the liquid and Brownian motion of the gas
Why was the term inert gases dropped? Why was the term inert gases dropped? The term "inert" was dropped because hydrogen was addviced to be included to "inert" gases. The term "inert" was dropped because it no longer described all the group 8A elements. The term "inert" was dropped because some organic gases such as methane were named "inert", too, and confusion happened. The term "inert" was dropped because author's rights for the term ended and scientists obtained possibility to change it.
Answer:
B) The term "inert" was dropped because it no longer described all the group 8A elements.
Explanation:
Inert elements in chemistry simply refers to elements that are chemically inactive and are not expected to form any compounds. this is the general belief for the group 8 elements as they all have complete duplet/octet configurations (and ideally, they ought to be very stable with no tendency to form compounds by participating in the loss and gain of electrons). However the discovery of compounds like xenon tetrafluoride (XeF4) proved this to be wrong.
Again, the reason the term - inert gses was droppedis beacause this term is not strictly accurate because several of them do take part in chemical reactions.
After dropping the term - Inert gases, they are now referred to as noble gases.
The term 'inert gases' was dropped because it was found that these gases can participate in chemical reactions under certain circumstances. Consequently, they were not entirely 'inert'. Additionally, to avoid confusion with other gases referred to as 'inert' in organic chemistry, the term was replaced with 'noble gases'.
Explanation:The term 'inert gases' was primarily used to describe the group 8A elements in the periodic table, now commonly referred to as the noble gases. Initially, these gases were called inert due to their low reactivity. However, as science progressed, it was discovered that these gases could indeed take part in chemical reactions under certain conditions - hence, they weren't completely 'inert'. This led to the term being dropped in favor of 'noble gases'. Additionally, the term 'inert' began to cause confusion as it was also used in organic chemistry to refer to gases like nitrogen and methane that do not readily react.
Learn more about Inert Gases here:https://brainly.com/question/35378111
#SPJ12
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
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
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
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
#SPJ11
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
Learn more:
https://brainly.com/question/5214674?referrer=searchResults
What is the molecular geometry of the left carbon atom (circled) in acetic acid? The left carbon atom is attached to three hydrogen atoms and one carbon atom through single bonds. The right carbon atom is attached to a carbon atom and an oxygen atom through single bonds and a second oxygen atom through a double bond.
Answer: Tetrahedral geometry
Explanation: The structure of the acetic acid is shown in the image below.
The left carbon atom is the one which is attached to three hydrogen atoms through single bonds and to one carbon atom through single bond as well.
Thus the orbitals which are used in the process of the formation of the chemical bond between these 4 are sp3 orbitals. And these orbitals results in the formation of the tetrahedral geometry.
The right carbon atom that is attached to a carbon atom and an oxygen atom through single bonds and a second oxygen atom through a double bond has trigonal planar geometry which involves the sp2 orbitals for the formation of the bond.
Final answer:
The left carbon atom in acetic acid has a tetrahedral molecular geometry with sp3 hybridization.
Explanation:
The molecular geometry of the left carbon atom in acetic acid, which is attached to three hydrogen atoms and one carbon atom, is tetrahedral. This carbon atom has a sp3 hybridization, which means it forms four sp3 hybrid orbitals.
Each of these orbitals forms a single bond with either a hydrogen atom or the other carbon atom, resulting in a tetrahedral shape, with bond angles close to 109.5 degrees.
Acetic acid has two distinct carbon-oxygen bonds because the right carbon atom is part of a carboxylic acid group. In acetic acid, this carbon is double-bonded to one oxygen atom (forming a carbonyl group) and single-bonded to the oxygen atom in the hydroxyl group (OH). This leads to different bond lengths and strengths.
However, when acetic acid loses a hydrogen ion and becomes the acetate ion, the negative charge is delocalized over the two oxygen atoms, resulting in equivalent resonance structures where both carbon-oxygen bonds have partial double-bond character, making them appear as one type of bond.
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.
Learn more about Michaelis-Menten equation here:https://brainly.com/question/30404535
#SPJ3
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.
The specific heat capacity of liquid mercury is 0.14 J/gK.
How many joules of heat are needed to raise the temperature of 5.00 g of mercury from 15.0°C to 36.5°C?
Answer:
There will be 15.05 Joules of heat released.
Explanation:
Step 1: Data given
Mass of mercury = 5.00 grams
The specific heat of mercury = 0.14 J/g*K = 0.14 J/g°C
Raise of temperature from 15.0°C to 36.5 °C
Step 2: Calculate heat released
q = m*c*ΔT
⇒ with q = The heat released in Joules
⇒ with m = the mass of mercury = 5.00 grams
⇒ with c = The specific heat of mercury = 0.14 J/g°C
⇒ with ΔT = The change of temperature = T2 - T1 = 36.5°C - 15.0°C = 21.5 °C
q = 5.00g * 0.14 J/g°C * 21.5 °C = 15.05 J
There will be 15.05 Joules of heat released.
The quantity of heat needed to raise 5g of mercury from 15.0°C to 36.5°C is 15.05 Joules
Specific heat CapacityGiven Data
Specific heat capacity of liquid mercury = 0.14 J/gK.Mass = 5.00 gT1 = 15.0°CT2 = 36.5°CWe know that the expression for the amount of heat is given as
Q =. mcΔT
Substituting our given data into the expression we have
Q = 5*0.14*(36.5 - 15)
Q = 0.7*21.5
Q = 15.05 Joules
Learn more about specific heat capacity here:
https://brainly.com/question/21406849
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.
Learn more about heat-sensitive glassware, here:
https://brainly.com/question/283915
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.
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.
Learn more about Heat Calculation here:https://brainly.com/question/32376262
#SPJ6
Ag+(aq) + e- → Ag(s) E° = +0.800 V AgBr(s) + e- → Ag(s) + Br-(aq) E° = +0.071 V Br2(l) + 2 e- → 2 Br-(aq) E° = +1.066 V Use some of the data above to calculate Ksp at 25°C for AgBr. Enter your answer in exponential format (sample 1.23E-4) with two decimal places and no units.
Answer:
[tex]k_{sp}=4.7 \times 10^{-13}.[/tex]
Explanation:
Now equation of tqo halves are:
Oxidation : [tex]Ag(s)-->Ag^+(aq)+e^-[/tex]
Reduction : [tex]AgBr(s)+e^--->Ag(s)+Br^-(aq)[/tex]
We know,
[tex]E^o_{cell}=0.071-(0.8)=-0.729\ V.[/tex]
[tex]\Delta G^o=-n\times F \times E^o= -R\times T\times ln(k_s_p)\\-1\times 96485\times (-0.729)=-8.314\times 298\times ln(k_s_p)\\k_s_p=4.7\times 10^{-13}.[/tex]
[tex]k_{sp}=4.7 \times 10^{-13}.[/tex]
Hence, this is the required solution.
Final answer:
To calculate Ksp for AgBr, use the given standard reduction potentials and the Nernst equation. The equilibrium constant, K, for the half-reaction can be found using the Nernst equation. The equilibrium constant Ksp can then be calculated from K.
Explanation:
To calculate Ksp for AgBr, we can use the given standard reduction potentials and the Nernst equation. The balanced half-reaction for the reduction of AgBr(s) to Ag(s) is:
AgBr(s) + e- → Ag(s) + Br-(aq)
The reduction potential for this half-reaction is +0.071 V. Using the Nernst equation, we can find the equilibrium constant, K, for this reaction:
K = [Ag(s)][Br-(aq)] / [AgBr(s)] = 10^(nE°/0.0592) = 10^((1)(0.071)/0.0592) = 1.44
Since the stoichiometry of the reaction is 1:1, the equilibrium constant Ksp for AgBr can be directly calculated from K:
Ksp = [Ag+(aq)][Br-(aq)] = 1.441
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!
The second-order rate constant for the dimerization of a protein (P) P + P → P2 is 6.2 × 10−3/M · s at 25°C. Part 1 out of 2 If the concentration of the protein is 1.6 × 10−4 M, calculate the initial rate (M/s) of formation of P2. rate = × 10 M/s (Enter your answer in scientific notation.)
Final answer:
The initial rate of formation of P2 in the dimerization reaction of a protein with a concentration of 1.6 × 10^-4 M is approximately 1.59 × 10^-10 M/s.
Explanation:
To calculate the initial rate of formation of P2 in the dimerization reaction of a protein, we can use the rate equation:
rate = k[P]^2
where k is the rate constant and [P] is the concentration of the protein.
Plugging in the given values:
rate = (6.2 × 10^-3 M^-1s^-1)(1.6 × 10^-4 M)^2
Simplifying:
rate = 6.2 × 10^-3 M^-1s^-1 * (1.6 × 10^-4 M)^2
rate = 6.2 × 10^-3 M^-1s^-1 * 2.56 × 10^-8 M^2
rate ≈ 1.59 × 10^-10 M/s
Therefore, the initial rate of formation of P2 is approximately 1.59 × 10^-10 M/s.
Germanium is a group 4A semiconductor. The addition of a dopant atom (group 3A element) that has fewer valence electrons than the host atom results in a p-type semiconductor. The addition of a dopant atom (group 5A elements) that has more valence electrons than the host atom results in an n-type semiconductor. Which of the following elements when used for doping germanium will yield p-type semiconductors, and which elements when used for doping germanium will yield n-type semiconductors?
Answer:
Doping with galium or indium will yield a p-type semiconductor while doping with arsenic, antimony or phosphorus will yield an n-type semiconductor.
Explanation:
Doping refers to improving the conductivity of a semiconductor by addition of impurities. A trivalent impurity leads to p-type semiconductor while a pentavalent impurity leads to an n-type semiconductor.
Germanium, being a group 4A semiconductor, has four valence electrons. When a group 3A element (with fewer valence electrons) is used as a dopant in germanium, it creates "holes" in the crystal lattice, resulting in p-type semiconductors.
When a group 5A element (with more valence electrons) is used as a dopant in germanium, it introduces extra electrons into the crystal lattice, creating an excess of negative charge carriers and resulting in n-type semiconductors.
For p-type semiconductors (group 3A dopants): Boron (B), a group 3A element, is commonly used to germanium, resulting in p-type germanium.For n-type semiconductors (group 5A dopants): Phosphorus (P) and Antimony (Sb), both group 5A elements, are commonly used to germanium, resulting in n-type germanium.To know more about semiconductors:
https://brainly.com/question/33275778
#SPJ3
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.
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.
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.
Learn more about Mass Calculation here:https://brainly.com/question/15743584
#SPJ3
The average bond energy (enthalpy) for a C=C double bond is 614 kJ/mol and that of a C−C single bond is 348 kJ/mol. Estimate the energy needed to break only the π bond of the double bond of 2-butene. Express your answer numerically in joules per molecule.
The energy required to break the π bond in 2-butene is approximately 4.42 x 10^-19 joules per molecule.
Explanation:The energy required to break only the π bond of a C=C double bond can be calculated by subtracting the energy of a C-C single bond from the energy of a C=C double bond. So, we need to subtract the bond energy of a single bond (348 kJ/mol) from that of a double bond (614 kJ/mol). Therefore, the π bond energy is 614 kJ/mol - 348 kJ/mol = 266 kJ/mol. However, as the student asked for the energy required to break the π bond in 2-butene, we must convert this energy into energy per molecule by using Avogadro's number (6.022 x 10^23 molecules/mol). Hence, the energy required would be 266 kJ/mol x 10^3 J/kJ / (6.022 x 10^23 molecules/mol) = approximately 4.42 x 10^19 joules per molecule.
Learn more about Bond Energy Calculation here:https://brainly.com/question/13859360
#SPJ3
To estimate the energy required to break the π bond in 2-butene, subtract the energy of a single C-C bond from that of a C=C double bond and divide by Avogadro's number, yielding approximately 4.42 x 10^-19 J/molecule.
Explanation:The student asked how to estimate the energy needed to break only the π bond of the double bond in 2-butene, expressed in joules per molecule. Given that the energy for a C=C double bond is 614 kJ/mol and the energy for a C-C single bond is 348 kJ/mol, we can calculate the energy associated with the π bond. First, we find the difference between the double bond and the single bond energies: 614 kJ/mol - 348 kJ/mol = 266 kJ/mol. This difference represents the π bond energy per mole.
Now, convert this energy to joules (since 1 kJ = 1000 J), we have 266 kJ/mol x 1000 J/kJ = 266,000 J/mol. To find the energy per molecule, we divide by Avogadro's number (approximately 6.022 x 1023 mol-1):
266,000 J/mol ÷ 6.022 x 1023 mol-1 = 4.42 x 10-19 J/molecule.
Note that this is an estimate based on average bond energies and actual energies may vary based on the molecular environment.
Learn more about Bond Energy here:https://brainly.com/question/34669684
#SPJ3
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.
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.
Which of the following acids will have the strongest conjugate base?HIHCNHNO3HClO4HCl
Answer:
HCN
Explanation:
Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.
When an acid donates a proton, it changes into a base which is known as its conjugate base.
Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.
The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.
Also, the strongest acid leads to the weakest conjugate base and vice versa.
Thus, Out of HI, HCN, [tex]HNO_3[/tex], [tex]HClO_4[/tex] and HCl , the weakest acid is:- HCN
Thus, HCN corresponds to the strongest conjugate base.
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) .
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.
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_____.
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]