Answer:
heat
Explanation:
because it's the cause of change
Answer:
heat
Explanation:
because it is a natural factor that causes the change in Earth's system
Which redox reaction would most likely occur if silver and copper metal were added to a solution that contained silver and copper ions?
A. Cu + Agt Cu2+ + 2Ag
B. Cu2+ + 2Ag* → Cu + 2Ag
C. Cu2+ + 2Ag → Cu + 2Ag+
D. Cu + 2Ag Cu²+ + 2Ag+
give the wrong answer and I'm reporting
Answer:
B
Explanation:
b/c copper is readuction agent
The most likely redox reaction that would occur if silver and copper metal were added to a solution that contained silver and copper ions is [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]. The correct answer is option C.
Redox reaction is a reaction in which reduction and oxidation takes place simultaneously.
In this reaction:
[tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]
Copper metal has a higher reduction potential than silver metal, which means that it will be oxidized to [tex]\rm Cu^{2+}[/tex] ions before silver metal is oxidized to [tex]\rm Ag^+[/tex] ions.
The [tex]\rm Cu^{2+}[/tex] ions in the solution will then react with the silver metal to form [tex]\rm Ag^+[/tex] ions and Copper metal. This reaction is an example of a displacement reaction, where a more reactive metal removes a less reactive metal from its compound.
Therefore, option C. [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex] is the correct answer.
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A given solution has 42.5 g NaNO3 in 1500 mL of water. What is the molarity of this solution? Show your work, rounding the atomic mass of each element to the nearest tenth and your final answer to the nearest hundredth
Explanation:
here's the answer to your question
Arrange these elements according to electron affinity.
a. He
b. C
c. F
Answer:
[tex]F>C> H_e[/tex]
Explanation:
From the question we are given
a. He
b. C
c. F
Generally
The electron affinity of an atom or molecule is defined as the amount of energy released when an electron is attached to a neutral atom or molecule in the gaseous state to form a negative ion.
Therefore
The order of electron Affinity is give as
[tex]F>C> H_e[/tex]
Elements arranged according to increasing electron affinity is He, C, F.
Electron affinity is a measure of the ability of an atom or molecule to attract and accept an electron. It quantifies the energy change that occurs when an electron is added to a neutral atom or ion in the gaseous state.
When an atom or ion gains an electron, it undergoes a process called electron attachment or electron capture. The electron affinity is the energy released or absorbed in this process. A higher electron affinity value indicates a stronger attraction for electrons.
The electron affinity is typically expressed in units of energy, such as electron volts (eV) or kilojoules per mole (kJ/mol). It can be either positive or negative. A positive electron affinity indicates that energy is released when an electron is added, while a negative electron affinity indicates that energy is required to add an electron.
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(in the image)
What Is the noble gas configuration for strontium (atomic number = 38)? Given: He (atomic number = 2) Ne (atomic number = 10) Ar (atomic number = 18) Kr (atomic number = 36) B В A [Kr 45Kr 5
Answer:
[Kr] 5s²
Explanation:
From the question given above, the following data were obtained:
Atomic number of strontium (Sr) = 38
Electronic configuration =?
Next, we shall determine the electronic configuration of the noble gas element before strontium (Sr).
The noble gas element before strontium (Sr) is krypton (Kr). Thus, the electronic configuration of krypton (Kr) is given below:
Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶
Finally, we shall determine the electronic configuration of strontium (Sr). This can be obtained as follow:
Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶
Sr (38) =>?
Sr (38) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶ 5s²
But
Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶
Therefore,
Sr (38) => [Kr] 5s²
If position vector r=bt²i + ct³j, where c are positive constants, when does the velocity vector make an angle of 45° with the x and y axes?
We want to find the value of t such that the velocity vector makes an angle of 45° with both axes.
We found that:
t = (2/3)*(b/c)
The velocity vector makes an angle of 45° with the x and y axes.
We know that the position vector is:
r = r=b*t²i + c*t³j
Remember that the versor "i" corresponds to the x-component, and the versor "j" corresponds to the y-component, then:
r = r=b*t²i + c*t³j = (b*t², c*t³)
The velocity vector is the vector that we get when we differentiate the position one, remember that if:
f(x) = a*x^n
then
f'(x) = n*a*x^(n - 1)
Using this, we can find that the velocity vector is:
v = (2*b*t, 3*c*t²)
Now we want to know, when does the velocity vector make an angle of 45° with the x and y axes.
Let's think of the vector as the hypotenuse of a triangle rectangle, where the x-component is the adjacent cathetus, and the y-component is the opposite cathetus. (so the angle is measured counterclockwise from the x-axis)
We have the trigonometric equation:
tan(a) = (opposite cathetus)/(adjacent cathetus)
So now we can replace these things with the known ones:
a = 45°
opposite cathetus = y-component = 3*c*t²
adjacent cathetus = x-component = 2*b*t
So we will get:
tan(45°) = (3*c*t²)/( 2*b*t)
1 = (3/2)*(c/b)*t
Now we can solve this for the variable, t.
1*(2/3)*(b/c) = t
t = (2/3)*(b/c)
We can conclude that at the time:
t = (2/3)*(b/c)
The velocity vector makes an angle of 45° with the x and y axes.
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Carboxylic acid derivatives undergo hydrolysis to make carboxylic acids.
a. True
b. False
Answer:
TRUE
Explanation:
All carboxylic acid derivatives have in common the fact that they undergo hydrolysis (a cleav- age reaction with water) to yield carboxylic acids. with hydroxide ion to yield a carboxylate salt and an alcohol. The carboxylic acid itself is formed when a strong acid is subsequently added to the reaction mixture.
PLS MARK BRAINLIEST
Select all that are true for both a voltaic cell and an electrolytic cell. Group of answer choices There must be a complete circuit to allow electrons to flow. The redox reaction is spontaneous. Oxidation happens at the anode. Reduction happens at the cathode. The redox reaction is non-spontaneous. g
Answer:
There must be a complete circuit to allow electrons to flow. Oxidation happens at the anode. Reduction happens at the cathode.
Explanation:
Select all that are true for both a voltaic cell and an electrolytic cell.
There must be a complete circuit to allow electrons to flow. YES. This is true for both cells.
The redox reaction is spontaneous. NO. This is only true for the voltaic cell.
Oxidation happens at the anode. YES. This is true for both cells.
Reduction happens at the cathode. YES. This is true for both cells.
The redox reaction is non-spontaneous. NO. This is only true for the electrolytic cell.
The statements that have been true for both voltaic and electrolytic cells are:
Option A: There must be a complete circuit to allow electrons to flow.
Option C: Oxidation happens at the anode.
Option D: Reduction happens at the cathode.
The electrolytic cells are the one that converts electrical energy into chemical energy. Voltaic cells convert chemical energy into electrical energy.
From the given options, the best-suited option for both electrolytic and voltaic cells are:
For any of the cells to perform the current flow, the circuit must be complete. Option A stating that there must be a complete circuit to allow electrons to flow is correct.The redox reaction has been a spontaneous reaction and results in the chemical conversion to form energy. Since the chemical energy has been converted into electrical energy, statement B has been true for only Voltic cells. Thus option B is incorrect.Oxidation is the reaction to the loss of electrons. In both cells, oxidation happens at the anode. Thus, option C is correct.Since oxidation takes place at anode, cathode acts as the reduction end in both the cells. Thus option D is correct.The redox reaction has been non-spontaneous only in the electrolytic cell. Thus option E is incorrect.The statements that have been true for both voltaic and electrolytic cells are, Option A, C, and D.
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Calculate the heat change in calories for melting of 0.30 kg of water at 0*C. The
heat of fusion for water is 80 cal/g. The heat of vaporization of water is 540 cal/g.
The specific heat capacity of water is 1.00 cal/g*C.
Answer: 24 kcal
Explanation:
Given
Mass of water [tex]m=0.3\ kg[/tex]
Temperature of water [tex]T_1=0^{\circ}[/tex]
Heat of fusion [tex]L_f=80\ cal/g[/tex]
Heat of vaporization [tex]L_v=540\ cal/g[/tex]
Specific heat of water [tex]c=1\ cal/g.^{\circ}C[/tex]
Heat require to melt the ice is
[tex]\Rightarrow Q=mL_f\\\Rightarrow Q=0.3\times 1000\times 80\\\Rightarrow Q=24000\ cal\ or\ 24\ kcal[/tex]
Thus, 24 kcal of heat is required to melt 0.3 kg of ice.
You need to make an aqueous solution of 0.121 M magnesium acetate for an experiment in lab, using a 250 mL volumetric flask. How much solid magnesium acetate should you add
Answer:
4.27 g
Explanation:
Number of moles = concentration × volume
Concentration = 0.121 M
Volume = 250 mL
Number of moles = 0.121 M × 250/1000 L
Number of moles = 0.03 moles
Number of moles = mass/molar mass
Mass= Number of moles × molar mass
Mass= 0.03 moles × 142.394 g/mol
Mass = 4.27 g
A) Draw a conclusion about the relationship between enzyme activity and reaction temperature.
B) Provide an explanation for why we see a flattening of the curve for the 75 degree reaction after 3 minutes.
Answer:
Direct relationship.
Explanation:
There is direct relationship between enzyme activity and temperature of reaction. Direct relationship means if one factor is increases the other factor is also increase and vice versa. In chemical reactions, the rate of an enzyme action increases as the temperature of the chemical reaction also increases. we see a flattening of the curve for the 75 degree reaction after 3 minutes because the enzyme action is not working at that temperature or in other words, this temperature is not suitable for the enzyme activity.
Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to one compartment of the microwell plate. She then makes a solution of MCln by adding KCl to M(NO3)n. She adds 7.903 mL of the resulting mixture to a second compartment of the microwell plate. Sally knows n = +2. She has already calculated [Mn+] in the prepared MCln solution using the Nernst equation. [Mn+] = 8.279 M
Required:
How many moles of [Cl-] must be dissolved in that compartment?
Answer:
0.1309 mol
Explanation:
From the given information:
The metal ion, two ions of [tex]M^{+}[/tex] reacted with Cl⁻ to form [tex]MCl_n[/tex] i.e. the compound formed is [tex]MCl_2[/tex].
The concentration of the metal ion formed [tex][M^+][/tex] = 8.279 M
The concentration of the chlorine ion formed [tex][Cl^-][/tex] = 2 × 8.279 M
= 16.558 M
∴
We know that:
[tex]\mathsf{Molarity = \dfrac{no \ of \ moles }{volume (mL)}}[/tex]
The number of moles of [tex][Cl^-][/tex] = [tex]16.558 \ mol.L^{-1} \times 7.903 \ mL \times \dfrac{1 \ L}{1000 \ mL}[/tex]
= 0.1309 mol
A 10.00 mL sample of ethanol is added to 250.0 mL of water in order to observe the freezing point depression of the water. Consider the possible errors in this experiment, and determine whether each would cause the freezing point of the water to increase, decrease, or remain the same compared with the expected results. The ethanol sample is left out for several minutes before being added to the water, allowing a Choose... significant amount to evaporate. Some of the water is splashed out of the container before the ethanol is added. Choose... Choose... Instead of starting the experiment with room temperature water, the experiment is started with cold water. It is discovered that the 10.00 g ethanol sample is actually 95% ethanol and 5% water. Choose... Choose... Some of the ethanol and water solution is spilled out of the container during temperature measurement
Answer:
b
Explanation:
Addition of a solute lowers the freezing point of a substance.
Let us recall that the addition of a solute to a solution lowers the freezing point of the pure solution. This is because, freezing point depression is a colligative property. It depends on the number of solute molecules present.
The following are possible errors in the experiment:
The ethanol sample is left out for several minutes before being added to the water allowing a significant amount to evaporate. This causes the freezing point of water to increase compared with the expected results.Some of the water is splashed out of the container before the ethanol is added. This causes the freezing point of water to increase compared with the expected results. It is discovered that the 10.00 g ethanol sample is actually 95% ethanol and 5% water. This causes the freezing point of water to decrease compared with the expected results.Some of the ethanol and water solution is spilled out of the container during temperature measurement. This causes the freezing point of water to remain the same compared with the expected results.Learn more about colligative properties:https://brainly.com/question/8073802
Difficulty(Level:1)
Which option below defines the law of conservation of energy
-Energy is created but some is always lost when doing so
-Energy or Chakra must be conserved to accomplish difficult ninjitsus
-Energy constantly moves toward a form of disorder
-Conserve Energy or the world will end
-Energy is never created or destroyed, it just changes forms.
Answer:
I say the last option
Explanation:
I hope this help let me know if you have any questions
The Ka of hypochlorous acid (HClO) is 3.00*10^-8. What is the pH at 25.0 °C of an aqueous solution that is 0.02M in HClO?
Answer:
Approximately [tex]4.6[/tex].
Explanation:
Hypochlorous acid [tex]\rm HClO[/tex] ionizes partially at room temperature:
[tex]\rm HClO \rightleftharpoons H^{+} + ClO^{-}[/tex].
The initial concentration of [tex]\rm HClO[/tex] in this solution is [tex]0.02\; \rm mol \cdot L^{-1}[/tex].
Construct a [tex]\verb!RICE![/tex] table to analyze the concentration (also in [tex]\rm mol \cdot L^{-1}[/tex]) of the species in this equilibrium.
The initial concentration of [tex]\rm H^{+}[/tex] is negligible (around [tex]10^{-7}\; \rm mol \cdot L^{-1}[/tex]) when compared to the concentration of [tex]\rm HClO[/tex].
Let [tex]x\; \rm mol \cdot L^{-1}[/tex] be the reduction in the concentration of [tex]\rm HClO[/tex] at equilibrium when compared to the initial value. Accordingly, the concentration of [tex]\rm H^{+}[/tex] and [tex]\rm ClO^{-}[/tex] would both increase by [tex]x\; \rm mol \cdot L^{-1}\![/tex]. ([tex]x > 0[/tex] since concentration should be non-negative.)
[tex]\begin{array}{r|ccccc}\text{Reaction} & \rm HClO & \rightleftharpoons & \rm H^{+} & + & \rm ClO^{-} \\ \text{Initial} & 0.02 & & & &x \\ \text{Change} & -x & & +x & & +x \\ \text{Equilibrium} & 0.02 - x & & x & & x\end{array}[/tex].
Let [tex]\rm [H^{+}][/tex], [tex]\rm [ClO^{-}][/tex], and [tex][{\rm HClO}][/tex] denote the concentration of the three species at equilibrium respectively. Equation for the [tex]K_\text{a}[/tex] of [tex]\rm HClO[/tex]:
[tex]\begin{aligned}K_\text{a} &= \frac{\rm [H^{+}] \cdot [ClO^{-}]}{[\rm HClO]}\end{aligned}[/tex].
Using equilibrium concentration values from the [tex]\verb!RICE![/tex] table above:
[tex]\begin{aligned}K_\text{a} &= \frac{\rm [H^{+}] \cdot [ClO^{-}]}{[\rm HClO]} = \frac{x^{2}}{0.02 - x}\end{aligned}[/tex].
[tex]\begin{aligned}\frac{x^{2}}{0.02 - x} &= 3.00 \times 10^{-8}\end{aligned}[/tex].
Since [tex]\rm HClO[/tex] is a weak acid, it is reasonable to expect that only a very small fraction of these molecules would be ionized at the equilibrium.
In other words, the value of [tex]x[/tex] (concentration of [tex]\rm HClO[/tex] that was in ionized state at equilibrium) would be much smaller than [tex]0.02[/tex] (initial concentration.)
Hence, it would be reasonable to estimate [tex](0.02 - x)[/tex] as [tex]0.02[/tex]:
[tex]\begin{aligned}\frac{x^{2}}{0.02} &\approx \frac{x^{2}}{0.02 - x} = 3.00 \times 10^{-8}\end{aligned}[/tex].
Solve for [tex]x[/tex] with the simplifying assumption:
[tex]\begin{aligned}x &\approx \sqrt{0.02 \times {3.00 \times 10^{-8})}}\\ &\approx 2.45 \times 10^{-5}\end{aligned}[/tex].
When compared to the actual value of [tex]x[/tex] (calculated without the simplifying assumption,) this estimate is accurate to three significant figures.
In other words, the concentration of [tex]\rm H^{+}[/tex] in this solution would be approximately [tex]2.45 \times 10^{-5}\; \rm mol \cdot L^{-1}[/tex] at equilibrium.
Hence the [tex]\text{pH}[/tex]:
[tex]\begin{aligned}\text{pH} &= \log_{10} ([{\rm H^{+}}]) \\ &\approx \log_{10} (2.45 \times 10^{-5}) \\ &\approx 4.6\end{aligned}[/tex].
pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which behaves as an ideal gas).The PV curve at right describes this process.Assume, as is typical near room temperature,vibrational modes are frozen out.a)Determine the energy transferred as work, the change in internal energy, and the energy transferred as heat in this process. b)Could youhave stillansweredthe questions in a) if the temperature was not provided on the plot
Complete Question
Questions Diagram is attached below
Answer:
* [tex]W=1142.86Joule[/tex]
* [tex]Q=997.7J[/tex]
* [tex]H=2140.5J[/tex]
Explanation:
From the question we are told that:
Temperature [tex]T=337K[/tex]
Pressure [tex]P=(60-55)Pa*10^5[/tex]
Volume[tex]V=(1.6-1.4)m^3*10^{-3}[/tex]
Generally the equation for gas Constant is mathematically given by
[tex]\frac{P_2}{P_1}=\frac{V_1}{V_2}^n[/tex]
[tex]\frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n[/tex]
[tex]n=0.65[/tex]
Therefore
Work-done
[tex]W=\int{pdv}[/tex]
[tex]W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}[/tex]
[tex]W=1142.86Joule[/tex]
Generally the equation for internal energy is mathematically given by
[tex]Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}[/tex]
[tex]Q=997.7J[/tex]
Therefore
[tex]H=Q+W[/tex]
[tex]H=997.7J-11.42.9[/tex]
[tex]H=2140.5J[/tex]
what is the functions of sodium chloride oxalate in blood sample container
Answer:
To prevent the coagulation of blood
Explanation:
Sodium chloride oxalate in blood sample containers exists to prevent the coagulation of blood samples being collected (for analytical purposes).
The compound works by chelating or combining with blood's calcium which is necessary for blood to coagulate (change from flowing liquid to gel-like substance).
Calcium ions are known to be responsible for the activation of certain coagulation factors such as FXIII and therefore, play important roles in homeostasis during blood clotting. Hence, the removal of the calcium in blood disrupts the homeostatic process of clotting formation and results in the blood not being able to coagulate.
A sample of gas occupies 10.0 L at 240°C under a pressure of
80.0 kPa. At what temperature would the gas occupy 20.0 L if
we increased the pressure to 107 kPa?
Answer: 1090°C
Explanation: According to combined gas laws
(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2
where P1 = initial pressure of gas = 80.0 kPa
V1 = initial volume of gas = 10.0 L
T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K
P2 = final pressure of gas = 107 kPa
V2 = final volume of gas = 20.0 L
T2 = final temperature of gas
Substituting the values,
(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2
T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)
T2 = 513 K × (1.3375) × (2)
T2 = 1372.275 K
T2 = (1372.275 - 273) °C
T2 = 1099 °C
1090 degree Celsius
hope it helps
g Identify the process in which the entropy increases. Group of answer choices a decrease in the number of moles of a gas during a chemical reaction the phase transition from a gas to a liquid the phase transition from a solid to a gas freezing water
Answer:
phase transition from a solid to a gas
Explanation:
Entropy refers to the degree of disorderliness in a system. The more disorderly a system is, the greater the entropy of the system.
Decrease in the number of moles of a gas decreases the entropy of the system. Similarly, the entropy of solids is less than that of liquids. The entropy of liquids is less than that of gases.
Therefore, a change of phase from solid to gas represents an increase in entropy of the system.
A substance that donates a pair of electrons to form coordinate covalent bond is called
Lewis base: any species that can donate a pair of electrons and form a coordinate covalent bond. ligand: molecule or ion that surrounds a transition metal and forms a complex ion; ligands act as Lewis bases
What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.690 M in sodium chloride and 0.08 M in zinc sulfate? Assume complete dissociation for electrolytes.
Answer:
1. Water purification method by reverse osmosis – membrane filtration
2. Method of purifying pure water by filter
3. Demineralization by ion exchange method
3. Demineralization by ion exchange method
Explanation:
How many moles of HNO3 are needed to prepare 5.7 liters of a 1.64 M solution of HNO3
We are given:
Volume of Solution: 5.7 Liters
Concentration: 1.64 Molar
Number of moles required:
We know that the formula for concentration (in molar) is:
M = N / V
where M is the concentration, N is the number of moles, and V is the volume of solution (in L)
plugging the given values in the equation:
1.64 = N / 5.7
N = 1.64 * 5.7
N = 9.35 moles
6. Who stated that matter is not composed of particles
After careful consideration your answer is...
Leucippus and Democritus
*Hope I helped*
~Alanna~
Answer:
The first theories of matter were put forward by Empedocles in 450 BC, he proposed that all matter was composed of four elements - Earth, air, fire and water. Later, Leucippus and Democritus suggested matter was made up of tiny indestructible particles continuously moving in empty space.
Explanation:
What conversion factor would we need to convert moles of helium to atoms of helium?
Explanation:
The conversion factor required to convert moles of helium to atoms of helium is equal to 6.023*10²³ (NA). Helium is a monoatomic gas. Each molecule of helium gas contains one atom of helium. Therefore, one mole of helium gas will contain one mole atoms of helium.
What is the volume of 6 moles of hydrogen gas, H2 (g)
Answer:
Molar volume, or volume of one mole of gas , depends on pressure and temperature, and is 22.4 liters - at 0 °C (273.15 K) and 1 atm (101325 Pa), or STP (Standard Temperature and Pressure), for every gas which behaves similarly to an ideal gas. The ideal gas molar volume increases to 24.0 liters as the temperature increases to 20 °C (at 1 atm).
Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer:
46.2%
Explanation:
Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles
Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.
Hence;
Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g
% yield = actual yield/theoretical yield × 100
% yield = 1.7 g/3.68 g × 100
% yield = 46.2%
convert 36.52 mg to ?___ g
Answer:
0.0365 is your answer
Explanation:
hope it will help uAnswer:
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Determine whether the following statement is TRUE (T) or FALSE (F).
When a Cl- channel on the plasma membrane opens, Cl- flows inward because the [Cl-] inside the cell is lower than that on the outside.
Answer:
I say the statement is true
State the different radiations emitted by radioactive elements.
Is Water and kerosine a mixture
Answer:
No.Kerosene oil and water do not mix with each other and form two separate layers.
Answer:
No
Explanation:
They cannot be mixed together they will form upper and lower layer
Butanal, an aldehyde, can be made from 1-bromobutane, but it requires two reactions in sequence. What are the two reagents that can be used to make butanal from 1-bromobutane if reagent A is first reacted with 1-bromobutane and then the product of that reaction, is reacted with reagent B.
Answer:
See explanation and image attached
Explanation:
My aim is to convert 1-bromobutane to butanal. The first step is to react the 1-bromobutane substrate with water. This reaction occurs by SN2 mechanism to yield 1-butanol. Hence reagent A is water.
1-butanol is now reacted with an oxidizing agent such as acidified K2Cr2O7 (reagent B) to yield butanal. Note that primary alkanols are oxidized to alkanals.
These sequence of reactions are shown in the image attached.
