Answer:
Al^3+
Explanation:
Solubility rules tell us what substances are soluble in water. Since NaCl was added and no precipitate was observed, the mercury II ion is absent.
Addition of Na2SO4 does not form a precipitate meaning that Ba^2+ is absent.
If a precipitate is formed when NaOH is added, the the ionic reaction is as follows;
Al^3+(aq) + 3 OH^-(aq) ------> Al(OH)3(s)
The reversible reaction: 2SO2(g) O2(g) darrow-tn.gif 2SO3(g) has come to equilibrium in a vessel of specific volume at a given temperature. Before the reaction began, the concentrations of the reactants were 0.060 mol/L of SO2 and 0.050 mol/L of O2. After equilibrium is reached, the concentration of SO3 is 0.040 mol/L. What is the equilibrium concentration of O2
Answer:
[tex][O_2]_{eq}=0.030M[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the mathematical expression for the concentration of oxygen at equilibrium, given the initial one and the change due to the reaction extent:
[tex][O_2]_{eq}=0.050M-x[/tex]
Whereas [tex]x[/tex] can be found considering the equilibrium of SO3:
[tex][SO_3]_{eq}=2x=0.040M[/tex]
Which means:
[tex]x=\frac{0.040M}{2} =0.020M[/tex]
Thus, the equilibrium concentration of oxygen gas turns out:
[tex][O_2]_{eq}=0.050M-0.020M=0.030M[/tex]
Regards!
How many atom in protons
Answer:
Its atomic number is 14 and its atomic mass is 28. The most common isotope of uranium has 92 protons and 146 neutrons. Its atomic number is 92 and its atomic mass is 238 (92 + 146).
In the given range,at what temperature does oxy gen have the highest solubility?
Si enfriamos mercurio de 100C. Calcular la cantidad de calor que se debe restar sabiendo que la masa de mercurio es de 1800gr
Answer:
I do not speak Spanish.
Explanation:
why beta carbon hydrogen is easily replaceable but not alpha carbon hydrogen
Answer:
Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.
Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.
The nucleophile in these reactions are new and called enols and enolates.
Explanation:
The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.
Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.
Aldehyde hydrogens not given Greek leters.
α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.
Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.
The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.
Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.
The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.
The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.
The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.
Aromaticity can also stabilize the enol tautomer over the keto tautomer.
Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.
Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.
How can a Bose-Einstein condensate be formed? A. B super-heating a gas. B. By super-cooling certain types of solid. C. By super-cooling certain types of plasma. D. By super-heating a plasma
Answer:
C. By super-cooling certain types of plasma.
Explanation:
Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.
Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.
When certain types of plasma are super cooled, Bose-Einstein condensate are formed.
Calculate the mass of sodium phosphate in aqueous solution to fully react with 37 g of chromium nitrate(III) an aqueous solution?(report answer in grams and only three Sigg figs do not put the unit)
Answer:
41 g
Explanation:
The equation of the reaction is;
Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)
Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles
1 mole of sodium phosphate reacts with 1 mole of chromium nitrate
x moles of sodium phosphate react as with 0.25 moles of chromium nitrate
x= 1 × 0.25/1
x= 0.25 moles
Mass of sodium phosphate = 0.25 moles × 163.94 g/mol
Mass of sodium phosphate = 41 g
g A high altitude balloon is filled with 1.41 x 104 L of hydrogen gas (H2) at a temperature of 21oC and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is -48oC and the pressure is 63.1 torr
Answer:
1.27 × 10⁵ L
Explanation:
Step 1: Given data
Initial pressure (P₁): 745 TorrInitial volume (V₁): 1.41 × 10⁴ LInital temperature (T₁): 21 °CFinal pressure (P₂): 63.1 TorrFinal volume (V₂): ?Final temperature (T₂): -48 °CStep 2: Convert the temperatures to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 21 °C + 273.15 = 294 K
K = -48 °C + 273.15 = 225 K
Step 3: Calculate the final volume of the balloon
We will use the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂/ T₁ × P₂
V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr
V₂ = 1.27 × 10⁵ L
Write a balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution. Be sure to add physical state symbols where appropriate.
Answer:
O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e
Explanation:
An oxidation reaction reaction refers to a reaction in which electrons are lost. In this case, we are about to see the full balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution.
The full equation is;
O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e
So, two electrons were lost in the process.
When solid Ni metal is put into an aqueous solution of Pb(NO3)2, solid Pb metal and a solution of Ni(NO3)2 result. Write the net ionic equation for the reaction.
Answer:
[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to write the complete molecular equation as shown below:
[tex]Pb(NO_3)_2(aq)+Ni(s)\rightarrow Ni(NO_3)_2(aq)+Pb(s)[/tex]
Now, we can separate the nitrates in ions as they are aqueous to obtain:
[tex]Pb^{2+}(aq)+2(NO_3)^-(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+2(NO_3)^-(aq)+Pb(s)[/tex]
And then, we cancel out the nitrate ions as the spectator ones, for us to obtain the net ionic equation:
[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]
Best regards!
A molecule of acetone and a molecule of propyl aldehyde are both made from 3 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom. The molecules differ in their arrangement of atoms. How do formulas for the two compounds compare? Both compounds have the same molecular formula, but have unique structural formulas. Both compounds have unique molecular formulas and structural formulas. Both compounds have the same structural formula, but have unique molecular formulas.
Explanation:
The structures of both acetone and propanal are shown below:
In the formula of propanal there is -CHO functional group at the end.
In acetone -CO- group is present in the middle that is on the second carbon.
The molecular formula is C3H6O.
Both have same molecular formula but different structural formulas.
A student prepares a aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource.
Answer:
10.71%
Explanation:
The dissociation of acetic acid can be well expressed as follow:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:
Then:
The I.C.E Table is expressed as follows:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Initial 0.0014 0 0
Change - x +x +x
Equilibrium (0.0014 - x) x x
Recall that:
Ka for acetic acid CH₃COOH = 1.8×10⁻⁵
∴
[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]
By rearrangement:
[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]
Multiplying through by (-) and solving the quadratic equation:
[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]
[tex](-0.00015 + x) (0.000168 + x) =0[/tex]
x = 0.00015 or x = -0.000168
We will only consider the positive value;
so x=[CH₃COO⁻] = [H⁺] = 0.00015
CH₃COOH = (0.0014 - 0.00015) = 0.00125
However, the percentage fraction of the dissociated acetic acid is:
[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]
= 10.71%
what is the hybridisation of the central carbon in CH3C triple bonded to N
Explanation:
the carbon would be sp3 hybridized, and it doesn't matter which carbon, since either of them have a full octet
For each of the following compounds, indicate the pH at which 50% of the compound will be in a form that possesses a charge and at which pH more than 99% of the compound will be in a form that possesses a charge.
ClCH2COOH (pKa = 2.86)
CH3CH2NH+3 (pKa = 10.7)
Express your answer using two decimal places
a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.
Answer:
a. 2..86 b. 4.86 c. 10.7 d. 8.7
Explanation:
a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA]
where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.
At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1
So, pH = pKa + log[A⁻]/[HA]
pH = pKa + log1
pH = pKa = 2.86
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.
pH = pKa + log0.99x/0.01x
pH = pKa + log0.99/0.01
pH = 2.86 + log99
pH = 2.86 + 1.996
pH = 4.856
pH ≅ 4.86
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA]
where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.
At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1
So, pH = pKa + log[A⁻]/[HA]
pH = pKa + log1
pH = pKa = 10.7
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.
Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.
pH = pKa + log0.01x/0.99x
pH = pKa + log1/99
pH = 10.7 - log99
pH = 10.7 - 1.996
pH = 8.704
pH ≅ 8.7
You will observe a weak acid-strong base titration in this experiment. Select all statements that are true about weak acid-strong base titrations.
A. Weak acid-strong base titrations always start at a higher pH than strong acid-strong base titrations, no matter the initial concentration.
B. The pH is less than 7 at the equivalence point.
C. The pH is greater than 7 at the equivalence point.
D. Half way to the equivalence point, a buffer region is observed.
Answer:
The pH is greater than 7 at the equivalence point.
Explanation:
Equivalence point is the point where the acid reacts with the base as stipulated in the equation of the reaction.
When a weak acid and a strong base are titrated, the pH of the solution at equivalence point is actually found to be around about pH ~ 9.
Hence, for a weak acid and strong base titration, The pH is greater than 7 at the equivalence point.
A titration between a weak acid and a strong base yields a solution whose pH is greater than 7 at the equivalence point.
What are weak acids?Weak acids are acids which only ionize partially in aqueous solutions.
When weak acids are dissolved in water, they produce only few hydrogen ions.
A strong base on the other hand ionizes completely to produce hydroxide ions in aqueous solutions.
The titration of a weak acid and a strong base gives a solution whose pH is greater than 7 at equivalence point.
Learn more about equivalence point at: https://brainly.com/question/18933025
How many atoms are in .45 moles of P4010
Answer:
5×6.02×1023
Explanation:
there are 5×6.02×1023 molecules of p4010 in 5mole. there are four P atoms in a single molecule of p4010
complete the following steps.
Remember to follow lower numbered rules first.
Na2CO3(aq) + Pb(OH)2(aq) → NaOH (?) + PbCO3(?)
a. Write a balanced chemical equation. (1 pt)
b. If a reaction occurs, write the balanced
chemical equation with the proper states of matter
(i.e. solid, liquid, aqueous) filled in. If no reaction
occurs, write “No reaction.” (1 pt)
c. If a reaction occurs, write the net ionic equation
for the reaction. If no reaction occurs, write "no
reaction.” (1 pt)
Answer:
See explanation
Explanation:
a) The balanced reaction equation is;
Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH + PbCO3
b) When we include states of matter;
Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH(aq) + PbCO3 (s)
c) Complete ionic equation;
2Na^+(aq) + CO3^2-(aq) + Pb^2+(aq) + 2OH^-(aq) ----> 2Na^+(aq) + 2OH^-(aq) + PbCO3(s)
Net Ionic equation;
Pb^2+(aq) + CO3^2-(aq) ----> PbCO3(s)
Help!!!!!!!!!
I'm using plato
Answer:
- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.
- One black ball and two black balls: they represent a compound formed by two different elements.
- One gray ball and two black balls: they represent a compound formed by two different elements.
- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.
Explanation:
Hey there!
In this case, according to the given information, we can firstly bear to mind the fact that each ball color represents a different element, for that reason we can tell the following:
- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.
- One black ball and two black balls: they represent a compound formed by two different elements.
- One gray ball and two black balls: they represent a compound formed by two different elements.
- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.
Regards!
Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.
half-reaction identification
Cu+(aq)--->Cu2+(aq) + e- _________
I2(s) + 2e--->2I-(aq) _________
Answer:
Cu+(aq)--->Cu2+(aq) + e- : oxidation
reason: there is loss of electrons.
I2(s) + 2e--->2I-(aq) : reduction
reason: There is reduction of electrons.
Question 9 of 25
How many hydrogen atoms are in a molecule of table sugar (C12H,2011)?
O A. 12
B. 45
C. 11
D. 22
SUBMIT
D.22
is my answer than welcome
2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4 g. Calculate the density of the liquid in Lbs/ in3.
Answer:
[tex]d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:
[tex]d=\frac{m}{V}[/tex]
Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:
[tex]m=552.4g-464.7g=87.7g[/tex]
So that we are now able to calculate the density in g/mL first:
[tex]d=\frac{87.7g}{27.8mL}=3.15g/mL[/tex]
Now, we proceed to the conversion to lb/in³ by using the following setup:
[tex]d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Regards!
A hypothetical A-B alloy of composition 53 wt% B-47 wt% A at some temperature is found to consist of mass fractions of 0.5 for both and phases. If the composition of the phase is 92 wt% B-8 wt% A, what is the composition of the phase
Answer:
the composition of the ∝ phase C∝ = 14 or [ 14 wt% B-86 wt% A ]
Explanation:
Given the data in the question;
Co = 53 or [ 53 wt% B-47 wt% A ]
W∝ = 0.5 = Wβ
Cβ = 92 or [ 92 wt% B-8 wt% A ]
Now, lets set up the Lever rule for W∝ as follows;
W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]
so we substitute our given values into the expression;
0.5 = [ 92 - 53 ] / [ 92 - C∝ ]
0.5 = 39 / [ 92 - C∝ ]
0.5[ 92 - C∝ ] = 39
46 - 0.5C∝ = 39
0.5C∝ = 46 - 39
0.5C∝ = 7
C∝ = 7 / 0.5
C∝ = 14 or [ 14 wt% B-86 wt% A ]
Therefore, the composition of the ∝ phase C∝ = 14 or [ 14 wt% B-86 wt% A ]
Balance the following chemical equation.
CCl4 -> ___ C+ ___ Cl2
Answer:
Explanation:
CCl4 => C + 2Cl2
What enzyme below is an exoenzyme?
A. Casease
B. Citrase
C. Catalase
D. Oxidase
A uniform plastic block floats in water with 50.0 % of its volume above the surface of the water. The block is placed in a second liquid and floats with 23.0 % of its volume above the surface of the liquid.
What is the density of the second liquid?
Express your answer with the appropriate units.
Answer:
density of second liquid = 650 kg/m³
Explanation:
Given that:
The volume of the plastic block submerged inside the water = 0.5 V
The force on the plastic block = [tex]\rho_1V_1g[/tex]
[tex]= 0.5p_1 V_g[/tex]
when the block is floating, the weight supporting the force (buoyancy force) is:
W [tex]= 0.5p_1 V_g[/tex]
[tex]\rho Vg = 0.5p_1 V_g[/tex]
[tex]\rho = 0.5 \rho _1[/tex]
where;
water density [tex]\rho _1[/tex] = 1000
[tex]\rho = 0.5 (1000)[/tex]
[tex]\rho = 500 kg/m^3[/tex]
In the second liquid, the volume of plastic block in the water = (100-23)%
= 77% = 0.7 V
The force on the plastic block is:
[tex]= 0.77p_2 V_g[/tex]
when the block is floating, the weight supporting the force (buoyancy force) is:
[tex]W = 0.77p_2 V_g[/tex]
[tex]\rho Vg = 0.77 \rho_2 V_g \\ \\ \rho = 0.77 \rho_2 \\ \\ 500 = 0.77 \rho_2 \\ \\ \rho_2 = 500/0.77[/tex]
[tex]\mathbf{ \rho_2 \simeq 650 \ kg/m^3}[/tex]
1. Most of the chemicals included in your General Chemistry Lab kit can be discarded down a drain. Describe a situation in which you would need to neutralize a chemical before discarding down a drain.
Answer: Chemicals like acids and bases are harmful and must be neutralized before draining.
Explanation:
A strong acid or strong base is required to be diluted or neutralized before it is discarded in the drain as if is discarded without diluting and neutralization it can spill and splash from sink or drain and can harm people in chemistry lab, moreover the fumes of the discarded chemical on spilling can cause respiratory tract burning and can even cause fire hazard so it must be converted into less harmful form and then must be drained.
2. For each of the ionic compounds in the table below, name the compound and explain the rule that you
used in formulating your name for the compound.
Name:
Rule for naming compound:
-PbF4
-NH4NO3
-Li2S
Answer:
2
Explanation:
Lead(|V) fluoride
Ammonium Nitrate
Lithium sulfide
For the rules, I don't know what you were taught. I just do it intuitively since I have done so much chemistry.
The first one the roman numerals represents the charge of the lead which much match the 4- charge from the 4 fluorides.
The second one is just two polyatomic ions which you just have to remember.
The last one is the typical ionic compound naming technique i guess.
Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon.
The question is incomplete, the complete question is;
Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon
a) electronic excitation
b) bond breakage
c) molecular vibration
d) molecular rotation
Answer:
molecular rotation
Explanation:
Microwaves are part of the electromagnetic spectrum. They are lower energy, lower frequency radiation.
When molecules absorb infrared radiation, they transition between the rotational states of the molecule.
Hence, the highest energy molecular process that occurs when a molecule absorbs a microwave photon is molecular rotation.
Identify the oxidation half-reaction for this reaction:
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
A. Fe2+ + 2e → Fe(s)
O B. H2(g) → 2H+ + 2e
O C. Fe(s) → Fe2+ + 2e
O D. 2H+ + 2e → H2(9)
Answer:
Fe(s)->Fe2+2e-
Explanation:
A.p.e.x
The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct
What is Oxidation reaction ?Oxidation reaction is a chemical reaction which can be described as follows ;
Addition of oxygen Removal of hydrogen Loss of ElectronAddition of electronegative atomRemoval of Electropositive elementIn the given reaction ;
Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)
Fe at RHS got converted to Fe²⁺ state at LHS which shows the gain of electron by Fe with in the reaction.
Therefore,
The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct
Learn More about redox reaction here ;
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CAN HF USED TO CLEAVE ETHERS EXPLAIN
Answer:
no
Explanation:
Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.
