Answer:
Let’s break down all the steps in the following SN1 reaction looking at the energy diagram: Step Breaking the C – LG bond. In this rate-determining step, a carbocation intermediateis formed: Step A nucleophilic attack.The carbocation is highly electron-deficient and the nucleophile attacks as a Lewis base usin…
Image
Explanation:
Which of the following could not be a resonance structure of CH3NO2?
a)
H
H-C-NO
H
b)
H .0:
H-C-N
H
c)
H:03
H-C-NC2
H:06
d)
H
H-C=N
H :9-H
e) Both c and d
Answer:
the answer is b.CH3NO2 I guess I'm correct
8. An experiment requires a solution that is 80%
methyl alcohol by volume. What volume of
methyl alcohol should be added to 200 mL of
water to make this solution?
Answer:
[tex]v_{solute}=160mL[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the volume of methyl alcohol solute by using the definition of by-volume percentage:
[tex]\%v=\frac{v_{solute}}{v_{solution}} *100\%[/tex]
Whereas we solve for the volume of the solute as shown below:
[tex]v_{solute}=\frac{\%v*v_{solution}}{100\%} \\\\v_{solute}=\frac{80\%*200mL}{100\%}\\\\ v_{solute}=160mL[/tex]
Regards!
If 0.3250 L of 0.125 M NaOH base were used in a titration, what were the moles of base?
Answer:
Explanation:You can download the ly/3fcEdSxans[tex]^{}[/tex]wer here. Link below!
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According to molar concentration, there are 0.0406 moles of base used in the titration.
What is molar concentration?Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.
The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.
In terms of moles, it's formula is given as molar concentration= number of moles /volume of solution in liters.Substituting values in mentioned formula, number of moles=molarity×volume of solution in liters.
∴number of moles=0.125×0.3250=0.0406 moles
Thus, 0.0406 moles are used in the titration.
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Which statement best describes the formula equation Cl2(g) + 2KBr(aq) Right arrow. 2KCl(aq)+ Br2(l)?
Carbon iodide reacts with potassium bromide to form potassium carbon iodide and bromine.
Bromine gas reacts with a solution of potassium chloride to form potassium bromide and chlorine gas.
Potassium bromine gas reacts with liquid chlorine to form potassium chloride in solution and bromine gas.
Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.
Answer:
Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.
Explanation:
The balanced equation of the chemical reaction is given below:
Cl2(g) + 2KBr(aq) → 2KCl(aq)+ Br2(l)
According to the above equation, it can be said that chlorine in its gaseous form (Cl2) reacts with pottasium bromide (reactants) to form pottasium chloride (KCl) and bromine, which is a liquid at room temperature.
Answer:
D: Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.
Element compound2. Select all the compounds from the following elements
HE
Fe2O3
O2
P4
C2H4O2
Answer: [tex]Fe_2O_3[/tex] and [tex]C_2H_4O_2[/tex] are the compounds.
Explanation:
A chemical compound is defined as a chemical substance that is formed by the combination of two or more atoms of different elements which cannot be separated by any physical means but when chemically treated, they decompose into their parent elements.
For example, water is made up of hydrogen and oxygen. This compound is a liquid and its individual components are gases. When water is decomposed, it forms hydrogen and oxygen gas.
For the given options:
He(Helium) is an element formed by the combination of only type of atoms.
[tex]O_2[/tex] and [tex]P_4[/tex] are molecules of same element.
[tex]Fe_2O_3[/tex] is a compound fomed by the combination of iron and oxygen atoms.
[tex]C_2H_4O_2[/tex] is a compound fomed by the combination of carbon, hydrogen, and oxygen atoms.
Hence, [tex]Fe_2O_3[/tex] and [tex]C_2H_4O_2[/tex] are the compounds.
Zinc sulfate is a 2-ion electrolyte,
dissociating 40% in a
certain concentration. Calculate its
dissociation (i) factor.
On the basis of 40% dissociation, 100
particles of zinc sulfate
will yield:
40zinc ions
40 sulfate ions
60undissociated particles
Jo 11:03
Answer: The value of i is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
Explanation:
The equation used to calculate the Vant' Hoff factor in dissociation follows:
[tex]\alpha =\frac{i-1}{n-1}[/tex]
where,
[tex]\alpha [/tex] = degree of dissociation = 40% = 0.40
i = Vant' Hoff factor
n = number of ions dissociated = 2
Putting values in above equation, we get:
[tex]0.40=\frac{i-1}{2-1}\\\\0.40=i-1\\\\i=1.4[/tex]
The equation used to calculate the degee of dissociation follows:
[tex]\alpha =\frac{\text{Number of particles dissociated}}{\text{Total number of particles taken}}[/tex]
Total number of particles taken = 100
Degree of dissociation = 40% = 0.40
Putting values in above equation, we get:
[tex]0.40=\frac{\text{Number of particles dissociated}}{100}\\\\\text{Number of particles dissociated}=(0.40\times 100)=40[/tex]
This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.
Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
calculate the pH of a 0.2 M * 4 solution for which Kb = 1.8*10^-5 at 26 c . The equation for the reaction
Nh3+H2O->NH4+oh
Answer: The pH of the solution is 11.24
Explanation:
We are given:
Molarity of ammonia = 0.2 M
[tex]K_b=1.8\times 10^{-5}[/tex]
The given chemical equation follows:
[tex]NH_3+H_2O\rightleftharpoons NH_4^++OH^-[/tex]
I: 0.2
C: -x +x +x
E: 0.2-x x x
The expression for equilibrium constant follows:
[tex]K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]
Putting values in above expression, we get:
[tex]1.8\times 10^{-5}=\frac{x^2}{0.2-x}\\\\1.8\times 10^{-5}(0.2-x)=x^2\\\\x^2+(1.8\times 10^{-5}x)-(0.36\times 10^{-5})=0\\\\x=1.88\times 10^{-3}, 1.9\times 10^{-3}[/tex]
Neglecting the negative value of x as concentration cannot be negative.
So, [tex][OH^-]=x=1.88\times 10^{-3}M[/tex]
pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution.
[tex]pOH=-\log [OH^-][/tex]
Putting values in above equation, we get:
[tex]pOH=-\log (1.88\times 10^{-3})\\\\pOH=2.76[/tex]
We know:
[tex]pH+pOH=14\\\\pH=14-2.76\\\\pH=11.24[/tex]
Hence, the pH of the solution is 11.24
Ideally, the van 't Hoff factor should be equal to the number of ions that make up a compound. In reality, van 't Hoff factors tend to be lower due to ion-pairing. Select all conditions that would increase the effect of ion- pairing and decrease the observed van 't Hoff factor.
A. Lower charge on the ions
B. High concentration
C. Low concentration
D. Higher charge on the ions
Answer:
High concentration
Higher charge on the ions
Explanation:
Ideally, the Van't Hoff factor is defined as the ratio between the actual concentration of particles produced when the substance is dissolved in a solution and the concentration of a substance as calculated from its mass(Wikipedia).
The Van't Hoff factor is influenced by the concentration of ions in solution as well as the magnitude of charge on the ions in solution.
Highly charged ions tend to remain paired in solution thereby leading to a lower value of Van't Hoff factor. Also, in highly concentrated solutions, the Van't Hoff factor is lower than in dilute solutions due to the pairing of ions
The conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff factor are a high concentration of solute and a higher charge on the ions.
Explanation:
A number of particles are formed by dissolving a unit molecule of a solute in a solvent.The higher concentration of solution lowers the value of the van 't Hoff factor as ionic compounds do not get completely dissociated in their aqueous solution.Also, some ions form ion pairs (pairing of opposite charges in the aqueous state) and exist as a single particle which decreases the value of the van 't Hoff factor.The ions with higher charges have a greater tendency to form ion-pair due to their high electrostatic interaction.So, from this, we can conclude that conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff factor are a high concentration of solute and a higher charge on the ions.
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Calculate ΔG° for the following reactions at 25 °C using the provided appendix
b. The following reaction takes place in a basic solution. (7 points)
MnO4–(aq) + NO2–(aq) MnO2(s) + NO3–(aq)
The half-reactions (balanced only for atoms) are the following:
MnO4– + 2H2O MnO2 + 4OH–
NO2– + 2OH– NO3– + H2O
Use the following steps to finish balancing the equation.
i. Balance each half-reaction for charge. (2 points)
ii. Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions. (2 points)
iii. Add the equations and simplify to get a balanced equation. (2 points)
iv. How can you tell from this equation that the reaction occurred in a basic solution? (1 point)
Answer: The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.
Explanation:
Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.
The given redox reaction follows:
[tex]MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)[/tex]
To balance the given redox reaction in basic medium, there are few steps to be followed:
Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electronsOxidation half-reaction: [tex]NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-[/tex]
Reduction half-reaction: [tex]MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-[/tex]
Multiply each half-reaction by the correct number in order to balance charges for the two half-reactionsOxidation half-reaction: [tex]NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-[/tex] ( × 3)
Reduction half-reaction: [tex]MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-[/tex] ( × 2)
The half-reactions now become:
Oxidation half-reaction: [tex]3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-[/tex]
Reduction half-reaction: [tex]2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-[/tex]
Add the equations and simplify to get a balanced equationOverall redox reaction: [tex]3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-[/tex]
As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution
Electrode A has a standard reduction potential of -0.21 volts and electrode B has a standard reduction potential of -0.15 volts. What electrode is the anode?
will give brainiest to person with the best answer ♡
So, we know the reduction potentials of the electrodes :
Electrode A = -0.21 V
Electrode B = -0.15 V
We want our cell to be spontaneous - this means that the voltage has to be positive. In order to do so, we need to turn the more negative reduction half reaction to be an oxidation half reaction.
-0.21 is more negative than -0.15, so electrode A will be the oxidation half reaction.
Anode is where the oxidation half reaction takes place, so electrode A is the correct answer . A fun way to remember this is An(ode)Ox(idation) = AnOx and Red(uction)Cat(hode) = RedCat
The E° cell will be 0.21 -0.15 V = 0.6 V
Help please with this question
Picture above
Answer:
3:372-88U:771-772828
g Suppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read the base titrant volume as 1.84 mL. After running the titration and reaching the endpoint, you read the base titrant volume as 27.30 mL. What volume, in mL, of base was required for the titration?
Answer:
25.46 mL
Explanation:
In a titration we use the volume and concentration of a solution to determine the previously unknown concentration of other solution. Let's consider the titration of an acid of unknown concentration with a standardized base (a base whose concentration we know). The generic neutralization reaction is:
HA + BOH ⇒ BA + H₂O
The base is in the buret and we will add it to the acid until the equivalence point is reached. The volume of base used is equal to the difference between the final reading of the buret and the initial reading of the buret.
V = 27.30 mL - 1.84 mL = 25.46 mL
2 AICI3 + 3 Ca - 3 CaCl2 + 2 Al
You react aluminum chloride with calcium metal. You want to produce 40.00 grams of aluminum. How many grams of calcium do
you need?
Answer:
50 gram calcium do you need
Explanation:
please make me brainlist answer
Chemosynthesis and photosynthesis are both processes that produce food.
True or false?
Answer:
true
Explanation:
probably true
Which of the following would eat a hamburger with lettuce on it?
Omnivore
Detritivore
Herbivore
Carnivore
Answer:
Omnivore
Explanation:
i know
Answer:
Omnivore
Explanation:
The answer is Omnivore
calculate the mass of water vapor that is produced by the reaction: 
1.4 g of CO2 and 2.2 g of KOH in the reaction: CO2 + 2KOH → K2CO3 + H20
Please show work, will give brainliest
Answer:
[tex]m_{H_2O}=0.353gH_2O[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to identify the required limiting reactant by calculating the moles of water vapor produced by each reactant, CO2 and KOH, as shown below:
[tex]1.4gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molH_2O}{1molCO_2}=0.0318mol H_2O\\\\2.2gKOH*\frac{1molKOH}{56.11gKOH}*\frac{1molH_2O}{2molKOH}=0.0196mol H_2O[/tex]
In such a way, since 2.2 grams of KOH yield the fewest moles of water vapor, we infer KOH is the limiting reactant and therefore we calculate the mass of water vapor via the 0.0196 moles we obtained:
[tex]m_{H_2O}=0.0196molH_2O*\frac{18.02gH_2O}{1molH_2O}=0.353gH_2O[/tex]
Regards!
WILL MAKE BRAINLIEST IF ANSWERED ASAP
Choose true or false
1. Acetone cannot be used as a solvent because the Grignard reagent will react with its carbonyl, instead of reacting with the planned synthesis carbonyl.
2. Tetrahydrofuran is not a suitable solvent for the Grignard reaction because his oxygen may form complexes with the Mg, deactivating the Grignard.
3. Phenol can be used as a solvent in Grignard reaction as long as is anhydrous and moisture is kept out of the system with a drying tube with drierite.
4. Syringes are used in the Grignard experiment to avoid mixing all the reagents.
5. We use syringes to inject the reagents through a septum preventing moisture to enter the system.
Explanation:
Grignard reagent reacts with ketones. Upon chemical reaction of acetone and Grignard reagent there will be formation of tertiary alcohol.
1). So the statement, acetone cannot be used as a solvent because the Grignard reagent will react with its carbonyl, instead of reacting with the planned synthesis carbonyl is true.
2). Tetrahydrofuran is not a suitable solvent for the Grignard reaction because his oxygen may form complexes with the Mg, deactivating the Grignard is a false statement.
3). Phenol can be used as a solvent in Grignard reaction as long as is anhydrous and moisture is kept out of the system with a drying tube with drierite is false statement.
4). Syringes are used in the Grignard experiment to avoid mixing all the reagents is false statement.
5). We use syringes to inject the reagents through a septum preventing moisture to enter the system is true statement.
7.A sample of oxygen gas, O2 weighs 28.4 grams. How many molecules of O2 and how
many atoms of O are present in this sample?
Answer:
5.34275*10²³
Explanation:
Molar mass of O2 is 32g/mol
mass of the sample is 28.4g
number of moles = (mass of the sample) / (molar mass of O2)
=28.5g / 32g/mol
=0.8875mol
number of molecules = number of moles * Avogadro's Constant
= 0.8875 * (6.02*10²³)
= 5.34275 molecules
HCIO4 is identified as what acid
How many atoms of nitrogen are in Fe2(NO4)2?
the number of nitrogen atoms in the compound is two
A straight chain hydrocarbon with the formula C5H8_____
Answer:
I has 2 double carbon carbon bonds
Calculate [OH - ] given [H 3 O + ] = 5.69x10 -5 M.
Answer:
7
Explanation:
Write the empirical formula for at least four ionic compounds that could be formed from the following ions:
a). PO3−4
b). NH+4
c). Fe3+
d). ClO−3
Answer:
a. Na₃PO₄
b. NH₄Cl
c. FeCl₃
d. KClO₃
Explanation:
a. Sodium tetraoxophosphate(V) Na₃PO₄
3Na⁺ + PO₄³⁻ → Na₃PO₄
b. Ammonium Chloride NH₄Cl
NH₄⁺ + Cl⁻ → NH₄Cl
c. Iron(III)chloride
Fe³⁺ + Cl⁻ → FeCl₃
d. Potassium trioxochlorate(V) KClO₃
K⁺ + ClO₃⁻ → KClO₃
Name a machine used to reap, thresh and clean the crop.
i will give brainliest who answers first
Answer:
combine harvester, or a combiner.
4. Cuanto electrones se necesitan para
pesarlo mismo que un proton?
Answer:
aproximadamente dos mil
Explanation:
masa_proton/masa_electron=
1.673e-24 / 9.11e-28 = 1836.443468715697
Which of the following ionization energies indicates an atom is most likely to gain electrons and form an anion or not form an ion at all?
Group of answer choices
578 kJ/mol
9460 kJ/mol
496 kJ/mol
786 kJ/mol
Answer:
Explanation:
578kj/mol
Please help me, it’s my last try
Answer:
Group 1A: alkali metals, or lithium family.
Group 2A: alkaline earth metals, or beryllium family.
Group 7A: the manganese family.
Group 8A: the iron family.
Explanation:
Answer:
1A: Alkali Metals
2A: Alkaline Earth Metals
7A: Halogens
8A: Noble Gases
ii Chlorine trifluoride is an extremely reactive substance and, for
instance, reacts explosively with water. The products of the reaction
are hydrofluoric acid, hydrochloric acid and oxygen gas. Write
an equation, including state symbols, for the reaction of chlorine
trifluoride with water.
Answer:
ClF₃(g) + 2 H₂O(l) ⇒ 3 HF(aq) + HCl(aq) + O₂(g)
Explanation:
Let's consider the unbalanced equation for the reaction between chlorine trifluoride and water to form hydrofluoric acid, hydrochloric acid and oxygen gas.
ClF₃(g) + H₂O(l) ⇒ HF(aq) + HCl(aq) + O₂(g)
We will balance it using the trial and error method.
First, we will balance F atoms by multiplying HF by 3.
ClF₃(g) + H₂O(l) ⇒ 3 HF(aq) + HCl(aq) + O₂(g)
Then, we will balance H atoms by multiplying H₂O by 2.
ClF₃(g) + 2 H₂O(l) ⇒ 3 HF(aq) + HCl(aq) + O₂(g)
This equation is balanced.