Answer:
The products have a higher heat content than the reactants.
Explanation:
The statement above is not true for an exothermic reaction because in an exothermic reaction heat is released to the surroundings. This simply means that the total energy of the products is less than that of the reactants.
What is the osmolarity of a 0.20 M solution of KCI?
A) 0.40 Osmol
B) 0.30 Osmol C) 0.20 Osmol D) 0.80 Osmol
E) 0.10 Osmol
Answer:
Osmolarity of solution of KCI = 0.40 osmol
Explanation:
Given:
KCL ⇒ K⁺ + Cl⁻
Find:
Osmolarity of solution of KCI
When M = 0.20 M
Computation:
1 mole of KCL = 2 osmol
1 M of KCl = 2 Osmolarity
So,
Osmolarity of solution of KCI = 2 × 0.20
Osmolarity of solution of KCI = 0.40 osmol
6. Potassium hydrogen phthalate (KHP, KHC8H4O4) is also a good primary standard. 20 mL of NaOH was titrated with 0.600 M KHC8H4O4 solution. The data was graphed and the equivalence point was found when 15.5 mL of the standard 0.600 M KHP solution was added. The reaction equation is: a. What is the molar ratio of NaOH:KHC8H4O4? b. What is the molarity of the NaOH solution?
Answer:
a. 1
b. 0.465M NaOH
Explanation:
KHP reacts with NaOH as follows:
KHP + NaOH → KP⁻ + Na⁺ + H₂O
a. Molar ratio represents how many moles of NaOH reacts per mole of KHP. As you can see in the reaction, 1 mole of NaOH reacts with 1 mole of KHP. Molar ratio is:
1/1 = 1
b. With volume and molar concentration of the KHP solution you can find how many moles of KHP were added until equivalence point, thus:
15.5mL = 0.0155L ₓ (0.600 moles KHP / L) = 0.0093 moles of KHP
In equivalence point, moles of NaOH = Moles KHP. That means moles of NaOH titrated are 0.0093 moles NaOH.
The volume of the NaOH solution was 20mL = 0.020L. Molarity of the solution is:
0.0093 moles NaOH / 0.020L =
0.465M NaOHa. The balanced equation shows a 1:1 molar ratio between NaOH and KHC₈H₄O₄. This means that for every 1 mole of NaOH, we require 1 mole of KHC₈H₄O₄. Therefore, the molar ratio of NaOH:KHC₈H₄O₄ is 1:1.
The balanced equation for the reaction:
NaOH + KHC₈H₄O₄ → NaKC₈H₄O₄ + H₂O
b. Molarity of KHP solution × volume of KHP solution = Molarity of NaOH solution × volume of NaOH solution at the equivalence point
Molarity of KHP solution = 0.600 M
Volume of KHP solution = 15.5 mL = 0.0155 L
Volume of NaOH solution at the equivalence point = 20 mL = 0.0200 L
Molarity of NaOH solution = (Molarity of KHP solution × volume of KHP solution) / volume of NaOH solution at the equivalence point
Molarity of NaOH solution = (0.600 M × 0.0155 L) / 0.0200 L
Molarity of NaOH solution ≈ 0.465 M
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write down structures for three isomers of hexane
Answer:
Three isomers of hexane are: 2-methylpentane, 3-methylpentane, and 2,3-dimethylbutane.
They are constitutional isomers because they each contain exactly the same number and type of atoms, in this case, six carbons and 14 hydrogens and no other atoms.
Explanation:
Hope it helps.
If a radioactive isotope of thorium (atomic number 90, mass number 232) emits 6 alpha particles and 4 beta particles during the course of radioactive decay, what is the mass number of the stable daughter product?
Answer:
The mass number of the stable daughter product is 208
Explanation:
First thing's first, we have to write out the equation of the reaction. This is given as;
²³²₉₀Th → 6 ⁴₂α + 4 ⁰₋₁ β + X
In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.
There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.
Mass Number
Reactant = 232
Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x
since reactant = product
232 = 24 + x
x = 232 - 24 = 208
Atomic Number
Reactant = 90
Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x
since reactant = product
90 = 8 + x
x = 90 - 8 = 82
Fe3+(aq)+6H2O(l)⇌Fe(H2O)63+(aq) : F e 3 + ( a q ) + 6 H 2 O ( l ) ⇌ F e ( H 2 O ) 6 3 + ( a q ) : blank is the Lewis acid and blank is the Lewis base. is the Lewis acid and F e 3 + ( a q ) + 6 H 2 O ( l ) ⇌ F e ( H 2 O ) 6 3 + ( a q ) : blank is the Lewis acid and blank is the Lewis base. is the Lewis base.
Answer:
Lewis acid- Fe3+
Lewis base- water molecule
Explanation:
Acids and bases have been defined in diverse ways. There have been definitions put forward by Arrhenius, Brownstead and Lowry as well as Lewis. Each definition his useful in its own way.
Lewis acids are lone pair acceptors such as metal ions. This implies that in the particular instance of this reaction, Fe3+ is the lewis acid.
Similarly, a Lewis base is a lone pair donor, all ligands are lone pair donors since they donate one or more lone pairs of electrons to Lewis acids. In the particular instance of this reaction, the Lewis base is the water molecule.
The branch of science which deals with the chemicals bond is called chemistry.
The correct answer is Lewis acid [tex]Fe^{3+[/tex] and the lewis base is a water molecule.
Those chemicals which release the H+ ion when reacting with water are called acids. Those chemicals which release OH- ion, when reacted with the water is called a base. These definitions are given by the Lewis
Lewis acids are lone pair acceptors such as metal ions. This implies that in the particular instance of this reaction, Fe3+ is the lewis acid.
Similarly, a Lewis base is a lone pair donor, all ligands are lone pair donors since they donate one or more lone pairs of electrons to Lewis acids. In the particular instance of this reaction, the Lewis base is the water molecule.
Hence, the correct answer is Fe3+.
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A chemistry student weighs out of formic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.
The given question is incomplete, the complete question is:
A chemistry student weighs out 0.0349g of formic acid HCHO2 into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1500M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.
Answer:
The correct answer is 5.06 ml.
Explanation:
Based on the given information, the weight of formic acid given is 0.0349 grams. The volume of formic acid of V1 given is 250 ml. The molecular mass of formic acid is 46 grams per mole. Now the molarity of formic acid will be,
[HCOOH] = weight * 1000 / molecular mass * volume (ml)
= 0.0349 * 1000 / 46 * 250
= 0.003035 M or M1
The molarity of NaOH given is 0.1500 M or M2
Let us assume that the volume needed to attain equivalence point is V2 ml. The volume V2 can be determined by using the dilution equation,
M1V1 = M2V2
V2 = M1V1/M2
V2 = 0.003035 * 250 / 0.1500
V2 = 5.06 ml.
Hence, the volume of NaOH needed is 5.06 ml.
The decomposition of H2O2 is first order in H2O2 and the rate constant for this reaction is 1.63 x 10-4 s-1. How long will it take for [H2O2] to fall from 0.95 M to 0.33 M?
Answer:
It will take 6486.92 minutes for [H2O2] to fall from 0.95 M to 0.33 M
Explanation:
The order of reaction is defined as the sum of the powers of the concentration terms in the equation. Order of a reaction is given by the number of atoms or molecule whose concentration change during the reaction and determine the rate of reaction.
In first order reaction;
[tex]In \dfrac{a}{a_o-x}= k_1 t[/tex]
where;
a = concentration at time t
[tex]a_o[/tex] = initial concentration
and k = constant.
[tex]In (\dfrac{0.33}{0.95})= -1.63 \times 10^{-4} \times t[/tex]
[tex]-1.05736933 = -1.63 \times 10^{-4} \times t[/tex]
[tex]t = \dfrac{-1.05736933}{ -1.63 \times 10^{-4} }[/tex]
t = 6486.92 minutes
Find the [OH−] of a 0.32 M methylamine (CH3NH2) solution. (The value of Kb for methylamine (CH3NH2) is 4.4×10−4.) Express your answer to two significant figures and include the appropriate units.
Answer:
[tex][OH^-]=0.01165M[/tex]
Explanation:
Hello,
In this case, for the dissociation of methylamine:
[tex]CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)[/tex]
We can write the basic dissociation constant as:
[tex]Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}[/tex]
That in terms of the reaction extent [tex]x[/tex], turns out:
[tex]Kb=\frac{x*x}{[CH_3NH_2]_0-x}[/tex]
[tex]4.4x10^{-4}=\frac{x^2}{0.32M-x}[/tex]
That has the following solution for [tex]x[/tex]:
[tex]x_1=-0.01209M\\x_2=0.01165M[/tex]
Yer 0.01165M is valid only as no negative concentrations are eligible. It means that it is the concentration of hydroxyl ions in the solution:
[tex][OH^-]=0.01165M[/tex]
Best regards.
1. Methanol is a high-octane fuel used in high performance racing engines. 2 CH3OH(l) + 3O2(g) → 2CO2(g) + 4 H20(g) a) Calculate ∆H० and ∆S० using thermodynamic data, and then ∆G
Answer:
The reaction given in the question is:
2CH₃OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 4H₂O (g)
The values of ΔH°formation and ΔS° of the reactants and products given in the reaction based on the thermodynamics data is:
ΔH°formation values of CH3OH (l) is -238.4 kJ/mol, CO2(g) is -393.52 kJ/mol, H2O (g) is -241.83 kJ/mol and O2 (g) is 0.
The S° values of CH3OH (l) is 127.19 J/molK, CO2(g) is 213.79 J/molK, H2O (g) is 188.84 J/moleK, and O2 (g) is 205.15 J/molK.
Now the values of ΔH° and ΔS° are,
ΔH°rxn = 2 * ΔH°formation CO2 (g) + 4 * ΔH°formation H2O (g) - 2*ΔH°formation CH3OH (l)
ΔH°rxn = 2 * (-393.52) + 4 (-241.83) -2 * (-238.4)
ΔH°rxn = -1277.56 kJ/mole
ΔS°rxn = 2 * S° CO2 (g) + 4 * S° H2O (g) - 2*S° CH3OH (l) - 3 * S° O2 (g)
ΔS°rxn = 2 * 213.79 + 4 * 188.84 - 2 * 127.19 - 3*205.15
ΔS°rxn = 313.11 J/mole/K
Now the formula for calculating ΔG°rxn is,
ΔG°rxn = ΔH°rxn - TΔS°rxn
ΔG°rxn = -1277.56 * 1000 J/mole - 298 * 313.11 J/mole
ΔG°rxn = -1370.86 kJ/mol
Write a balanced equation for the double-replacement precipitation reaction described, using the smallest possible integer coefficients. A precipitate forms when aqueous solutions of ammonium bromide and silver(I) nitrate are combined. (Use the lowest possible coefficients. Omit states of matter.)
Answer:
NH4Br + AgNO3 —> AgBr + NH4NO3
Explanation:
When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:
NH4Br(aq) + AgNO3(aq) —>
In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:
NH4Br(aq) —> NH4+(aq) + Br-(aq)
AgNO3(aq) —> Ag+(aq) + NO3-(aq)
The double displacement reaction will occur as follow:
NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)
NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)
Enough of a monoprotic weak acid is dissolved in water to produce a 0.01660.0166 M solution. The pH of the resulting solution is 2.532.53 . Calculate the Ka for the acid.
Answer:
Explanation:
Let the monoprotic acid be HX
HX ⇄ H⁺ + X⁻
pH = 2.53
Hydrogen ion concentration
[tex][ H^+]=10^{-2.53}[/tex]
[tex][ X^-]=10^{-2.53}[/tex]
Concentration of undissociated acid will remain almost the same as it is a weak acid
So
Ka = concentration of H⁺ x concentration of Cl⁻ / concentration of acid
= [ H⁺] x [Cl⁻ ] / [ HX]
[tex]k_a=\frac{10^{-2.53}\times 10^{-2.53}}{.0166}[/tex]
[tex]k_a=\frac{.00295^2}{.0166}[/tex]
= 5.24 x 10⁻⁴ M .
Predict the order of acid strengths in the following series of cationic
species: CH3CH2NH3
+, CH3CH=NH2
Answer:
CH3CH=NH2+>CH3CH2NH3+
Explanation:
If we look at the both species under review, we will realize that they are both amines hence they possess the polar N-H bond.
Electrons are ordinarily attracted towards the nitrogen atom hence making both compounds acidic. It is worthy of note that certain features of a compound may make it more acidic than another of close structural proximity. 'More acidic' simply means that the proton is more easily lost.
CH3CH=NH2+ contains an sp2 hybridized carbon atom which is highly electronegative and further withdraws electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ compared to CH3CH2NH3+
A reaction is performed in a lab whereby two solutions are mixed together. The products are a liquid and a solid precipitate. What procedures would facilitate measurement of actual yield of the solid
Answer:
filtration, drying, and weighing
Explanation:
The procedures that would facilitate the measurement of the actual yield of the solid would be filtration of the precipitate, drying of the precipitate, and weighing of the precipitate respectively.
The liquid/solid mixture resulting from the reaction can be separated by the process of filtration using a filter paper. The residue in the filter paper would be the solid while the filtrate would be the liquid portion of the reaction's product.
The residue can then be allowed to dry, and then weighed using a laboratory-grade weighing balance. The weight of the solid represents the actual yield of the solid.
What is the oxidation number change for the iron atom in the following reaction? 2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g)
Answer:
[tex]\boxed{From \ +6 \ to \ 0}[/tex]
Explanation:
2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g)
In the given reaction, Iron in the reactants side have the oxidation number of +6. This is because [tex]O_{3}[/tex] with [tex]Fe_{2}O_{3}[/tex] has oxidation state -6, So any atom with it would have an oxidation state of +6 to give the resultant of zero.
In the products side, Iron acts as a free element reacting with no other atom. So, as per the rule of oxidation states, the oxidation state of Iron in the products side will be zero.
So, the oxidation number changes from +6 to 0 .
Extra Info: Decrease in oxidation state is Reduction , So Iron is being reduced here.
The change in the oxidation number of the iron atom in the reaction is from +3 to 0
Oxidation is simply defined as the the loss of electron. However, Oxidation number simply talks about the number of electrons that is either gained or lossed during bond formation.
The change in the oxidation number of iron in the reaction can be obtained as follow:
2Fe₂O₃(s) + 3C(s) → 4Fe(s) + 3CO₂(g)
Oxidation number of Fe in Fe₂O₃Oxidation number of Fe₂O₃ = 0 (ground state)
Oxidation number of oxygen = –2
Oxidation number of Fe =?Fe₂O₃ = 0
2Fe + 3O = 0
2Fe + 3(–2) = 0
2Fe – 6 = 0
Collect like term
2Fe = 6
Divide both side by 2
Fe = 6/2
Fe = +3Thus, the oxidation number of Fe in Fe₂O₃ is +3
Oxidation number of Fe (ground state) is zeroTherefore, the change in the oxidation number of the iron, Fe, atom in the reaction is from +3 to 0
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what is the osmotic pressure of pure water
Answer:
The osmotic pressure of ocean water is about 27 atm.
Explanation:
Pure water is water that contains no impurities. Ocean water is 96.5% pure with only about 3.5% of its content, salt water.
Osmotic pressure occurs when solutions that have different concentrations are isolated by a membrane. This osmotic pressure makes water move towards the solution that has the highest concentration, which means that if the concentration or temperature of the solution is high, the osmotic pressure becomes higher.
The equation for osmotic pressure is pi = iMRT.
Consider Zn + 2HCl → ZnCl2 + H2 (g). If 0.30 mol Zn is added to HCl, how many mol H2 are produced?
Answer:
0.3 mol
Explanation:
Assuming HCl is in excess and Zn is the limiting reagent,
from the balanced equation, we can see the mole ratio of Zn:H2 = 1:1,
which means, each mole of zinc reacted gives 1 mole of H2.
So, if 0.30 mol Zn is added, the no. of moles of H2 produced will also be 0.3 mol, since the ratio is 1:1.
If the distance between two objects increased, what would happen to the force of gravity between them? It would increase. It would stay the same. It would depend on the speed. It would decrease.
Answer:
it will decrease
Explanation:
force of gravity is inversely proportional to the distance
Answer:
It shall decrease! -w-ll
Explanation:
because of the gravitational pull
15. Ammonium nitrate, NH4NO3, and aluminum powder react explosively producing nitrogen gas, water vapor and aluminum oxide. Write the balanced equation and calculate the enthalpy change for this reaction.
Answer:
[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]
[tex]Entalpy=-2861.9~KJ[/tex]
Explanation:
In this case, we have to start with the reagents:
[tex]Al~+~NH_4NO_3[/tex]
The compounds given by the problem are:
-) Nitrogen gas = [tex]N_2[/tex]
-) Water vapor = [tex]H_2O[/tex]
-) Aluminum oxide = [tex]Al_2O_3[/tex]
Now, we can put the products in the reaction:
[tex]Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_) [/tex]
When we balance the reaction we will obtain:
[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]
Now, for the enthalpy change, we have to find the standard enthalpy values:
[tex]Al_(_S_)=0~KJ/mol[/tex]
[tex]NH_4NO_3_(_a_q_)=-132.0~KJ/mol[/tex]
[tex]N_2_(_g_)=0~KJ/mol[/tex]
[tex]H_2O_(_g_)=~-~241.8~KJ/mol[/tex]
[tex]Al_2O_3_(_S_)=~-~1675.7~KJ/mol[/tex]
With this in mind, if we multiply the number of moles (in the balanced reaction) by the standard enthalpy value, we can calculate the energy of the reagents:
[tex](0*2)~+~(-132*3)=~-396~KJ[/tex]
And the products:
[tex](0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ[/tex]
Finally, for the total enthalpy we have to subtract products by reagents :
[tex](-3125.9~KJ)-(-396~KJ)=-2729.9~KJ[/tex]
I hope it helps!
What does the state symbol (aq) mean when written after a chemical
compound in a chemical equation?
A. It means the compound is in the liquid phase.
B. It means the compound is dissolved in water.
C. It means the compound is in the gas phase.
D. It means the compound is in the solid phase.
B. it means the compound is dissolved in water
Answer:
b
Explanation:
a p e x :)
How many milliliters of a 0.250 MNaOHMNaOH solution are needed to completely react with 500. gg of glyceryl tripalmitoleate (tripalmitolein)
Answer:
[tex]7.48X10^3~mL[/tex]
Explanation:
For this question we have:
-) A solution NaOH 0.25 M
-) 500 g of glyceryl tripalmitoleate (tripalmitolein)
We can start with the reaction between NaOH and tripalmitolein. NaOH is a base and tripalmitolein is a triglyceride, therefore we will have a saponification reaction. The products of this reaction are glycerol and (E)-hexadec-9-enoate.
Now, with the reaction in mind, we can calculate the moles of NaOH that we need if we use the molar ratio between NaOH and tripalmitolein (3:1) and the molar mass of tripalmitolein (801.3 g/mol). So:
[tex]500~g~tripalmitolein\frac{1~mol~tripalmitolein}{801.3~g~tripalmitolein}\frac{3~mol~NaOH}{1~mol~tripalmitolein}=1.87~mol~NaOH[/tex]
With the moles of NaOH we can calculate the volume (in litters) if we use the molarity equation and the Molarity value:
[tex]M=\frac{mol}{L}[/tex]
[tex]0.25~M=\frac{1.87~mol~NaOH}{L}[/tex]
[tex]L=\frac{1.87~mol~NaOH}{0.25~M}[/tex]
[tex]L=7.48[/tex]
Now we can do the conversion to mL:
[tex]7.48~L~\frac{1000~mL}{1~L}=~7.48X10^3~mL[/tex]
I hope it helps!
Which compound is composed of oppositely charged ions? A. SCl2 B. OF2 C. PH3 D. Li2O
Answer:
D.
Explanation:
If a compound is composed of oppositely charged ions, it has to be formed by metal and non-metal.
Li2O
Li - metal
O - non-metal
This pluton occurs deep in Earth and does not cause any changes to the surface of Earth . True or False
Answer:
The given statement is false.
Explanation:
However, if the pluton exists beneath the ground, this could be conveniently shown in the illustration something from the peak such pluton appears convex in form resembling a lopolith and perhaps diapir, which would be a particular form of statistically significant pluton recognized as the sill.Mostly from the figure it could also be shown that subsurface sheets are lined or curved, throughout the pluton mold. And therefore it is inferred that such a pluton creates adjustment to something like the ground atmosphere by altering the form of the levels above it.So that the given is incorrect.
If 10.4 grams of iron metal react with 28.4 grams of silver nitrate, how many grams of iron nitrate can be formed and how many grams of the excess reactant will be left over when the reaction is complete? Show all of your work. unbalanced equation: Fe + AgNO3 Fe(NO3)3 + Ag
Answer:
71.1
Explanation:
1 mol Fe = 10.4 g/55.85 g/mol = 0.186
1 mol AgNo3 = 28.4 g/169.87 g/mol = 0.178 mol AgNo3
then since Ag:Fe is 1:3, AgNo3 is the limiting reactant
So now
0.178 moles * 1/3 * 241.83 g/mol Fe(NO3)3 = 14.35 g Fe(NO3)3
Excess reactant: 0.178 moes AgNO3 * 1/3 = 0.059
0.186 - 0.059 = 0.127 moles Fe * 55.85 g/mol Fe = 7.1 g Fe excess
Calculate the mass of feso4 that would be produced by 0.5mole of Fe
Answer:76 grams
Explanation:
Fe+H₂SO₄-->FeSO₄+H₂
For one mole of Fe we get 1 mole of feso4, therefore for 0.5 moles of Fe we get 0.5 moles of feso4.
The molar mass of feso4 is AFe+AS+4AO(A is atomic mass)
56+32+4*16=152grams/mole
Now, we need to multiply the number of moles by the molar mass to get the mass that reacts
152*0.5=76 grams
For the reaction system, 2SO2(g) + O2(g) <--> 2SO3(g), the equilibrium concentrations are: SO3: 0.120M SO2: 0.860M O2: 0.330M Calculate the value of Kc for this reaction.
Answer:
0.0590 M⁻¹
Explanation:
Kc represents the equilibrium constant. It is given as;
Kc = [products] / [reactants]
For the reaction; 2SO2(g) + O2(g) <--> 2SO3
Products = SO3
Reactants = SO2 and O2
Kc is given as;
Kc = [SO3]² / [SO2]² [O2]
Kc = 0.120² / (0.860)² (0.330)
Kc = 0.0144 / 0.2440 = 0.0590 M⁻¹
The specific rotation of (S)-carvone (at 20°C) is +61. A chemist prepared a mixture of (R)-carvone and its enantiomer, and this mixture had an observed rotation of -55°.
A) What is the specific rotation of (R)-carvone at 20°C?
B) Calculate the % ee of this mixture.
C) What percentage of the mixture is (S)-carvone?
Answer:
a) Specific Rotation of (R)-carvone = -61°
b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%
c) The percentage of (S)-carvone in the mixture = 4.9%
Explanation:
a) The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.
Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.
Specific Rotation of (R)-carvone = -(61°) = -61°
b) Enantiometic excess is used to measure the optical purity of an enantiomeric mixture.
The enantiomeric excess is given mathematically as
ee% = (Observed rotation × 100)/(Specific rotation)
Hence, to calculate the enantiomeric excess of (R)-carvone,
Observed rotation of the mixture = -55°
Specific Rotation of (R)-carvone = -61°
ee% = (-55×100)/(-61) = 90.16% = 90.2%
c) An enantiomeric excess of 90.2% for (R)-carvone indicates that it's actual percentage is 90.2% more than the percentage of its enantiomeric partner, (S)-carvone, in the mixture.
Let the percentage of (R)-carvone in the mixture be x
Let the percentage of (S)-carvone in the mixture be y
x + y = 100
x - y = 90.2
2x = 190.2
x = (190.2/2) = 95.1%
y = 100 - x = 100 - 95.1 = 4.9%
Hence, the percentage of (R)-carvone in the mixture = 95.1%
The percentage of (S)-carvone in the mixture = 4.9%
Hope this Helps!!!
a) Specific Rotation of (R)-carvone = -61°
b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%
c) The percentage of (S)-carvone in the mixture = 4.9%
a) Calculation of Specific Rotation:The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.
Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.
Specific Rotation of (R)-carvone = -(61°) = -61°
b) Calculation for Enantiomeric excess:
The enantiomeric excess is given mathematically as
ee% = (Observed rotation × 100)/(Specific rotation)
Hence, to calculate the enantiomeric excess of (R)-carvone,
Observed rotation of the mixture = -55°
Specific Rotation of (R)-carvone = -61°
ee% = (-55×100)/(-61) = 90.16% = 90.2%
c) Calculation of percentage:
Let the percentage of (R)-carvone in the mixture be x
Let the percentage of (S)-carvone in the mixture be y
x + y = 100
x - y = 90.2
2x = 190.2
x = (190.2/2) = 95.1%
y = 100 - x = 100 - 95.1 = 4.9%
Hence, the percentage of (R)-carvone in the mixture = 95.1%
The percentage of (S)-carvone in the mixture = 4.9%
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Which of the following statements about metal elements is correct?
A. Metals tend to easily gain more valence electrons.
B. Metal elements are always heavier than non-metal elements.
C. Metals tend to easily lose their valence electrons.
D. A metal atom can take an electron from a non-metal atom.
Answer: C. Metals tend to easily lose their valence electrons.
Explanation:
Metals are those substances which have tendency to loose their valence electrons to attain noble gas configuration and forms positive ions called as cations.
Example: Gold, potassium etc
[tex]M\rightarrow M^++e^-[/tex]
Non metals are those substances which have tendency to gain valence electrons to attain noble gas configuration and form negative ions called as anions.
Example: Sulphur, Chlorine
[tex]N+e^-\rightarrow N^-[/tex]
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 1.5 L flask with 0.59 atm of sulfur dioxide gas and 2.9 atm of oxygen gas at 35.0 °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.53 atm.
Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.
Kp=_______.
Answer:
P SO₂ = 0.06atm
P O₂ = 2.635atm
P SO₃ = 0.53atm
Kp = 29.6
Explanation:
The reaction of Sulfur dioxide and oxygen react to form sulfur trioxide is as follows:
2SO₂(g) + O₂(g) ⇄ 2SO₃(g)
And Kp is defined as:
[tex]Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}[/tex]
Where P represents the pressure at equilibrium of each reactant.
If you add, in the first, 0.59atm of SO₂ and 2.9atm of O₂, the equilibrium pressures will be:
P SO₂ = 0.59atm - 2X
P O₂ = 2.9atm - X
P SO₃ = 2X
Where X represents the reaction coordiante.
As equilibrium pressure of SO₃ is 0.53atm:
0.53atm = 2X
0.265atm = X
Replacing, equilibrium pressures of each species will be:
P SO₂ = 0.59atm - 2×0.265atm
P O₂ = 2.9atm - 0.265atm
P SO₃ = 2×0.265atm
P SO₂ = 0.06atmP O₂ = 2.635atmP SO₃ = 0.53atmAnd Kp will be:
[tex]Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}[/tex]
[tex]Kp = \frac{0.53^2}{{0.06}^2*{2.635}}[/tex]
Kp = 29.6Which is most likely to happen during a precipitation reaction?
A. A solid substance will break down into two new substances that
are gases.
B. An insoluble solid will form when ions in dissolved compounds
switch places.
C. A substance will react with oxygen to form water and carbon
dioxide.
D. A gas will form when positive ions switch places to form new
compounds.
Answer:
I think its B
Explanation:
Precipitation reactions leave a solid behind. The solid is called a precipitate.
Answer:
B
Explanation:
An insoluble solid will form when ions in dissolved compounds switch places.
Which one of the following is most likely to gain electrons when forming an ion, based on the natural tendency of the element?
A Ni
B S
C Na
D Cr
E Be
Answer:
Option B. S
Explanation:
All of the options except sulphur, S is metal.
Metals tend to lose electron in order to form ion. Non metals on the other hand gain electron to form ion.
Sulphur, S has atomic number of 16 with electronic configuration as:
S (16) => 1s² 2s²2p⁶ 3s²3p⁴
From the above illustration, we can see that sulphur needs two more electrons to complete it's octet configuration.
Therefore, sulphur, S will gain two electrons to form ion.
As stated earlier, the rest option given are all metals which will form ion by losing electron(s).
Answer
B) Sulphur (S)
Explanation
Here in the options we have been provided with elements like Nickel (Ni), Sulphur (S), Sodium (Na), Chromium (Cr) and Beryllium (Be) but except for Sulphur all the other ones are metals.
Now, let us understand what is a metal and a non-metal.
Metal- electron donors are called as metal.Non-metal- electron acceptors are called non-metals.So, sulphur being the only non metal will accept electron to complete its octate and to stablize itself and form a Anion.
Now let us also look at the electronic configuration of Sulphur to get the picture more clearly
atomic no. of sulphur would be = 16[tex]S\rightarrow 1s^2\; 2s^2\;2p^6\;3s^2\;3p^4[/tex]
so here the p-subshell is incomplete and is in need of 2 electrons.
Therefore the element which is most likely to gain electrons, forming an Anion will be sulphur.
To learn more about Ion Formation
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