Answer:
b. Al3+ + 3Br– → Al + (3/2)Br2
Explanation:
If we look at this reaction closely, aluminum was reduced while bromine was oxidized. The reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Recall that the more negative the redox potential of a chemical specie, the greater its tendency to function as a reducing agent donating electrons in an electrochemical reaction.
However, in this reaction, aluminium is found to accept rather than donate electrons. Therefore, the process is non spontaneous and can only occur in an electrolytic cell, hence the answer.
b. [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]
Electrolytic Cell v/s Electrochemical Cell:
Electrochemical cells convert chemical energy into electrical energy or vice versa while an electrolytic cell is a type of electrochemical cell in which electrical energy is converted into chemical energy.Electrochemical cells consist of a cathode (+) and an anode(-) while Electrolytic cells consist of a positively charged anode and a negatively charged cathode.Out of the given reactions, [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex] is carried out in an electrolytic cell rather than in an electrochemical cell.
As we know, In electrolytic cells, like galvanic cells, are composed of two half-cells
one is a reduction half-cell, the other is an oxidation half-cell.In the given reaction, aluminum is being reduced while bromine gets oxidized and the value of reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Therefore, the process is non spontaneous and can only occur in an electrolytic cell.
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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.
How many grams of sodium phosphate are needed to have 1.67 moles of sodium ion?
Answer:
91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.
Explanation:
The formula for sodium phosphate is Na₃PO₄
Molar mass of sodium phosphate = 164 g/mol
The dissociation of one mole of sodium phosphate produces 3 moles of sodium ions;
Na₃PO₄ ------> 3Na⁺ + PO₄³
Number of moles of Na₃PO₄ that will produce 1.67 moles of Na⁺ = 1/3 * 1.67 = 0.556 moles of Na₃PO₄
Mass of 0.556 moles of Na₃PO₄ = 0.556 moles * 164 g/mol = 91.29 g
Therefore, 91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.
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.
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
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
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.
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)
. Calculate ΔG for the following reaction at 25°C. Will the reaction occur (be spontaneous)? How do you know? NH3(g) + HCl(g) → NH4Cl(s) ΔH = -176.0 kJ ΔS = -284.8 J·K-1
Answer:
[tex]\triangle G = -911.296 \ kJ[/tex]
Explanation:
ΔG = ΔH-TΔS
Where ΔH = -176 kJ = -176000 J , T = 25°C + 273 = 298 K , ΔS = -284.8 JK⁻¹
=> [tex]\triangle G =-176000 - (298)(-284.8)[/tex]
=> [tex]\triangle G = -176000+84870.4[/tex]
=> [tex]\triangle G = -91129.6 \ J[/tex]
=> [tex]\triangle G = -911.296\ kJ[/tex]
Since the value is negative, the reaction is spontaneous under standard conditions at 298 K and the reactants have more free energy than the products.
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
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 :)
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.
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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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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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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.
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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Air contains nitrogen, oxygen, argon, and trace gases. Ifthe partial pressure of nitrogen is 592 mm Hg, oxygen is160 mm Hg, argon is 7 mm Hg, and trace gas is 1 mm Hg,what is the atmospheric pressure
Answer:
760 mmHg
Explanation:
Step 1: Given data
Partial pressure of nitrogen (pN₂): 592 mmHgPartial pressure of oxygen (pO₂): 160 mmHgPartial pressure of argon (pAr): 7 mmHgPartial pressure of the trace gas (pt): 1 mmHgStep 2: Calculate the atmospheric pressure
Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.
P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg
1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en la siguiente reacción: CH3NH2 + O2 → CO2 + H2O + N2 Si reaccionan 0,5 mol de metil amina (CH3NH2) con 25,6 g de O2. Determine: a) Balancee la ecuación. (2 ptos) b) ¿Cuántos gramos de nitrógeno (N2) se pueden producir? (4 ptos) c) Si experimentalmente se obtuvieron 3,5 gramos de N2. Determine el porcentaje de rendimiento de la reacción. (4 ptos) Por favor es urgente!!!
Answer:
a) 4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂
b) m = 5,043 g
c) % = 69,4 %
Explanation:
a) La ecuación balanceada es la siguiente:
4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂
En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.
b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:
[tex]n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles[/tex]
[tex]n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles[/tex]
De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.
Ahora, podemos calcular la masa de nitrógeno producida:
[tex]n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles[/tex]
[tex]m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g[/tex]
Por lo tanto, se pueden producir 5,043 g de nitrógeno.
c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:
[tex] \% = \frac{R_{r}}{R_{T}}*100 [/tex]
Donde:
[tex]R_{r}[/tex]: es el rendimiento real
[tex]R_{T}[/tex]: es el rendimiento teórico
[tex]\% = \frac{3,5}{5,043}*100 = 69,4[/tex]
Entonces, el procentaje de rendimiento de la reacción es 69,4%.
Espero que te sea de utilidad!
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.
Which of the following elements is in the same family as fluorine?
a. silicon
b. antimony
O c. iodine
O d. arsenic
e. None of these.
Answer:
c iodine
Explanation:
fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride
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
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
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
The SNArreaction requires 1 mmol methyl 4-fluoro-3-nitrobenzoateand 2 mmol of 4-chlorobenzyl amine. Calculate the mass (mg) of both reagents.(2
Answer: Mass of methyl 4-fluoro-3-nitrobenzoate = 199 mg;
Mass of 4-chlorobenzylamine = 282 mg
Explanation: The mass and mol of a molecule is related by its molar mass, which is given in g/mol.
The molar mass of methyl 4-fluoro-3-nitrobenzoate, which has molecular formula: [tex]C_{8}H_{6}FNO_{4}[/tex] is:
[tex]C_{8}H_{6}FNO_{4}[/tex] = 12.8 + 6.1 + 19 + 14 + 16.4 = 199 g/mol
Since it is asking in mg: MM = 199.10³mg/mol
For 4-chlorobenzylamine, with molecular formula [tex]C_{7}H_{8}ClN[/tex]:
[tex]C_{7}H_{8}ClN[/tex] = 12.7 + 8.1 + 35 + 14 = 141 g/mol
In mg: MM = 141.10³mg/mol
The reaction requires 1 mmol of [tex]C_{8}H_{6}FNO_{4}[/tex] , then its mass is:
m = 1.10⁻³ mol * 199.10³mg/mol = 199 mg
For [tex]C_{7}H_{8}ClN[/tex], it requires 2mmol:
m = 2.10⁻³ mol * 141.10³ mg/mol = 282 mg
For the SNAr reaction, it is necessary 199 mg of methyl 4-fluoro-3-nitrobenzoate and 282 mg of 4-chlorobenzylamine
The mass of 1 mmol methyl 4-fluoro-3-nitrobenzoate is 199.14 mg and the mass of 2 mmol of 4-chlorobenzyl amine is 283.2 mg
To determine the masses of both reagents,
First we will determine their molar masses
For methyl 4-fluoro-3-nitrobenzoate (C₈H₆FNO₄)Molar mass = 199.14 g/mol
Now, for the mass of 1 mmol of methyl 4-fluoro-3-nitrobenzoate
Using the formula
Mass = Number of moles × Molar mass
Mass = 1 mmol × 199.14 g/mol
Mass = 199.14 mg
For 4-chlorobenzyl amine (C₇H₈ClN)Molar mass = 141.6 g/mol
Now, for the mass of 2 mmol of 4-chlorobenzyl amine
Mass = 2 mmol × 141.6 g/mol
Mass = 283.2 mg
Hence, the mass of 1 mmol methyl 4-fluoro-3-nitrobenzoate is 199.14 mg and the mass of 2 mmol of 4-chlorobenzyl amine is 283.2 mg
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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!
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.
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 = 29.6Enough 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 .
Nitric acid is often manufactured from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming of natural gas, in a two-step process. In the first step, nitrogen and hydrogen react to form ammonia:
Answer:
[tex]N_2~+~3H_2~+~4O_2~->~2HNO_3~+~2H_2O[/tex]
Explanation:
We can start with the reaction of hydrogen and nitrogen to produce ammonia, so:
[tex]N_2~+~H_2~->~NH_3[/tex]
When we balance the reaction we will obtain:
[tex]N_2~+~3H_2~->~2NH_3[/tex]
Now, the production of nitric acid with oxygen would be:
[tex]NH_3~+~O_2~->~HNO_3~+~H_2O[/tex]
If we balance the reaction we will obtain:
[tex]NH_3~+~2O_2~->~HNO_3~+~H_2O[/tex]
Now, if we put the reactions together we will obtain:
[tex]N_2~+~3H_2~->~2NH_3[/tex]
[tex]NH_3~+~2O_2~->~HNO_3~+~H_2O[/tex]
We can multiply the second reaction by "2":
[tex]N_2~+~3H_2~->~2NH_3[/tex]
[tex]2NH_3~+~4O_2~->~2HNO_3~+~2H_2O[/tex]
We have "[tex]2NH_3[/tex]" on both sides. In the first reaction is in the right in the second reaction is on the left. Therefore we can cancel out this compound and we will obtain:
[tex]N_2~+~3H_2~+~4O_2~->~2HNO_3~+~2H_2O[/tex]
On this reaction, we will have 2 nitrogen atoms on both sides, 6 hydrogen atoms on both sides, and 8 oxygen atoms on both sides. So, this would be the net reaction for the production of nitric acid.
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+