Answer:
A) [tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex].
B) [tex]Kc=0.0933[/tex].
C) 0.9 mol.
D) Increasing both temperature and pressure.
Explanation:
Hello,
In this case, given the information, we proceed as follows:
A)
[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]
B) For the calculation of Kc, we rate the equilibrium expression:
[tex]Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]
Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent [tex]x[/tex], we have:
[tex][NH_3]=0.6M=2*x[/tex]
[tex]x=\frac{0.6M}{2}=0.3M[/tex]
Next, the concentrations of nitrogen and hydrogen at equilibrium are:
[tex][N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M[/tex]
[tex][H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M[/tex]
Therefore, the equilibrium constant is:
[tex]Kc=\frac{(0.6M)^2}{(0.7M)*(1.77M)^3}\\ \\Kc=0.0933[/tex]
C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.
D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.
Best regards.
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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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
How many moles of HNO3 will be produced 3 NO2+H2O=2HNO3+ NO
Answer:
2 moles of HNO3
Explanation:
The equation seems to be balanced correctly. The problem is we done know what you started with. We will assume it is 3 moles of NO2.
If that is the case then 2 moles of HNO3 will be produced.
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.
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
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+
Prepare 10.00 mL of 0.010 M NaOH by diluting the NaOH solution used in Trial 1. (Your procedure should clearly show your calculations and the glassware used to perform the dilution.)
Answer:
= 0.2 mL.
Explanation:
Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:
[tex]C_{1} V_{1} = C_{2} V_{2}[/tex]
where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.
Let C1 = 0.5M, V2 = ?
C2 = 0.010M; V2 = 10mL
⇒Volume of stock solution to be diluted, V2
= [tex]\frac{10}{0.5}[/tex] × 0.010
= 0.2 mL.
Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.
I hope this was a detailed explanation given the missing details of "Trial 1" in the question.
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.
. 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.
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 = 29.6Drag each image to the correct location on the model. Each image can be used more than once. Apply the rules and principles of electron configuration to draw the orbital diagram of aluminum. Use the periodic table to help you.
Answer:
The answer to your question is given below.
Explanation:
Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:
Al (13) —› 1s² 2s²2p⁶ 3s²3p¹
The orbital diagram is shown on the attached photo.
Answer: screen shot
Explanation:
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
Stearic acid (C18H36O2) is a fatty acid, a molecule with a long hydrocarbon chain and an organic acid group (COOH) at the end. It is used to make cosmetics, ointments, soaps, and candles and is found in animal tissue as part of many saturated fats. In fact, when you eat meat, you are ingesting some fats containing stearic acid. ( of C18H36O2 = –948 kJ/mol, CO2=-393.5kJ/mol, H2O=-241.826kJ/mol).
Calculate the heat (q) released in kcal when 2.831 g of stearic acid is burned completely.
The molar mass of stearic acid is 18*AC+36*AH+2*AO=18*12+36*1+16*2=284g/mol.
n=m/M=2.831/284=0.01 moles
C18H36O2+27O2-->18CO2+18H2O
we have 18*0.01=0.18 moles of CO2
18*0.01=0.18 moles of H2O
0.01*948=9.48kJ from stearic acid
0.18*393.5=70.83kJ from CO2
0.18*241.826=43.52kJ from H2O
9.48+70.83+43.52=123.83kJ
123.83*4.184=518.10kcal
If the H+ concentration is 0.00001 M, what is the OH- concentration?
Answer:
1.00x10^-9
Explanation:
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.
What is an ideal gas?
Answer:
a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
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 .
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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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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A diode has IS = 10−17 A and n = 1.05. (a) What is the diode voltage if the diode current is 70 μA? (b) What is the diode current for VD = 0.1 mV?
Answer:
(a) The diode voltage, [tex]V_D =[/tex] 0.776 V
(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A
Explanation:
Given;
saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A
nonideality factor, n = 1.05
(a) the diode voltage
Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A
Diode voltage is calculated as;
[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]
Where;
[tex]V_T[/tex] is thermal voltage at 25°C = 0.025
[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]
b) the diode current for VD = 0.1 mV
[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]
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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Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use the method of successive approximations in your calculations or the quadratic formula.
Answer:
[tex][H^+]=0.000285[/tex]
[tex]pH=3.55[/tex]
Explanation:
In this, we can with the ionization equation for the hydrazoic acid ([tex]HN_3[/tex]). So:
[tex]HN_3~<->~H^+~+~N_3^-[/tex]
Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:
[tex]Ka=\frac{[H^+][N_3^-]}{[HN_3]}[/tex]
For each mol of [tex]H^+[/tex] produced we will have 1 mol of [tex]N_3^-[/tex]. So, we can use "X" for the unknown values and replace in the Ka equation:
[tex]Ka=\frac{X*X}{[HN_3]}[/tex]
Additionally, we have to keep in mind that [tex]HN_3[/tex] is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:
[tex]Ka=\frac{X*X}{0.004-X}[/tex]
Finally, we can put the ka value and solve for "X":
[tex]2.2X10^-^5=\frac{X*X}{0.004-X}[/tex]
[tex]2.2X10^-^5=\frac{X^2}{0.004-X}[/tex]
[tex]X= 0.000285[/tex]
So, we have a concentration of 0.000285 for [tex]H^+[/tex]. With this in mind, we can calculate the pH value:
[tex]pH=-Log[H^+]=-Log[0.000285]=3.55[/tex]
I hope it helps!
The [H+] and pH of a 0.0040 M hydrazoic acid solution is 0.000296648 and 3.527759
pH based problem:What information do we have?
Hydrazoic acid solution = 0.0040 M
Ka of hydrazoic acid = 2.20 × 10⁻⁵
We know that weak acids
[H+] = √( Ka × C)
[H+] = √( 2.2 × 10⁻⁵ × 0.0040)
[H+] = 0.000296648
So,
pH = -log [H+]
pH = -log [0.000296648]
Using log calculator
pH = 3.527759
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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
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!
Using appropriate chemical reactions for illustration, show how calcium present as the dissolved bicarbonate salt in water is easier to remove than other forms of hardness such dissolved Calcium chloride
Answer:
Explanation:
Calcium bicarbonate dissolved in hard water can easily be removed by heating the hard water . On heating , it decomposes to give calcium carbonate which is insoluble and therefore can be filtered out .
Ca( HCO₃)₂ = CaCO₃ + CO₂ + H₂O.
In this way hardness of water is removed .
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.
Applied Exercises (40 points) Answer the following questions in complete sentences. 1) In molecules with the same number of electron groups but different molecular geometries, discuss what happens to the bond angle? 2) What happens to the bond angle as you increase the number of bonding groups? 3) In 5 electron group molecules, what is the difference between axial and equatorial positions? Which groups are removed as lone pairs are added? 4) What is the difference between tetrahedral bent and
Answer:
See explanation
Explanation:
In molecules with the same number of electron groups but different molecular geometries, the bond angles differ markedly owing to the presence of lone pairs on the central atom. Recall that lone pairs of electrons take up more space around the central atom and causes more repulsion thus squeezing the bond angle and making it less than the value expected on the basis of the Valence Shell Electron Pair Repulsion Theory.
As the number of bonding groups increases, the bond angle increases since the repulsion due to lone pairs of electrons is being progressively removed by increase in the number of bonding groups.
For 5 electron group molecules, the axial groups are oriented at a bond angle of 90° while the equatorial groups are oriented at a bond angle of 120°. In the presence of lone pairs, the equatorial bonds are removed because the equatorial bonds often have a greater bond length than the axial bonds.
In the tetrahedral geometry, four groups are bonded to the central atom while in a bent molecular geometry, only two groups are bonded to the central atom with two lone pairs present in the molecule.
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!
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.
Be sure to answer all parts. Arrange the following substances in order of increasing strength of intermolecular forces. Click in the answer box to open the symbol palette.
a. H2
b. Ne
c. O2
d. NH3
Answer:
H2<Ne<O2<NH3
Explanation:
Intermolecular forces refer to the force of attraction between molecules of a substance in any given state of matter whether solid, liquid or gas. Molecules in a substance must be held together by intermolecular forces of attraction. The magnitude of these intermolecular forces of attraction depends on many factors.
For H2, He and O2, the intermolecular force present in these gases are London forces. As the relative molecular mass of individual gas molecules becomes greater, London forces increases significantly with molecular mass. This explains the sequence shown in the answer.
NH3 has the strongest intermolecular interaction because it contains hydrogen bonds since nitrogen is an electronegative element. This a greater intermolecular interaction than dispersion forces.
Intermolecular forces are those forces that bind the molecules of the substance and the polarity of molecules. These forces range from the strongest to the weakest in ion-dipole, hydrogen bonding, and dipole to dipole.
H2 being a noble gas has a weak dispersion or a weak dipole force. Ne has an intermolecular force being a noble gas it increases the molecular weight and thus has a modest increase of dipole bounding. The O2 has a strong dipole force than Ne and is stronger due to the two Oxygen molecules. The NH3 has the strongest dipole and intermolecular interaction force. The nitrogen atom strongly pulls the electrons.Hence the form the strongest to the weakest is NH3, 02, Ne and H2.
Learn more about the substances in order of increasing the strength of intermolecular forces.
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