Answer:
since 1 mole = 6.022×10^23
hence in 2.32 moles no. of atoms= 6.022×10^23 × 2.32
=
13.97104 ×10^23
plz mark it as brainliest
How many moles are in 18.2 g of CO2?
41.4 moles
801 moles
0.414 moles
0 2.42 moles
Answer:
0.414 mole (3 sig. figs.)
Explanation:
Given grams, moles = mass/formula weight
moles in 18.2g CO₂(g) = 18.2g/44g/mole = 0.413636364 mole (calc. ans.)
≅ 0.414 mole (3 sig. figs.)
Answer this please t
Lol
Answer: trial b
Explanation:
Which best illustrates the way in which radiation transfers thermal energy?
O
Warr
Cool
o
Warm
Cool
Warm
Cool
Warm
H11
Cool
Answer:
It is so because heat is flowing from hot body to cold body, and there is no direct contact between the body. It explains correctly the mode of transmission of thermal energy through the process of radiations.
Explanation:
Fun fact:
How does thermal energy transfer by radiation?
Radiation. All objects transfer energy to their surroundings by infrared radiation . The hotter an object is, the more infrared radiation it gives off. No particles are involved in radiation, unlike conduction.
17. Which of the following is a device that generates electricity using a chemical reaction?
O A. Fuel cell
B. Battery
C. Charging station
O D. Solar panel
Answer:
Hydrogen and fuel cell technologies power cars, buildings and more. But how ... Test your knowledge with this quiz! ... How do fuel cells generate electricity
Answer:
A
Explanation:
fuel cell is a device that converts the chemical energy from fuel into electricity via a chemical reaction with oxygen or another oxidizing agent. Batteries work in a closed system, while fuel cells require their reactants to be replenished.
Place the following in order of increasing molar entropy at 298 K.
a. C3H8 < SO < CO2
b. CO2 < C3H8 < SO
c. C3H8 < CO2 < SO
d. SO < CO2 < C3H8
e. CO2 < SO < C3H8
Answer:
SO < CO2 < C3H8
Explanation:
Entropy refers to the degree of disorderliness of a system. The standard molar entropy of a substance refers to the entropy of 1 mole of the substance vunder standard conditions.
The molar entropy depends on the number of microstates in the system which in turn depends on the number of atoms in the molecule.
C3H8 has 11 atoms and hence the highest number of microstates followed by CO2 having three atoms and least of all SO having only two atoms.
7. Which shows a way to represent a single covalent bond between atoms?
Ο Η + Η
Ο H/H
Ο HH
Explanation:
A single covalent bond can be represented by a single line between the two atoms. For instance, the diatomic hydrogen molecule, H2, can be written as H—H to indicate the single covalent bond between the two hydrogen atoms.
Consider these two cases.
Case 1: An electron jumps from energy level 1 to energy level 2 in an atom.
Case 2: An electron jumps from energy level 1 to energy level 3 in an atom.
For case 1, what happens when an electron jumps from energy level 1 to energy level 3 in an atom?
A. A photon is absorbed by the atom.
B. A photon is emitted by the atom.
C. A proton is absorbed by the atom.
D. A proton is emitted by the atom.
Assuming that both cases describe Hydrogen-like atoms with one electron, for which case is more energy emitted or absorbed?
A. The energy is the same for both cases.
B. More energy is emitted or absorbed for case 2
C. It is impossible to tell.
D. More energy is emitted or absorbed for case 1
Answer:
A photon is absorbed by the atom.
More energy is emitted or absorbed for case 2
Explanation:
According to the Bohr model of the atom, electrons occur in energy levels. The energy of each level is fixed. However, electrons can absorb photons and move from a lower to higher energy level or emit photons and move from a higher to a lower energy level.
In each case, the energy absorbed or emitted is equal to the difference in energy between the two energy levels.
Since energy level 3 is much higher than energy level 2, the electron absorbs more energy in moving from energy level 1 to energy level 3 than it absorbs when moving from energy level 1 to energy level 2.
Which reaction produces an insoluble product?
A
2KI + Pb(NO )2 → → PbI, + 2KNO
B
2AgNO3 + Ca(CH,02)2 + Ca(NO3)2 + 2AgC,H,O,
C
3BaCl2 + ALLS
→ 2AlCl3 + 3BaS
.
D
SrBr, + Mg(OH)2 → MgBr, + Sr(OH)2
Answer: The correct option is A).
Explanation:
Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.
For the given chemical reactions:
A): [tex]2KI+Pb(NO_3)_2\rightarrow PbI_2+2KNO_3[/tex]
The iodide of lead is generally insoluble in water. Thus, lead iodide is a precipitate.
B): [tex]2AgNO_3+Ca(CH_3COO)_2\rightarrow Ca(NO_3)_2+2CH_3COOAg[/tex]
The nitrates and acetates of all metals are soluble in water.
C): [tex]3BaCl_2+Al_2S_3\rightarrow 2AlCl_3+3BaS[/tex]
The sulfide of barium is soluble in water.
D): [tex]SrBr_2+Mg(OH)_2\rightarrow MgBr_2+Sr(OH)_2[/tex]
The hydroxide of strontium is soluble in water.
Hence, the correct option is A).
g Suppose 0.0350 g M g is reacted with 10.00 mL of 6 M H C l to produce aqueous magnesium chloride and hydrogen gas. M g ( s ) + 2 H C l ( a q ) → M g C l 2 ( a q ) + H 2 ( g ) What is the limiting reactant in this reaction?
Answer:
Mg will be the limiting reagent.
Explanation:
The balanced reaction is:
Mg + 2 HCl → MgCl₂ + H₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
Mg: 1 moleHCl: 2 molesMgCl₂: 1 moleH₂: 1 moleBeing the molar mass of each compound:
Mg: 24.3 g/moleHCl: 36.45 g/moleMgCl₂: 95.2 g/moleH₂: 2 g/moleBy reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
Mg: 1 mole* 24.3 g/mole= 24.3 gHCl: 2 moles* 36.45 g/mole= 72.9 gMgCl₂: 1 mole* 95.2 g/mole= 95.2 gH₂: 1 mole* 2 g/mole= 2 g0.0350 g of Mg is reacted with 10.00 mL (equal to 0.01 L) of 6 M HCl.
Molarity being the number of moles of solute that are dissolved in a certain volume, expressed as:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
in units [tex]\frac{moles}{liter}[/tex]
then, the number of moles of HCl that react is:
[tex]6 M=\frac{number of moles of HCl}{0.01 L}[/tex]
number of moles of HCl= 6 M*0.01 L
number of moles of HCl= 0.06 moles
Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl react with 24.3 grams of Mg, 0.06 moles of HCl react with how much mass of Mg?
[tex]mass of Mg=\frac{0.06 moles of HCl* 24.3 grams of Mg}{2 moles of HCl}[/tex]
mass of Mg= 0.729 grams
But 0.729 grams of Mg are not available, 0.0350 grams are available. Since you have less mass than you need to react with 0.06 moles of HCl, Mg will be the limiting reagent.
The limiting reactant in the reaction is Magnesium (Mg)
From the question,
We are to determine the limiting reactant in the reaction.
The given balanced chemical equation for the reaction is
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
This means
1 mole of Mg is required to react completely with 2 moles of HCl
Now, we will determine the number of moles of each reactant present
For Magnesium (Mg)Mass = 0.0350 g
Using the formula
[tex]Number\ of\ moles = \frac{Mass}{Atomic\ mass}[/tex]
Atomic mass of Mg = 24.305 g/mol
∴ Number of moles of Mg present = [tex]\frac{0.0350}{24.305}[/tex]
Number of moles of Mg present = 0.00144 mole
For HClConcentration = 6M
Volume = 10.00 mL = 0.01 L
Using the formula
Number of moles = Concentration × Volume
∴ Number of moles HCl present = 6 × 0.01
Number of moles HCl present = 0.06 mole
Since,
1 mole of Mg is required to react completely with 2 moles of HCl
Then
0.00144 mole of Mg is required to react completely with 2×0.00144 mole of HCl
2×0.00144 = 0.00288
∴ The number of moles of HCl required to react completely with the Mg is 0.00288 mole
Since the number of moles of HCl present is more than 0.00288 mole, then HCl is the excess reactant and Mg is the limiting reactant.
Hence, the limiting reactant in the reaction is Magnesium (Mg)
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define reaction rate
Answer:
The reaction rate or rate of reaction is the speed at which a chemical reaction takes place, defined as proportional to the increase in the concentration of a product per unit time and to the decrease in the concentration of a reactant per unit time. Reaction rates can vary dramatically.
You pre-weigh a glass vial to hold your sample and find its mass to be 5.010 g. You add your sample to the vial and reweigh it on the same balance and find that the mass has increased to 6.130 g. What is the mass of the sample in grams
When we pre-weigh a glass vial to hold our sample and find its mass to be 5.010 g. Then we add our sample to the vial and reweigh it on the same balance and find that the mass has increased to 6.130 g. The mass of the sample in grams is 1.12 g.
What is mole concept?Avogadro's number is the number of units in one mole of any substance and equals to 6.02214076 × 10²³. The units can be electrons, atoms, ions, or molecules.
No. of moles is defined as a particular no. of particles that we can calculate with the help of Avogadro’s number.
Mass of a particular product is also find out by stoichiometry of a reaction as per the no. of mole given in the reaction.
Mass is generally can be represented by units like Kg, g etc.
Given,
weigh of glass vial = 5.010 g
weigh of glass vial with sample = 6.130 g
Therefore, When we pre-weigh a glass vial to hold our sample and find its mass to be 5.010 g. Then we add our sample to the vial and reweigh it on the same balance and find that the mass has increased to 6.130 g. The mass of the sample in grams is 1.12 g.
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At a given temperature, K = 1.3x10^-2 for the reaction:
N2(g) + 3H2(g) ⇌ 2NH3(g)
Calculate values of K for the following reactions at this temperature.
a. 1/2N2 + 3/2H2(g) ⇌ NH3(g)
b. 2NH3(g) ⇌ N2(g) + 3H2(g)
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)
Answer:
a) 0.11
b)76.9
c) 8.8
d) 1.7*10^-4
Explanation:
Step 1: Data given
K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 2: Formula of K
aA(g) + bB(g) ⇌ cC(g) + dD(g)
K = [C]^c *[D]^d / [A]^a * [B]^b
K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]
Step 3:
a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
1/2N2 + 3/2H2(g) ⇌ NH3(g) =>K' = [tex]\sqrt{K}[/tex]
K' = [tex]\sqrt{1.3*10^-2}[/tex] = 0.11
b. 2NH3(g) ⇌ N2(g) + 3H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
2NH3(g) ⇌ N2(g) + 3H2(g) =>K' = 1/K
K' = 1/(1.3*10^-2) = 76.9
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
=>K' = [tex]\frac{1}{\sqrt{K} }[/tex]
K' = [tex]\frac{1}{\sqrt{1.3*10^-2} }[/tex]
K' = 8.8
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
2N2(g) + 6H2(g) ⇌ 4NH3(g)
K' = K²
K' = (1.3*10^-2)²
K' = 1.7 *10 ^-4
Values of equilibrium constant at given temperature for the following reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.
How we calculate equilibrium constant?Equilibrium constant is define as the ration of the concentrations of product to the concentrations of reactant with respect to the exponent of their coefficients.
Given chemical reaction is:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Equilibrium constant for this reaction is:
K = [NH₃]² / [N₂][H₂]³
K = 1.3 × 10⁻² (given)
Equilibrium constant K₁ for below reaction will be written as:1/2N₂(g) + 3/2H₂(g) ⇌ NH₃(g)
K₁ = √K
Because concentration of all given species is 1/2 of the given reaction, so value of K₁ will be written as:
K₁ = √(1.3 × 10⁻²) = 0.11
2NH₃(g) ⇌ N₂(g) + 3H₂(g)
K₂ = 1/K
Because concentration of reactant and products are reciprocal from the concentration of original given reaction, so value of K₂ will be written as:
K₂ = 1/1.3 × 10⁻² = 76.9
NH₃(g) ⇌ 1/2N₂(g) + 3/2H₂(g)
K₃ = 1/√K
Because concentrations of given species is reciprocal as well as half of the given original reaction, so value of K₃ will be written as:
K₃ = 1/√(1.3 × 10⁻²) = 8.8
2N₂(g) + 6H₂(g) ⇌ 4NH₃(g)
K₄ = K²
Because concentrations of given species is double of the given original reaction, so value of K₄ will be written as:
K₄ = (1.3 × 10⁻²)² = 1.7 × 10⁻⁴
Hence, the value of K for given reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.
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Phosphorus pentachloride, PCl5, a white solid that has a pungent, unpleasant odor, is used as a catalyst for certain organic reactions. Calculate the number of moles in 38.7 g of PCl5.
Answer:
0.186 moles
Explanation:
In order to convert grams of PCl₅ into moles, we need to use its molar mass:
Molar Mass of PCl₅ = Molar mass of P + (Molar mass of Cl)*5Molar Mass of PCl₅ = 208.24 g/molThen we proceed to calculate the number of moles:
38.7 g ÷ 208.24 g/mol = 0.186 molThere are 0.186 moles of PCl₅ in 38.7 g of PCl₅.In the given question Phosphorus pentachloride is used as a catalyst for certain chemical reaction. 38.7 g of [tex]\rm PCl_5[/tex], there are 0.186 moles of [tex]\rm PCl_5[/tex].
A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy required for the reaction to occur.
To calculate the number of moles in 38.7 g of [tex]\rm PCl_5[/tex], we need to divide the given mass of [tex]\rm PCl_5[/tex] by its molar mass.
The molar mass of [tex]\rm PCl_5[/tex] can be calculated by adding the atomic masses of one phosphorus atom and five chlorine atoms:
Molar mass of [tex]\rm PCl_5[/tex] = (1 x atomic mass of P) + (5 x atomic mass of Cl)
= (1 x 30.97 g/mol) + (5 x 35.45 g/mol)
= 208.22 g/mol
Now, we can calculate the number of moles of [tex]\rm PCl_5[/tex]:
Number of moles = mass / molar mass
= 38.7 g / 208.22 g/mol
= 0.186 moles
Therefore, there are 0.186 moles of [tex]\rm PCl_5[/tex] in 38.7 g of [tex]\rm PCl_5[/tex].
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Question 6
Based on your observations, what conclusion can you draw about the lengths of AD, DB, AE, and EC?
Answer: The ratio of AD to DB is equal to the ratio of AE to EC. In other words, the pairs of lengths are proportional.
Explanation:
Sample answer from plato
Answer:
The pairs of lengths are proportional, because, the ratios of AD and DB are the exact same, so they are equal to the ratios of the set AE and EC.
Explanation:
It is rewritten from a sample answer from Plato just to be safe from plagiarism.
The equilibrium constant, Kc, for the following reaction is 77.5 at 600 K. CO(g) Cl2(g) COCl2(g) Calculate the equilibrium concentrations of reactant and products when 0.555 moles of CO and 0.555 moles of Cl2 are introduced into a 1.00 L vessel at 600 K.
Answer:
[CO] = 0.078M
[Cl2] = 0.078M
[COCl2] = 0.477M
Explanation:
Based on the reaction:
CO(g) Cl2(g) ⇄ COCl2(g)
Where equilibrium constant, kc, is:
kc = 77.5 = [COCl2] / [CO] [Cl2]
[] represents the equilibrium concentration of each gas. The initial concentration of each gas is:
[CO] = 0.555mol/1.00L = 0.555M
[Cl2] = 0.555M
And equilibrium concentrations are:
[CO] = 0.555M - x
[Cl2] = 0.555M - x
[COCl2] = x
Where x is reaction coordinate
Replacing in kc expression:
77.5 = [x] / [0.555M - x] [0.555M - x]
77.5 = x / 0.308025 - 1.11 x + x²
23.8719 - 86.025 x + 77.5 x² = x
23.8719 - 87.025 x + 77.5 x² = 0
x = 0.477M. Right answer
x = 0.646M. False answer. Produce negative concentrations
Replacing:
[CO] = 0.555M - 0.477M = 0.078M[Cl2] = 0.078M[COCl2] = 0.477MAnd those concentrations are the equilibrium concentrations
a weak acid undergoes _ ionization in water
Answer:
Partial
Explanation:
A strong acid will completely ionize in water while a weak acid will only partially ionize.
All of the different types of electromagnetic radiation (light, x-rays, ultraviolet
radiation, and so on) make up the
atomic spectrum
electromagnetic spectrum.
sunlight
spectral lines,
Answer:
bleh
Explanation:
does light appears to travel in straight lines. travelling from light sources until it hits the surface of an object?? (Truer or False) if your answer is true then what is the reason why does light appears to travel in straight lines??.
Answer:
true once light has been produce it will keep travelling jn straight parts until it hits something else
Why is there a huge diversity of substances?
A. Atoms can combine in many ways to form various compounds
and molecules.
B. Any element can combine with any other element.
C. Atoms break apart to form new types of atoms during chemical
reactions.
D. There is an unlimited number of elements.
A. Atoms can combine in many ways to form various compounds
and molecules.
1. When the following oxidation-reduction reaction in acidic solution is balanced, what is the
lowest whole-number coefficient for Rb*(aq)?
Rb(s) + Sr?+(aq) → Rb+ (aq) + Sr(s)
Consider the following reaction at 298 K.
2 SO2(g) + O2(g) → 2 SO3(g)
An equilibrium mixture contains O2(g) and SO3(g) at partial pressures of 0.43 atm and 2.6 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO2 in the mixture.
______atm.
Answer and Explanation:
The reaction is in the gas phase, so the equilibrium constant is expressed in terms of the partial pressures (P) of the products and reactants, as follows:
[tex]Kp = \frac{P^{2}_{SO_{3} } }{P_{SO_{2}} ^{2}P_{O_{2}} }[/tex]
We have the following data:
P(SO₃) = 2.6 atm
P(O₂) = 0.43 atm
We need Kp for this reaction. We can assume that in Appendix 4 we found that Kp = 7 x 10²⁴.
Then, we introduce the data in the equilibrium constant expression to calculate the partial pressure f SO₂ (PSO₂), as follows:
[tex]P_{SO_{2} } = \sqrt{\frac{P_{SO_{3} } ^{2} }{Kp P_{O_{2} } } } = \sqrt{\frac{(2.6 atm)^{2} }{(7 x 10^{24)}(0.43 atm) } } = 1.5 x 10^{-12} atm[/tex]
Therefore, the partial pressure of SO₂ is 1.5 x 10⁻¹² atm (for the given Kp).
name hydrogen ion
what the symbolotom
Answer:
H+
Explanation:
it's H+
as you see hydrogen ion it could H+
A hot ballon with mass of 400 kilograms moves across the aky with 3,200 joules of kinetic energy. The velocity of the ballon is
Answer:
4 m/s
Explanation:
formula is v = (KE/.5m)^1/2
there is a calculator
https://www.calculatorsoup.com/calculators/physics/kinetic.php
A 11.79 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 14.151 g . Add subscripts to correctly identify the empirical formula of the new oxide.
Answer:
MoO₃
Explanation:
To solve this question we must find the moles of molybdenum in Mo2O3. The moles of Mo remain constant in the new oxide. With the differences in masses we can find the mass of oxygen and its moles obtaining the empirical formula as follows:
Moles Mo2O3 -Molar mass: 239,878g/mol-
11.79g * (1mol / 239.878g) = 0.04915 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.09830 moles Mo
Mass Mo in the oxides:
0.09830 moles Mo * (95.95g/mol) = 9.432g Mo
Mass oxygen in the new oxide:
14.151g - 9.432g = 4.719g oxygen
Moles Oxygen:
4.719g oxygen * (1mol/16g) = 0.2949 moles O
The ratio of moles of O/Mo:
0.2949molO / 0.09830mol Mo = 3
That means there are 3 moles of oxygen per mole of Molybdenum and the empirical formula is:
MoO₃The pH of a solution with a hydrogen-ion concentration of 4.90 x 10-'Mi
Please answe I’ll give you brainliest
Answer:
pH < 7; pH = 7; pH > 7
Explanation:
when dealing with an acidic base, you have the formula for [H3O+] > [OH-] which yields a pH < 7.
when you have a neutral base, the reactive ion concentration would be [H3O^+] = [OH^-] which yields a pH = 7.
finally, when dealing with a basic classification, the formula would be [H3O^+] < [OH^-] yields a pH > 7.
Part B
[H3O+] = 2 x 10-6 M
Express your answer using one decimal place.
Answer:
pH = 5.7
Explanation:
Which is the pH of the solution?
The pH is a measurement widely used in chemistry in quality assurance of products and another analysis. Is defined as the -log [H3O+]. That means, the pH of the solution that is [H3O+] = 2x10-6 M is:
pH = -log [H3O+]
pH = -log [2x10-6 M]
pH = 5.7For each amino acid, the name, three-letter abbreviation, or one-letter abbreviation is given. Complete the missing information name: proline three-letter abbreviation: one-letter abbreviation: Select the class (side chain) for proline. name: three-letter abbreviation: Phe name: three-letter abbreviation: Phe one-letter abbreviation: Select the class (side chain) for Phe. name: three-letter abbreviation: name: three-letter abbreviation: one-letter abbreviation: D Select the class (side chain) for D. nathe: lysine three-letter abbreviation: latihan aidantului one-letter abbreviation: Select the class (side chain) for lysine. name: three-letter abbreviation: Gin one-letter abbreviation: Select the class (side chain) for Gln.
Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . What is the theoretical yield of sodium chloride formed from the reaction of 0.73g of hydrochloric acid and 1.3g of sodium hydroxide?
Round your answer to 2 significant figures.
i think its 2.0
Why do i think this-If you add 0.73g to 1.3g it comes to 2.0g
A beaker contains a 25 mL solution of an unknown monoprotic acid that reacts in a 1:1 stochiometric ratio with NaOH. Titrate the solution with NaOH to determine the concentration of the acid.Perform a titration by setting the concentration of the NaOH solution and adding it to the acid solution using the different Add Base buttons.The equivalence point of the titration is passed when the solution color changes.The unknown sample can be titrated multiple times by pressing the Retitrate button and starting over.Enter the concentration of the unknown acid solution.The base is 20.05 mL with 1.000 M
Answer:
0.80 M
Explanation:
Step 1: Write the generic neutralization reaction
HA + NaOH ⇒ NaA + H₂O
Step 2: Calculate the reacting moles of NaOH
20.05 mL of 1.000 M NaOH react.
0.02005 L × 1.000 mol/L = 0.02005 mol
Step 3: Calculate the reacting moles of HA
The molar ratio of NaOH to HA is 1:1. The reacting moles of HA is 1/1 × 0.02005 mol = 0.02005 mol.
Step 4: Calculate the concentration of HA
0.02005 moles of HA are in 25 mL.
[HA] = 0.02005 mol/0.025 L = 0.80 M
At 35°C, K = 1.6 × 10^-5 for the reaction
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
Calculate the concentrations of all species at equilibrium for each of the following original mixtures.
a. 2.0 mol pure NOCl in a 2.0 L flask
b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask
Answer:
a) [NOCl] = 0.968 M
[NO] = 0.032M
[Cl²] = 0.016M
b) [NOCl] = 1.992M
[NO] = 0.008 M
[Cl2] = 1.004 M
Explanation:
Step 1: Data given
Temperature = 35°C = 308K
K = 1.6 × 10^-5
Step 2: The reaction
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2
Step 3
a. 2.0 mol pure NOCl in a 2.0 L flask
Concentration at the start:
Concentration = mol / volume
[NOCl] = mol / volume
[NOCl] = 2.0 / 2.0 L
[NOCl] = 1.0 M
[NO] = 0 M
[Cl] = 0M
Concentration at the equillibrium
[NOCl] = 1.0M - 2x
[NO] = 2x
[Cl2]= x
K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5
1.6*10^-5 = ((2x)² * x) / (1.0-2x)²
x = 0.016
[NOCl] = 1.0 - 2*0.016 = 0.968 M
[NO] = 2*0.016 = 0.032M
[Cl²] = 0.016M
b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask
Concentration at the equillibrium
[NOCl] = 2.0 mol / 1.0 L = 2.0 M
[NO] = 0 M
[Cl2]= 1.0 mol / 1.0 L = 1.0 M
Concentration at the equillibrium
[NOCl] = 2.0M - 2x
[NO] = 2x
[Cl2]= 1.0 + x
K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5
1.6 *10^-5 = (2x)²*(1.0+x) / ((2.0-2x)²)
1.6 *10^-5= (2x)² * 1 )/2.0²
1.6 *10^-5= 4x² / 4 = x²
x = [tex]\sqrt{1.6 *10^-5}[/tex] = 4.0*10^-3
[NOCl] = 2.0 - 2*0.004 = 1.992M
[NO] = 2*0.004 = 0.008 M
[Cl2] = 1+ 0.004M = 1.004 M