The equilibrium concentrations of reactant and products when 0.388 moles of PCl5(g) are introduced into a 1.00 L vessel at 500 K is 0.063 M.
Equilibrium concentration can be defined as when the both the reactants and the products of the reaction are in a concentration that is not in change with time. This is called as the state of chemical equilibrium. In the equilibrium state, the rate of forward reaction is same as the rate of backward reaction when the reaction is at equilibrium. Equilibrium concentration being the ratio of the concentrations raise to the stoichiometric coefficients.
Kc = [PCl3][Cl2]/[PCl5] = 1.2x10-2
Kc = [PCl3][Cl2]/[PCl5] = (x)(x)/(0.388-x) = 1.20x10-2 =
x2 + 1.2x10-2x - 4.66x10-3 = 0
x = -b ± √b2 - 4ac2a
x = 0.0626
[PCl5] = 0.388 - x = 0.388 - 0.0626 = 0.325 M
[PCl3] = [Cl2] = x = 0.063 M
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The complete question is,
Calculate the equilibrium concentrations of reactant and products when 0.388 moles of PCl5(g) are introduced into a 1.00 L vessel at 500 K.
which of the following statements correctly describe standard electrode potentials? in what way must half-reactions and/or electrode potentials be manipulated when writing a balanced equation for a redox reaction? multiple select question. by convention, standard electrode potentials are quoted as reduction potentials. the half-reaction for the anode must be reversed when writing the balanced equation for the overall reaction. the sign for the anode potential must be reversed in order to to use the equation ecell
To accomplish this, the half-reaction with the smallest number of electrons may be multiplied. When writing the balanced equation, the half-reaction for the anode should be reversed to account for the oxidation occurring at the anode. Finally, to use the equation ecell, the sign of the anode potential must be reversed.
When answering questions on the Brainly platform, it is important to be factually accurate, professional, and friendly. One should be concise and avoid providing extraneous amounts of detail. Typos and irrelevant parts of the question should be ignored. The answer to the given question is as follows:By convention, standard electrode potentials are quoted as reduction potentials. The half-reaction for the anode must be reversed when writing the balanced equation for the overall reaction. The sign for the anode potential must be reversed to use the equation ecell.Standard electrode potentials are measured for half-reactions in their standard states, such as solutions of 1 mol/L and gases at a pressure of 1 atm. It indicates the ability of a half-reaction to accept electrons, with the half-reaction with the greatest reduction potential being the strongest oxidizing agent.When writing a balanced equation for a redox reaction, half-reactions and/or electrode potentials must be manipulated in order to balance the number of electrons transferred. Since the electrons must cancel out in the overall reaction, one half-reaction should be multiplied to match the number of electrons in the other half-reaction.
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If a 454.4 g sample of chlorine gas (MM = 71.0 g/mol) was
reacted with excess hydrogen at 565 K and 2.30 atm, how
many grams of hydrogen chloride gas (MM = 36.5 g/mol) are
produced?
Answer:
Isabelle is taking a survey to find the most popular music group of students in her community. Which of these is not a way for her to get a representative sample of this information?
ask every tenth student she sees ac a concert
ask every fifth student entering her school in the morning
ask every third student she encounters at the mall
ask every student at a local movie theater
Explanation:
estimate the maximum conversion of ethylene to ethanol by vapor-phase hydration at 523.15 k and 35 bars for an initial steam-to-ethylene ratio of 5. at these conditions, the fugacity coefficients of ethylene, ethanol, and water are 0.977, 0.827, and 0.887 respectively. a) write the balanced chemical reaction and specify the stoichiometric coefficient of all the species. b) write the mol fraction of each species at any time in terms of x (molar extent of reaction), assuming that initially there are 1 mol of ethylene, 5 mol of steam, and no ethanol. c) calculate the equilibrium constant, k, under these conditions. d) calculate the equilibrium conversion of ethylene to ethanol.
The equilibrium conversion of ethylene to ethanol is:
x = 0.581 or 58.1% (rounded to one decimal place)
The balanced chemical reaction is as follows:
C2H4 + H2O → C2H5OH
The stoichiometric coefficient of ethylene is 1, and the stoichiometric coefficient of water is 1.
The stoichiometric coefficient of ethanol is also
1.b)The mole fraction of ethylene is given by (1-x)/6.
The mole fraction of steam is given by (5-3x)/6.
The mole fraction of ethanol is given by x/6.c)The expression for the equilibrium constant, K is given by the following formula:
K = yethanol / (yethylene * ywater)
K = (x/6) / [(1-x)/6 * (5-x)/6]
K = x / [(1-x) * (5-x)]d)
The equilibrium conversion of ethylene to ethanol is given by the following formula:x = K / (1+K)At the given conditions of 523.15 K and 35 bars, the value of K is 1.389. Therefore, the equilibrium conversion of ethylene to ethanol is:
x = 1.389 / (1+1.389)
x = 0.581 or 58.1% (rounded to one decimal place)
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which factor has more impact on the strength of an acid, the polarity of the bond or the length of the bond
Both polarity and bond length have an impact on the strength of an acid. But bond length has an effect on limited number of acids. So while comparing polarity has more impact on strength of the acid.
Strength of an acid is determined by how easily the acid dissociates and ionizes in water. Dissociation of acid is governed by the following factors.
1. Size of the atom
2. Electronegativity differences
3. Charge on the acid
4. Oxidation state of the central atom.
As the size increases, the bond length increases. The bond becomes weaker and dissociates easily. So the acid become stronger. This trend is usually seen in hydro halides. HCl will be a stronger acid than HF. But the bond length determines the strength of an acid only to a certain extend.
As difference in electronegativity increases, the polarity between atoms increases. So dissociation will become easier. So higher the polarity, greater the acid strength. In almost all acids, polarity determines its strength.
So polarity has more impact than bond length.
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Our atmosphere is made up of 78% N2, 21% O2, and 1% other gases. What is the partial pressure of N2 when atmospheric pressure is 0.980 atm?
Answer: 0.7644 atm
Explanation:
Given,
Total atmospheric pressure= 0.980atm
Percentage of N₂=78%=0.78
Partial Pressure of N₂=Total atmospheric pressure*Percentage of N₂
=0.980 atm × 0.78
=0.7644 atm
you are asked to write your name on a suitable surface, using a piece of chalk that is pure calcium carbonate, caco3. how could you calculate the number of carbon atoms in your signature?
You can calculate the molar mass of CaCO3, which is equal to the sum of the atomic masses of calcium, carbon, and three oxygen atoms.
The atomic mass of calcium is 40.078 g/mol, carbon is 12.011 g/mol, and oxygen is 15.999 g/mol.
So, the molar mass of CaCO3 is:
Molar mass of CaCO3 = (1 × 40.078 g/mol) + (1 × 12.011 g/mol) + (3 × 15.999 g/mol)
= 100.086 g/mol
The molar mass of CaCO3 can also be calculated using the atomic weights of the elements, which are found on the periodic table. Once you know the molar mass of CaCO3, you can calculate the number of moles of CaCO3 used to write your name, based on the mass of chalk used.
Method 2:You can use the Avogadro constant to convert the number of moles of CaCO3 used to write your name into the number of formula units of CaCO3. Since each formula unit of CaCO3 contains one carbon atom, you can then determine the number of carbon atoms in your signature.
Method 3:Alternatively, you can use the stoichiometry of the reaction that occurs when CaCO3 is used to write on a surface. When CaCO3 is used to write on a surface, it reacts with carbon dioxide (CO2) in the air to form calcium oxide (CaO) and carbon dioxide (CO2).
The balanced chemical equation for this reaction is:
CaCO3(s) + CO2(g) → CaO(s) + CO2(g) + Heat
From this equation, you can see that each formula unit of CaCO3 reacts with one molecule of CO2 to produce one carbon atom in the form of CO2. Therefore, the number of carbon atoms in your signature is equal to the number of molecules of CO2 produced during the reaction. To calculate the number of molecules of CO2 produced, you need to know the mass of CaCO3 used to write your name and the volume of CO2 produced. The volume of CO2 can be measured using a gas syringe or a gas collection method. Once you know the volume of CO2, you can convert it to moles of CO2 using the ideal gas law, and then to molecules of CO2 using Avogadro's number.
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Which feature of a molecule can be determined from its NMR spectrum?
Among the features that can be determined from its NMR spectrum are: number of distinct proton environments, relative number of protons in each environment, chemical shift values for each environment, and splitting patterns.
The proton (1H) nuclear magnetic resonance (NMR) spectrum of a molecule helps to identify several features of that molecule.
The number of distinct proton environments is equivalent to the number of different kinds of protons in the molecule. For example, in a molecule with three different types of protons, such as [tex]CH3CH2OH[/tex] (ethanol), there will be three separate peaks in the spectrum.
The relative number of protons in each environment, which is proportional to the area under each peak, provides information on the molecule's composition.
The chemical shift values for each environment are a measure of the strength of the magnetic field felt by the protons in that environment. The magnetic field strength is influenced by the neighboring atoms and functional groups.
Therefore, the chemical shift can be used to infer information about the molecule's electronic structure and functional groups. The splitting patterns in a peak provide information about the number of neighboring protons. This helps to infer the connectivity of the molecule.
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what is the enthalpy change when a 6.00 g-sample of table sugar (c12h22o11) is oxidized? c12h22o11(s) 12 o2(g) 12 co2(g) 11 h2o(l) h
The enthalpy change for the combustion of 6.00 g of table sugar is -268,228 kJ/mol.
The enthalpy change for the combustion of one mole of table sugar (C12H22O11) can be calculated using the standard enthalpies of the formation of the reactants and products.
The balanced chemical equation for the combustion of table sugar is:
C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)
The standard enthalpies of formation of C12H22O11(s), CO2(g), and H2O(l) are -1274.9 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively. The standard enthalpy of the formation of O2(g) is 0 kJ/mol.
To calculate the enthalpy change for the combustion of 6.00 g of table sugar, we need to convert the mass to moles:
moles of C12H22O11 = 6.00 g / 342.3 g/mol = 0.0175 mol
Using the stoichiometric coefficients in the balanced equation, we can determine that 12 moles of O2 are required to completely react with 1 mole of C12H22O11. Therefore, the number of moles of O2 required to react with 0.0175 mol of C12H22O11 is:
moles of O2 = 12 × 0.0175 mol = 0.21 mol
The enthalpy change for the combustion of 0.21 mol of O2 can be calculated using the standard enthalpies of formation:
ΔH = (12 × -393.5 kJ/mol) + (11 × -285.8 kJ/mol) - (-1274.9 kJ/mol) + (0 kJ/mol)
ΔH = -4694.4 kJ/mol
To calculate the enthalpy change for the combustion of 6.00 g of table sugar, we need to divide by the number of moles of C12H22O11:
ΔH = -4694.4 kJ/mol / 0.0175 mol
ΔH = -268,228 kJ/mol
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Answer the following question in the attachment please for homework that is due tomorrow
Answer:
Details on attachment.
Explanation:
See attached worksheet.
compound has a molar mass of and the following composition: element mass % phosphorus 43.64% oxygen 56.36% write the molecular formula of .
The molecular formula is 2(P2O5), or P4O10.
To determine the molecular formula of the compound, we first need to find the empirical formula. We can assume 100 g of the compound, which means that there are 43.64 g of phosphorus and 56.36 g of oxygen.
We can convert the masses of each element to moles by dividing by their respective atomic masses:
moles of P = 43.64 g / 30.97 g/mol = 1.41 mol
moles of O = 56.36 g / 16.00 g/mol = 3.52 mol
Next, we can divide each number of moles by the smallest number to get the mole ratio:
P:O = 1.41 mol / 1.41 mol = 1
O:O = 3.52 mol / 1.41 mol = 2.49
We can round the mole ratio to the nearest whole number to get the empirical formula: P2O5
To find the molecular formula, we need to know the molar mass of the compound. Let's assume it is 284 g/mol (a multiple of the empirical formula mass of 142 g/mol).
We can divide the molar mass by the empirical formula mass to get the integer multiple:
n = 284 g/mol / 142 g/mol = 2
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which salt is produced by the neutralization of hydrobromic acid with magnesium hydroxide? group of answer choices mgbr2 mg3br2 mg2br mg2br3 mgbr
The salt produced by the neutralization of hydrobromic acid (HBr) with magnesium hydroxide (Mg(OH)₂) is magnesium bromide (MgBr₂).
During a neutralization reaction, an acid and a base react to form a salt and water. In this case, hydrobromic acid is the acid, and magnesium hydroxide is the base. The balanced chemical equation for this reaction is:
HBr + Mg(OH)₂ → MgBr₂ + 2H₂O
In this equation, the H+ ions from hydrobromic acid and the OH- ions from magnesium hydroxide combine to form water (H₂O), while the Mg²+ ions from magnesium hydroxide and the Br- ions from hydrobromic acid combine to form magnesium bromide (MgBr₂).
To determine the correct formula for the resulting salt, it is essential to consider the charges of the ions involved. Magnesium (Mg) has a charge of +2, and bromide (Br) has a charge of -1. To form a neutral compound, the charges must balance, which is why the formula for magnesium bromide is MgBr₂, with two bromide ions to balance the +2 charge of magnesium.
Thus, the correct answer from the given choices is MgBr₂, magnesium bromide.
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if 2.00 mol of carbon dioxide and 1.5 mol of hydrogen are placed in a5.00 l vessel and equilibrium is established, what will be the concentration of carbonmonoxide?7)
The concentration of carbon monoxide is also [tex]x = 0.0371\ \mathrm{M}$.[/tex]The balanced chemical equation for the reaction is:
[tex]$\mathrm{CO_2 + 4H_2 \rightleftharpoons CH_4 + 2H_2O}$[/tex]
The equilibrium expression for the reaction is:
[tex]$K_c = \dfrac{[CH_4][H_2O]^2}{[CO_2][H_2]^4}$[/tex]
At equilibrium, let the concentration of CO2 be $x$,, the concentration of CH4 be $y$, and the concentration of H2 be $z$.
Initial concentrations:
[tex]$[CO_2] = 2.00\ \mathrm{mol}/5.00\ \mathrm{L} = 0.400\ \mathrm{M}$[/tex]
[tex]$[H_2] = 1.50\ \mathrm{mol}/5.00\ \mathrm{L} = 0.300\ \mathrm{M}$[/tex]
[tex][CH_4] = 0\ \mathrm{M}$ (initially)[/tex]
[tex][H_2O] = 0\ \mathrm{M}$ (initially)[/tex]
At equilibrium, we know that:
[tex]y = [CH_4] = 2x$[/tex]
[tex]2y = [H_2O]$[/tex]
[tex]z = [H_2] - 4y = 0.300 - 4(2x) = 0.300 - 8x$[/tex]
Substituting these expressions into the equilibrium expression and solving for $x$:
[tex]K_c = \dfrac{(2x)(2y)^2}{x(0.300 - 8x)^4} = 3.80$[/tex]
Solving this equation gives:
[tex]x = 0.0371\ \mathrm{M}$[/tex]
Therefore, the concentration of carbon monoxide is also [tex]x = 0.0371\ \mathrm{M}$.[/tex]
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Equilibrium is established in a reversible reaction when:
the [product] = [reactants]
product is no longer produced
rate of reaction of products = rate of reaction of reactants
all the reactants dissolve or dissociate
A reversible chemical reaction is said to be in equilibrium when the forward and reverse reactions happen at the same rate and there is no total change in the concentrations of the reactants and products.
The concentrations of the reactants and products achieve a steady state at equilibrium, where the rates of the forward and reverse reactions are equal.
Although the concentration of the reactants and products is constant once the equilibrium is reached, this does not indicate that all of the reactants have been converted to products or that the production of products has ceased.
The concentrations of the reactants and products are no longer changing over time because the forward and backward reaction rates have instead equaled out.
In other words, when a system reaches equilibrium, it is in a dynamic state where both forward and backward reactions are still taking place but the concentrations of the reactants and products are constant. This indicates that the system is in equilibrium and the ratio of reactant to product concentrations is stable.
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Answer:
rate of reaction of products = rate of reaction of reactants
Explanation:
i got it right
What is the balanced equation for
Solid calcium fluoride decomposes to form calcium metal and fluorine gas?
The balanced equation for the given reaction is:
CaF2(s) → Ca(s) + F2(g)
This equation indicates that one molecule of calcium fluoride (CaF2) decomposes into one molecule of calcium (Ca) and one molecule of fluorine gas (F2). The equation is balanced because the number of atoms of each element is equal on both sides of the equation.
a solution is prepared by adding 0.10 mol of potassium chloride, kcl, to 1.00 l of water. which statement about the solution is correct?
The solution has a concentration of 0.10 M, is neutral with a pH of 7, and is electrically neutral.
At the point when 0.10 mol of potassium chloride (KCl) is added to 1.00 L of water, an answer is shaped. This arrangement has a grouping of 0.10 M, and that really intends that there are 0.10 moles of KCl per liter of water.The arrangement is impartial, as KCl is a salt that separates totally in water, delivering equivalent measures of potassium particles (K+) and chloride particles (Cl-), neither of which have acidic or essential properties. Subsequently, the pH of the arrangement is 7, which is unbiased.
This fixation is otherwise called the molarity of the arrangement. Furthermore, the arrangement is electrically nonpartisan, as the positive charges from the potassium particles balance out the negative charges from the chloride particles.
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What is a correct statement about the solution prepared by adding 0.10 mol of potassium chloride, KCl, to 1.00 L of water?
when the pressure of helium gas is increased to 4.1 atm, the volume is reduce to 1.2 l at a final temperature of 15 degrees celsius. what was the initial temperature of helium gas with a pressure of 3.5 atm and a volume 1.5 L?
Explanation:
Re-arrange the equation to :
T1 = P1V1 * T2 / (P2V2) and note that temp needs to be in K
15C = 288.15 K
T1 = 3.5 (1.5)(288.15) /(4.1 * 1.2) = 307.48 K which is 34.3 C
a student failed to notice a bubble in the tip of the burette before starting a titration. how does the bubble affect the final reading of the volume of naoh at the end point of the titration? biased high or biased low? explain your answer
The bubble in the tip of the burette will lead to an inaccurate final reading of the volume of NaOH at the end point of the titration.
The reading will be biassed high as a result of the bubble. This is so that, without really contributing to the measurement, the bubble will increase the volume of NaOH in the burette.
The bubble is not a component of the measurement, thus any volume it occupies will be added to the final value, making the reading higher than it should be.
As a result, the bubble will produce an unreliable and skewed high reading, which could result in false results.
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calculate the ph during the titration of 36.53 ml of 0.29 m hno3(aq) with 0.12 m naoh after 11.23 ml of the base have been added
During the titration of 36.53 ml of 0.29 m hno3(aq) with 0.12 m naoh after 11.23 ml of the base 0.71 of pH have been added.
During the titration of [tex]HNO_{3}[/tex] with NaOH, the reaction can be represented as:
[tex]HNO_{3}[/tex] + NaOH → [tex]NaNO_{3}[/tex] +[tex]H_{2}O[/tex]
To calculate the pH, first determine the moles of [tex]HNO_{3}[/tex] and NaOH.
moles of [tex]HNO_{3}[/tex] = volume (L) × concentration (M) = 0.03653 L × 0.29 M = 0.01059 mol
moles of NaOH = 0.01123 L × 0.12 M = 0.001348 mol
Now, find the moles of [tex]HNO_{3}[/tex] remaining after reaction with NaOH:
moles of [tex]HNO_{3}[/tex] remaining = 0.01059 mol - 0.001348 mol = 0.009242 mol
Calculate the new concentration of[tex]HNO_{3}[/tex]:
concentration of [tex]HNO_{3}[/tex] = moles of [tex]HNO_{3}[/tex] remaining / total volume
total volume = initial volume of [tex]HNO_{3}[/tex]+ volume of NaOH added = 0.03653 L + 0.01123 L = 0.04776 L
concentration of [tex]HNO_{3}[/tex] = 0.009242 mol / 0.04776 L = 0.1935 M
Finally, calculate the pH using the formula:
pH = -log[H+] = -log([[tex]HNO_{3}[/tex]]) = -log(0.1935) ≈ 0.71
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when the balancing of the equation for the reaction, taking place in acidic media is properly completed, what is the sum of all the coefficients in the equation?
The sum of all the coefficients of the reaction MnO⁴⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻ after balancing is found to be 16.
Here, is the unbalanced equation ('skeleton equation') of the chemical reaction. All reactants and products must be known. For a better result write the reaction in ionic form.
MnO⁴⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻
Firstly, we balanced the not oxygen and hydrogen compounds by adding the required coefficient in from the the respective element of the reaction in the acidic medium.
Now, then we balance the oxygen atom by adding the water molecules on the opposite side of the required side of the reaction.
It is to be noted that we are using H₂O molecule and OH⁻ for balancing hydrogen molecules because it is in the acidic medium.
Now, the final reaction is,
2MnO₄⁻ + 3Mn²⁺ + 4OH⁻ → 5MnO₂ + 2H₂O
Now, the sum of all coefficients is 16.
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Complete question - when the balancing of the equation for the reaction,
MnO⁴⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻ taking place in acidic media is properly completed, what is the sum of all the coefficients in the equation?
g when 15.5g of lead nitrate reacts with 15.5g of potassium iodide, 17.2g of lead iodide is produced. what is the percent yield?
The percent yield of the reaction is 79.6%.
To calculate the percent yield of the reaction, we need to first calculate the theoretical yield of lead iodide based on the given amount of lead nitrate;
1 mol Pb(NO₃)₂ = 331.2 g
15.5 g Pb(NO₃)₂ = 15.5/331.2 mol Pb(NO₃)₂ = 0.0469 mol Pb(NO₃)₂
According to the balanced chemical equation, 1 mole of Pb(NO₃)₂reacts with 2 moles of KI to produce 1 mole of PbI2. Therefore,
0.0469 mol Pb(NO₃)₂ x (1 mol PbI2/1 mol Pb(NO₃)₂) = 0.0469 mol PbI₂ (theoretical yield)
The molar mass of PbI₂ is 461.0 g/mol, so the theoretical yield of PbI₂ in grams is;
0.0469 mol PbI₂ x 461.0 g/mol = 21.6 g PbI₂
Now we can calculate the percent yield:
% yield=(actual yield/theoretical yield) x 100
The actual yield is given as 17.2 g PbI₂, so
% yield = (17.2 g/21.6 g) x 100
= 79.6%
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--The given question is incomplete, the complete question is
"When 15.5g of lead nitrate reacts with 15.5g of potassium iodide, 17.2g of lead iodide is produced. What is the percent yield? Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)"--
How many moles of O2 form when 1.0 mole of KCIO3 decomposes?
The balanced chemical equation for the decomposition of KCIO3 is:
2 KClO3 → 2 KCl + 3 O2
From the equation, it can be seen that for every 2 moles of KCIO3 that decompose, 3 moles of O2 are formed. Therefore, to determine the number of moles of O2 formed when 1.0 mole of KCIO3 decomposes, we need to use the mole ratio of KCIO3 to O2.
The mole ratio of KCIO3 to O2 is 2:3 (from the balanced chemical equation), which means that for every 2 moles of KCIO3 that decompose, 3 moles of O2 are formed. Therefore, to find the number of moles of O2 formed when 1.0 mole of KCIO3 decomposes, we can use the following proportion:
2 moles KCIO3 / 3 moles O2 = 1 mole KCIO3 / x moles O2
Solving for x, we get:
x = (3 moles O2)(1 mole KCIO3) / (2 moles KCIO3) = 1.5 moles O2
Therefore, when 1.0 mole of KCIO3 decomposes, 1.5 moles of O2 are formed.
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What is required to calculate the equilibrium constant of a chemical reaction?
Responses
the concentration of products and reactants
the change in entropy of the reaction
the energy of the reaction
the change in temperature of the reaction
Answer:
The concentration of products and reactants is required to calculate the equilibrium constant of a chemical reaction.
you perform an electrochemical reaction in which 0.800 mol of cu are reduced to solid cu . how many coulombs of charge are transferred?
The number of coulombs of charge transferred is 32220 C
When you perform an electrochemical reaction in which 0.800 mol of Cu are reduced to solid Cu, the number of coulombs of charge transferred can be calculated using Faraday's constant. The answer is 32220 C (coulombs).Explanation:
Given: The amount of Cu reduced to solid Cu = 0.800 mol
The amount of charge transferred can be calculated using Faraday's constant.
Faraday's constant = 96500 C mol^-1
Amount of charge transferred = n x FWhere,
n = Number of moles of electrons transferred.
F = Faraday's constant.Number of moles of electrons transferred
= 2 [Since the Cu ion gains 2 electrons to form Cu]Amount of charge transferred
= 2 x 96500 C mol^-1
= 193000 C [Or 1 F = 96500 C]Amount of charge transferred when 0.800 mol of Cu is reduced to solid Cu
= 193000 x 0.800 = 154400 C or 1.54 x 10^5 C (Approximately).
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which energy difference in the energy profile below corresponds to the activation energy for the forward reaction?
The activation energy of the forward reaction is represented by the energy difference between X and Y*, and the activation energy of the backward reaction is represented by the energy difference between Y and Y*. So, for the forward reaction, the correct response is X-Y*, and for the opposite reaction, it is Y-Y*.
The height from the valley to the apex serves as a visual cue in an energy profile graphic to indicate the activation energy. Based on the need, the valley may hold a reagent or a product.
Energy ––– A chain of reactions The activation energy of the component is represented by the red line, while the activation energy of the product is represented by the blue line.
As a result, the reactant's activation energy is x and the product's activation energy is x+y.
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The three mixtures below (a, b, and c) were prepared from three very narrow molar mass distribution of polystyrene samples with molar masses of 10000; 30000; and 100000 g mol−1
the number-average and weight-average molar masses for each are 47273 g/mol and 78872 g/mol.
To calculate the number-average and weight-average molar masses of each mixture, we need to use the following formulas:
Number-average molar mass (Mn) = (ΣNiMi) / ΣNi
Weight-average molar mass (Mw) = (ΣWiMi) / ΣWi
where Ni and Wi are the number and weight fractions of each component i, and Mi is the molar mass of component i.
(a) For the first mixture, where equal numbers of molecules of each sample were mixed, the number and weight fractions of each component are 1/3. The number-average molar mass is:
[tex]Mn = (\frac{1}{3} *10000) + (\frac{1}{3} * 30000)+(\frac{1}{3} *100000)= 46667g/mol.[/tex]
The weight-average molar mass is:
The number-average molar mass represents the average molar mass of the polymer chains in terms of their numbers, while the weight-average molar mass takes into account the relative abundance of each molar mass in terms of their weight. In this case, since the three samples have equal number and weight fractions, the number-average and weight-average molar masses are very close.
(b) For the second mixture, where equal masses of each sample were mixed, the number and weight fractions of each component are different due to their different molar masses. The number fraction of each component can be calculated as follows:
n1 = m1/M1 = 1/3
n2 = m2/M2 = 1/3
n3 = m3/M3 = 1/3
where mi is the mass of component i and Mi is the molar mass of component i.
The weight fraction of each component can be calculated as follows:
w1 = n1M1 / (n1M1 + n2M2 + n3M3) = 0.186
w2 = n2M2 / (n1M1+ n 2M2 + n3M3) = 0.294
w3 = n3M3 / (n1M1 + n2M2 + n3M3) = 0.520
where we have used the fact that the total mass of the mixture is the sum of the masses of each component, i.e. m1 + m2 + m3 = 1.
[tex]Me= (\frac{1}{3} *10000*10000)+(\frac{1}{3} *30000*30000)+(\frac{1}{3} *100000*100000)/ (\frac{1}{3} *10000) + (\frac{1}{3} * 30000)+(\frac{1}{3} *100000)= 47273 g/mol.[/tex]
Using these fractions, we can calculate the number-average and weight-average molar masses of the mixture:
[tex]Mn = (\frac{1}{3} *10000) + (\frac{1}{3} * 30000)+(\frac{1}{3} *100000)= 46667g/mol.[/tex]
Molecular weight= [tex](0.186 x 10,000 x 10,000) + (0.294 x 30,000 x 30,000) + (0.520 x 100,000 x 100,000) / (0.186 x 10,000) + (0.294 x 30,000) + (0.520 x 100,000) = 78,872 g mol-1[/tex]
In this case, the weight-average molar mass is significantly higher than the number-average molar mass. This indicates that the higher molar mass component (100,000 g mol-1) is contributing more to the weight of the mixture, while the lower molar mass components (10,000 and 30,000 g mol-1) are contributing more to the number of polymer chains
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What is the amount of heat required if 250.0 g of water is heated from 22.0 degrees C to 75.0 degrees C?
To calculate the amount of heat required to heat 250.0 g of water from 22.0 degrees C to 75.0 degrees C, we can use the formula:
Q = m × c × ΔT
where Q is the amount of heat, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature of the water.
The specific heat capacity of water is 4.184 J/(g·°C), which means that it takes 4.184 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
Substituting the given values, we get:
Q = 250.0 g × 4.184 J/(g·°C) × (75.0°C - 22.0°C)
Q = 250.0 g × 4.184 J/(g·°C) × 53.0°C
Q = 55,317.2 J or 55.32 kJ (to two decimal places)
Therefore, it requires 55.32 kJ of heat to raise the temperature of 250.0 g of water from 22.0°C to 75.0°C.
The solubility of a gas is 0.584 g/L at a pressure of 109 kPa. What is the solubility of the gas if the pressure is increased to 85 kPa, given that the temperature is held constant?
The solubility of the gas at a pressure of 85 kPa is 0.456 g/L.
What occurs to gas solubility as pressure increases?The relationship between pressure and a gas's solubility is straightforward. That is, it gets bigger as the strain gets bigger.
The combined gas law can be used to resolve this issue and says that:
(P1V1)/T1 = (P2V2)/T2
To solve for V2, which stands for the new volume at the reduced pressure, we can rearrange this equation as follows:
V2 = (P1V1T2)/(P2T1)
Since the temperature is held constant, T1 = T2, and this simplifies to:
V2 = (P1V1)/P2
Solubility2 = (Solubility1 x P2) / P1
where the solubility at greater pressure is denoted by Solubility1.
With numbers from the problem substituted, we obtain:
Solubility2 = (0.584 g/L x 85 kPa) / 109 kPa
Simplifying this expression, we get:
Solubility2 = 0.456 g/L.
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in a sample of 'a' found today, there are 128,000 atoms. if the halflife of 'a' is 5,000 years, in what year will the sample have 8,000 atoms?
In the year 20,002.5, the sample of 'a' will have 8,000 atoms, after passing through approximately 4 half-lives of 5,000 years each.
To solve the problem, we can use the equation for half-life:
T = (ln(N₀/N))/λ
where T is the half-life, N₀ is the initial number of atoms, N is the final number of atoms, and λ is the decay constant. Rearranging the equation to solve for N gives:
N = N₀e^(-λT)
We can use this equation to solve for the year when the sample will have 8,000 atoms.
Let's plug in the values we know:
N₀ = 128,000N = 8,000T = 5,000 years
λ = ln(2)/THalf-life (T) is 5,000 years.
Thus, decay constant λ is given by:
λ = ln(2)/T= ln(2)/5000= 0.00013862789.
Now we can plug in the values and solve for the year:N = N₀e^(-λT)8000 = 128,000e^(-0.00013862789T)
Divide both sides by 128,000:0.0625 = e^(-0.00013862789T)Take the natural logarithm of both sides:-
2.77259 = -0.00013862789TT = 20,002.5 years ago.
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mark the developed spot with pencil. calculate the rfvalues. determine the components of unknowns a and bpare these values to those reported in literature.what could bethe identity of sample b?
Based on the identity of the components present in the unknown sample and their properties, make an educated guess as to the identity of sample B.
In general, however, the steps involved in calculating Rf values and identifying unknown components in chromatography would be as follows:
Run the chromatography experiment using a known set of standards and the unknown sample.
Develop the chromatogram by visualizing the spots using UV light, ninhydrin spray, iodine vapor, or other suitable methods.
Mark the center of each spot with a pencil or other suitable marking tool.
Measure the distance traveled by each spot from the origin to the center of the spot (known as the "spot distance") and the distance traveled by the solvent front (known as the "solvent distance").
Calculate the Rf value of each spot using the formula Rf = spot distance / solvent distance.
Compare the Rf values of the unknown sample to those of the known standards and literature values to identify the components present in the sample.
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How many protons, neutrons, and electrons are in this ion?
Answer:
Ans C is the correct one.
As the element with 15 proton 15 electrons and 16 neutron is phosphorus