The E⁰, ΔG⁰ and K for the reaction 2Cu(aq) ⇌ Cu²⁺(aq) + Cu(s) is 1.04 V, -200,630 J/mol and 1.108 x 10³⁵ repectively.
To calculate E⁰, ΔG⁰, and K for the given reaction at 25°C, we can use the Nernst equation and the relationship between ΔG⁰, K, and E⁰.
The Nernst equation relates the cell potential (E) to the standard cell potential (E₀) and the reaction quotient (Q) as follows:
E = E⁰ - (RT / nF)ln(Q)
Where:
E = Cell potential
E⁰ = Standard cell potential
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
n = Number of electrons transferred in the reaction
F = Faraday's constant (96485 C/mol)
ln = Natural logarithm
Q = Reaction quotient
The relationship between ΔG⁰, K, and E⁰ is given by:
ΔG⁰ = -nFE⁰
[tex]K = e^{(- \triangle G^0/ (RT))}[/tex]
Let's calculate the values:
Given:
E⁰ = 0.52 V for the reduction Cu(aq) + e⁻ → Cu(s)
1. E⁰ for the reaction 2Cu(aq) ⇌ Cu²⁺(aq) + Cu(s):
Since the reaction involves the transfer of 2 electrons, the E₀ for the reaction will be:
E₀ = 2(0.52 V)= 1.04 V
2. ΔG⁰:
ΔG⁰ = -nFE⁰
Since the reaction involves the transfer of 2 electrons, n = 2.
ΔG⁰ = -2(96485 C/mol)(1.04 V)
ΔG⁰ = -200,630 J/mol
3. K:
[tex]K = e^{(- \triangle G^0/ (RT))}[/tex]
T = 25°C = 298 K
[tex]K = e^{\frac {(-200,630 J/mol)}{(8.314 J/(molK)298 K)}}[/tex])
[tex]K \approx e^{(80.37)} \approx 1.108 \times 10^35[/tex]
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A 1 liter solution contains 0.487 M acetic acid and 0.365 M sodium acetate.
Addition of 0.183 moles of hydrobromic acid will:
(Assume that the volume does not change upon the addition of hydrobromic acid.)
Raise the pH slightly
Lower the pH slightly
Raise the pH by several units
Lower the pH by several units
Not change the pH
Exceed the buffer capacity
A 1 liter solution contains 0.474 M hydrocyanic acid and 0.356 M potassium cyanide.
Addition of 0.391 moles of nitric acid will:
(Assume that the volume does not change upon the addition of nitric acid.)
Raise the pH slightly
Lower the pH slightly
Raise the pH by several units
Lower the pH by several units
Not change the pH
Exceed the buffer capacity
Multiple choices can be selected for the answer, could you also explain how this is done?
The addition of 0.183 moles of hydrobromic acid to a 1 liter solution containing 0.487 M acetic acid and 0.365 M sodium acetate will lower the pH slightly. On the other hand, the addition of 0.391 moles of nitric acid to a 1 liter solution containing 0.474 M hydrocyanic acid and 0.356 M potassium cyanide will lower the pH by several units.
In the first case, the acetic acid and sodium acetate form a buffer solution, which helps maintain a relatively stable pH. The acetic acid acts as a weak acid, while the sodium acetate is its conjugate base. When a small amount of hydrobromic acid is added, it reacts with the acetate ions in the buffer system, shifting the equilibrium towards the formation of more acetic acid and water. This results in an increase in the concentration of hydronium ions (H3O+), leading to a slight decrease in pH.
In the second case, the hydrocyanic acid and potassium cyanide also form a buffer solution. However, nitric acid is a strong acid that completely ionizes in water. The addition of nitric acid increases the concentration of hydronium ions significantly, overpowering the buffer system and causing a substantial decrease in pH.
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Which types of chemical reactions are
considered opposites of one another?
Answer:
Hi
Explanation:
Decomposition reactions are really the opposite of combination reactions. In decomposition reactions, a single compound breaks down into two or more simpler substances (elements and/or compounds).
Explanation:
Decomposition chemical reactions
Decomposition reactions are really the opposite of combination reactions. In decomposition reactions, a single compound breaks down into two or more simpler substances (elements and/or compounds).
The decomposition of water into hydrogen and oxygen gases
and the decomposition of hydrogen peroxide t form oxygen gas and water
are examples of decomposition reactions.
Question 4 (2 points)
Manganese-56 decays by beta emission and has a half life of 2.6 hours. If a sample
is analyzed and found to contain 6.25% parent isotopes and 93.75% daughter
isotopes, what must the age of the sample be?
Answer:
10.4 hours
Explanation:
Now from;
0.693/t1/2 = 2.303/t log (No/N)
t1/2= half life of the radioactive isotope
t = age of the radioactive isotope
No= Original amount of radioactive isotope
N= amount of radioactive isotope left after time t
Substituting values;
N = 0.0625No
0.693/2.6 = 2.303/t log (No/0.0625No)
0.693/2.6 = 2.303/t log (1/0.0625)
0.267 = 2.303/t * log 16
0.267 = 2.773/t
t = 2.773/0.267
t = 10.4 hours
If the concentration of OH in a saturated solution of Ca(OH)₂ is 5.2 x 10-² M, is the solubility of Ca(OH)2 greater than 0.1g/100.mL? (Molar mass of Ca(OH)2 is 74.1 g/mol) a. Same b. No C. Yes
Comparing the calculated solubility of Ca(OH)₂ (1.924 g/100 mL) to the given value of 0.1 g/100 mL, we can conclude that the solubility of Ca(OH)₂ is greater than 0.1 g/100 mL.
Option (c) is correct.
To determine if the solubility of Ca(OH)₂ is greater than 0.1 g/100 mL based on the given concentration of OH⁻, we need to calculate the solubility of Ca(OH)₂ using the provided concentration.
The balanced chemical equation for the dissociation of Ca(OH)₂ is:
Ca(OH)₂ ⇌ Ca²⁺ + 2OH⁻
From the equation, we can see that for every mole of Ca(OH)₂ that dissolves, two moles of OH⁻ ions are produced. Therefore, the concentration of OH⁻ is twice the concentration of Ca(OH)₂.
Given that the concentration of OH⁻ is 5.2 x 10⁻² M, the concentration of Ca(OH)₂ can be calculated by dividing the concentration of OH⁻ by 2:
Ca(OH)₂ concentration = (5.2 x 10⁻² M) / 2 = 2.6 x 10⁻² M
To determine the solubility of Ca(OH)₂ in grams per 100 mL, we can use the molar mass of Ca(OH)₂:
Solubility of Ca(OH)₂ = (2.6 x 10⁻² M) x (74.1 g/mol) = 1.924 g/100 mL
Comparing the calculated solubility of Ca(OH)₂ (1.924 g/100 mL) to the given value of 0.1 g/100 mL, we can conclude that the solubility of Ca(OH)₂ is greater than 0.1 g/100 mL.
Therefore, the correct answer is c) Yes.
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What is the mass of sulfuric acid, h2so4, that is contained in 2.0L of a 5.85 M solution ?
A- 1147.44 grams
B- 0.12 grams
C- 11.7 grams
D- 98.07 grams
Answer:
A. 1147.44 grams
Explanation:
Multiply volume by molarity to get the moles of solution.
5.85 M = mol/2.0 L
mol = (5.85 M)(2.0 L)
mol = 11.7
There is 11.7 moles of H2SO4.
Convert to grams with molar mass.
11.7 mol H2SO4 x (98.076 g/1 mol) = 1147.49 g
Closest answer is A, 1147.44 g.
I need help on this question
Answer:
I only
last answer
Explanation:
The critical angle for a certain liquid-air surface is 47.7 degrees. What is the index of refraction of the liquid?
The critical angle is given by sin ic = n2 / n1, where n1 and n2 are the indices of refraction for the media.
Using the formula sin ic = n2 / n1 for the critical angle for a liquid-air surface, we can calculate the index of refraction of the liquid.47.7° = sinic = n2 / n1, where n1 is the index of refraction of air, which is 1.00. The index of refraction of the liquid is therefore 1.33.
When light travels from a medium with a high refractive index (n1) to a medium with a low refractive index (n2), total internal reflection occurs if the angle of incidence is greater than the critical angle (ic).
The critical angle (ic) can be calculated using the formula sin ic = n2 / n1, where n1 and n2 are the indices of refraction of the media. The angle of incidence that causes total internal reflection is the critical angle.
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4. Identify What do scientists call a substance that forms during a chemical reaction?
5. Identify On what side of the equation are reactants found?
Answer:
4. Reactant
5. Left
I hope this helps ❤️❤️
Experiment 4C- solvent extraction 2: A three component mixture; an example of the separation of an acid, a base, and a neutral substance. During experiment 4C, you collect the following data: component Ethyl-4-amino benzoate Benzoic Acid 9-fluorenone Initial mass (g) 0.0498 0.0588 0.0508 Recovered (g) 0.0137 0.0322 0.0422 Melting point (C) 84.0 121.5 82.0 Calculate the percent recovery for each component (basic, acidic, and neutral) a. Basic b. Acidic c. Neutral Calculate the percent error for the melting point of each component a. basic b. acidic c. neutral
To calculate the percent recovery for each component (basic, acidic, and neutral) and the percent error for the melting point of each component in Experiment 4C, specific formulas and calculations are required. The percent recovery is calculated by dividing the mass of the recovered component by the initial mass and multiplying by 100. The percent error for the melting point is calculated by comparing the experimental melting point to the accepted literature value and expressing the difference as a percentage of the accepted value.
To calculate the percent recovery for each component, you need to divide the mass of the recovered component by the initial mass and multiply by 100. Let's perform the calculations for each component:
a. Basic (Ethyl-4-amino benzoate):
Percent Recovery = (Recovered mass of basic component / Initial mass of basic component) x 100
Percent Recovery = (0.0137 g / 0.0498 g) x 100 = 27.51%
b. Acidic (Benzoic Acid):
Percent Recovery = (Recovered mass of acidic component / Initial mass of acidic component) x 100
Percent Recovery = (0.0322 g / 0.0588 g) x 100 = 54.76%
c. Neutral (9-fluorenone):
Percent Recovery = (Recovered mass of neutral component / Initial mass of neutral component) x 100
Percent Recovery = (0.0422 g / 0.0508 g) x 100 = 83.07%
To calculate the percent error for the melting point of each component, you need to compare the experimental melting point to the accepted literature value. The percent error is calculated using the formula:
Percent Error = ((Experimental melting point - Accepted melting point) / Accepted melting point) x 100
Let's perform the calculations for each component:
a. Basic (Ethyl-4-amino benzoate):
Percent Error = ((84.0°C - Accepted melting point) / Accepted melting point) x 100
b. Acidic (Benzoic Acid):
Percent Error = ((121.5°C - Accepted melting point) / Accepted melting point) x 100
c. Neutral (9-fluorenone):
Percent Error = ((82.0°C - Accepted melting point) / Accepted melting point) x 100
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When H+ forms a Bond with H2O to form the Hydronium ion H3 plus this bond is called a coordinate covalent bond because
Answer:
Because both the bonding electrons come from the oxygen atom. Explanation: A coordinate covalent bond is formed when both the bonding electrons are coming from the same atom
Explanation:
here is a more complex redox reaction involving the permanganate ion in acidic solution: 5no2− 6h 2mno4− → 5no3− 2mn2 3h2o
The balanced equation of the redox reaction is:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2O.
Redox reaction involving the permanganate ion in acidic solution:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2OThe balanced redox reaction that involves the permanganate ion in acidic solution is given by:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2OFor balancing the given reaction in acidic solution, we follow the given steps:Step 1: Writing the unbalanced reaction in the ionic form:Step 2: Separating the half-reactions for oxidation and reduction:Oxidation half-reaction: 5NO2− → 5NO3−Reduction half-reaction: 2MnO4− → 2Mn2+Step 3: Balancing the number of atoms of elements in half-reactionsBalancing oxidation half-reaction:5NO2− → 5NO3− + 10H+ + 2e-Balancing reduction half-reaction: 16H+ + 2MnO4− → 2Mn2+ + 8H2O + 5e-Step 4: Balancing the number of electrons lost and gained in each half-reaction by multiplying by a coefficient.Oxidation half-reaction: 5NO2− → 5NO3− + 10H+ + 2e-Balancing reduction half-reaction: 16H+ + 2MnO4− → 2Mn2+ + 8H2O + 5e-Now, the electrons lost and gained are balanced.Step 5: Balancing the number of hydrogen ions in each half-reaction:Oxidation half-reaction: 5NO2− + 10H+ → 5NO3− + 2e-Balancing reduction half-reaction: 2MnO4− + 16H+ → 2Mn2+ + 8H2O + 5e-Step 6: Balancing the number of oxygen atoms in each half-reaction by adding H2O molecules to the side that needs it. For acidic solution, add H+ ions to balance the oxygen atoms.Oxidation half-reaction: 5NO2− + 10H+ → 5NO3− + 2e-Balancing reduction half-reaction: 2MnO4− + 16H+ → 2Mn2+ + 8H2O + 5e-Now, the number of oxygen atoms is balanced.Step 7: Adding half-reactions, multiplying by integers (if necessary) to equalize the electrons in each half-reaction, and canceling species common to both sides.5NO2− + 2MnO4− + 16H+ → 5NO3− + 2Mn2+ + 8H2OFinally, the balanced equation of the redox reaction is:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2O.
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True or false? Lymph vessels pick up a little bit of fluid that leaves your blood vessels every time you heart pumps.
Answer:
Explanation:
true.
Answer:
True. why?
Because the fluid that is lost through the blood vessels is restored but with no central pump or heart, this device releases and collects fluid.
Calculate the number of moles in 7.04 g of lead (II) nitrate, Pb(NO3)2
Can Someone help me covert this?!
Answer:
moles Pb(NO₃)₂ ≅ 0.0213 mole (three sig. figs.)
Explanation:
converting...
grams to moles => divide by formula weight
moles to grams => multiply by formula weight
for this problem ...
moles Pb(NO₃)₂ = 7.04g / 331.2g/mol = 0.021256 (calculator answer)
≅ 0.0213 mole (three sig. figs.)
Mole measure the number of elementary entities of a given substance that are present in a given sample. Therefore, the number of moles in 7.04 g of lead (II) nitrate, Pb(NO₃)₂ is 0.0212moles.
What is mole?The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity or amount of substance. We know one mole of any element contains 6.022×10²³ atoms which is also called Avogadro number.
Mathematically,
mole =given mass ÷ molar mass
Molar mass of 1 mole of Pb(NO₃)₂ = 331.2g/mol
given mass of Pb(NO₃)₂=7.04g
Substituting the values in above formula we get
moles of Pb(NO₃)₂ = 7.04g / 331.2g/mol
moles of Pb(NO₃)₂ = 0.0212moles
Therefore, the number of moles in 7.04 g of lead (II) nitrate, Pb(NO₃)₂ is 0.0212moles.
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A bicycle tire holds 1.50 L of air at 5atm and 20.0 °C. How many moles of air is this?
If the average mass of air is 29.0 g/mol, what is the mass of air in the tire?
Answer:
9.05 g
Explanation:
PV=nRT
Use the ideal gas equation. Substitute values.
P = 5 atm
V = 1.50 L
n = ?
R (gas constant) = 0.08206 L-atm/mol-K
T = 20.0°C
*Always convert °C to K.
T = 20.0° + 273 = 293K
Substitute values.
(5 atm)(1.50 L) = n(0.08206 L-atm/mol-K)(293K)
n = (5 atm)(1.50 L) / (0.08206 L-atm/mol-K)(293K)
n = 0.3119335... mol
Convert to grams with the given average mass of air.
0.3119335... mol x (29.0 g/1 mol) = 9.05 g
Help pls I don’t know the answer
Answer:
The first one is correct.
Explanation: Eh Are you cheating on quiz?
Describe the acid-base behavior of amphiproti substances Question Choose the options below that are amphiprotic, Select all that apply: a. H2SO4 b. HS- c. PO43- d. NH3
The amphiprotic substances among the options provided are HS⁻ and NH₃.
The term "amphiprotic" refers to substances that can act as both acids and bases, depending on the reaction conditions.
a. H₂SO₄ (sulfuric acid) is not considered amphiprotic. It is a strong acid that readily donates a proton (H+) but does not accept protons in typical acid-base reactions.
b. HS⁻ (hydrogen sulfide ion) is amphiprotic. It can act as a base by accepting a proton to form H₂S (hydrogen sulfide), and it can also act as an acid by donating a proton to form S²⁻ (sulfide ion).
c. PO₄³⁻ (phosphate ion) is not considered amphiprotic. It is a polyprotic base that can donate multiple protons, but it does not accept protons in typical acid-base reactions.
d. NH₃ (ammonia) is amphiprotic. It acts as a base by accepting a proton to form NH₄⁺ (ammonium ion), and it can also act as an acid by donating a proton to form NH₂⁻ (amide ion).
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which set of elements is arranged in order of increasing electronegativity? A. cl < f < se < as B. se < f < as < cl C. f < cl < se < as D. as < se < cl < f
f < cl < se < as (fluorine, chlorine, selenium, arsenic), which arranges the elements in increasing order of electronegativity.
Hence, the correct option is C.
Electronegativity is a measure of an atom's ability to attract electrons towards itself when it is part of a chemical bond. It generally increases as you move across a period from left to right and decreases as you move down a group in the periodic table.
A. cl < f < se < as (chlorine, fluorine, selenium, arsenic):
Chlorine (Cl) has a higher electronegativity than fluorine (F), selenium (Se), and arsenic (As). Therefore, it is arranged correctly that chlorine has the highest electronegativity, followed by fluorine, selenium, and then arsenic.
B. se < f < as < cl (selenium, fluorine, arsenic, chlorine):
This arrangement is not in increasing order of electronegativity. Fluorine (F) has the highest electronegativity among these elements, followed by chlorine (Cl), then arsenic (As), and selenium (Se) has the lowest electronegativity.
C. f < cl < se < as (fluorine, chlorine, selenium, arsenic):
This arrangement is correct. Fluorine (F) has the highest electronegativity, followed by chlorine (Cl), then selenium (Se), and arsenic (As) has the lowest electronegativity.
D. as < se < cl < f (arsenic, selenium, chlorine, fluorine):
This arrangement is not in increasing order of electronegativity. Fluorine (F) has the highest electronegativity among these elements, followed by chlorine (Cl), then selenium (Se), and arsenic (As) has the lowest electronegativity.
Therefore, the correct answer is option C: f < cl < se < as (fluorine, chlorine, selenium, arsenic), which arranges the elements in increasing order of electronegativity.
Hence, the correct option is C.
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What is the process of old crust sinking back into a mantle?
What would it mean if a forecaster claimed, “There will be east winds of 25 to 30 mph
Answer:
there will be winds moving from 25miles per hour to 30 miles per hour towards the east
What gas gets in and what gas get out of the body?
Answer:
carbon dioxide and oxygen
Explanation:
Answer:
This process produces methane and hydrogen
Explanation:
Which of the following explanations accounts for the fact that the ion-solvent interaction is greater for Li+ than for K+?
The ionization energy of Li is higher than that for K.
Li has a lower density than K.
Li+ is of a lower mass than K+
Li+has a smaller ionic radius than K+
Li reacts with water more slowly than K.
The most appropriate explanation among the given options for the fact that the ion-solvent interaction is greater for Li⁺ than for K⁺ is that "Li⁺ has a smaller ionic radius than K⁺." Option D is correct.
Ionic radius refers to the size of an ion, specifically the distance between the nucleus and the outermost electron shell. Smaller ions tend to have stronger interactions with solvent molecules because the charge is more concentrated in a smaller space, leading to a higher electrostatic attraction between the ion and the surrounding solvent molecules.
In this case, Li⁺ has a smaller ionic radius compared to K⁺. As a result, Li⁺ has a stronger interaction with the solvent molecules, which leads to a greater ion-solvent interaction.
Hence, D. is the correct option.
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--The given question is incomplete, the complete question is
"Which of the following explanations accounts for the fact that the ion-solvent interaction is greater for Li⁺ than for K⁺? A) The ionization energy of Li is higher than that for K. B) Li has a lower density than K. C) Li⁺ is of a lower mass than K⁺ D) Li⁺ has a smaller ionic radius than K⁺ E) Li reacts with water more slowly than K."--
.(Part A)
Write balanced net ionic equation for the following reaction:
Fe(OH)3(s)+H2SO4(aq)→?
Express your answer as a chemical equation. Identify all of the phases in your answer.
(Part B)
Write balanced net ionic equation for the following reaction:
HClO3(aq)+Na(OH)(aq)→?
Note that HClO3 is a strong acid.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Part A: [tex]Fe(OH)_{3} (s)[/tex] + 3H+(aq) → [tex]Fe^3+(aq)[/tex]+ [tex]3H_{2}O(l)[/tex](solid [tex]Fe(OH)_{2}[/tex] reacts with aqueous H+ to form [tex]Fe^3+[/tex] ions and water). Part B: H+(aq) + [tex]ClO_{3}^-[/tex](aq) + Na+(aq) +[tex]OH^-(aq)[/tex]→ Na+(aq) + [tex]ClO_{3}^-(aq)[/tex] + [tex]H_{2}O(l)[/tex](H+ reacts with [tex]OH^-[/tex] to form water in the presence of Na+ and [tex]ClO_{3}^-[/tex] ions).
Part A:
The balanced net ionic equation for the reaction between solid [tex]Fe(OH)_{3}[/tex] and aqueous [tex]H_{2}SO_{4}[/tex]is:
[tex]Fe(OH)_{3} (s)[/tex] + 3H+(aq) + [tex]3SO_{4}^2-(aq)[/tex] → [tex]Fe^3+(aq)[/tex] + [tex]3SO_{4}^2-(aq)[/tex] +[tex]3H_{2}O(l)[/tex].
In this reaction, the [tex]Fe(OH)_{3}[/tex] solid reacts with the H+ ions from [tex]H_{2}SO_{4}[/tex]to form[tex]Fe^3+[/tex] ions and water molecules. The sulfate ions [tex](SO_{4}^2-)[/tex]are spectator ions and do not participate in the net reaction. The phases are indicated by (s) for solid, (aq) for aqueous, and (l) for liquid.
Part B:
The balanced net ionic equation for the reaction between [tex]HClO_{3}(aq)[/tex] and NaOH(aq) is:
H+(aq) + [tex]ClO_{3}^-(aq)[/tex]+ Na+(aq) + [tex]OH^-(aq)[/tex] → Na+(aq) + [tex]ClO_{3}^-(aq)[/tex] + [tex]H_{2}O(l)[/tex].
Here, the H+ ion from [tex]HClO_{3}[/tex] reacts with the [tex]OH^-[/tex] ion from NaOH to form water molecules. Sodium ion (Na+) and chlorate ion ([tex]ClO_{3}^-[/tex]) are spectator ions. The phases are indicated by (aq) for aqueous and (l) for liquid.
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consider the effect of a volume change on this reaction at equilibrium. 2h2s(g)⇌2h2(g) s2(g)
Predict the effect (shift right, shift left, or no effect) of increasing the reaction volume.
For the reaction 2H2S(g) ⇌ 2H2(g) + S2(g), the effect of a volume change on this reaction at equilibrium can be considered. It can be predicted whether increasing the reaction volume will cause the reaction to shift to the right, shift to the left, or have no effect on the reaction.
In this reaction, two moles of gaseous reactants are being converted into two moles of gaseous products. Therefore, this is a reaction that involves a change in volume. According to Le Chatelier's principle, the equilibrium will shift in such a way as to counteract the effect of the volume change.
If the reaction volume is increased by adding inert gas (i.e., a gas that doesn't react with the reactants), then the total pressure of the system will increase, but the partial pressures of the reactants and products will remain the same. Since the partial pressures remain unchanged, the equilibrium will not shift in either direction.
However, if the reaction volume is decreased by removing some of the gaseous products, then the partial pressure of the products will decrease. In order to counteract this decrease, the equilibrium will shift to the right to produce more gaseous products (i.e., more H2 and S2). Therefore, decreasing the volume will cause the reaction to shift to the right.
Therefore, increasing the reaction volume will have no effect on this reaction at equilibrium.
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Calculate the mass in grams of 8.35 × 10²² molecules of CBr₄.
To calculate the mass in grams of 8.35 × 10²² molecules of CBr₄ (carbon tetrabromide), we need to use Avogadro's number to convert the given number of molecules to moles and then use the molar mass of CBr₄ to convert moles to grams.
The molar mass of CBr₄ can be calculated by adding up the atomic masses of carbon (C) and four bromine (Br) atoms. The atomic mass of carbon is approximately 12.01 g/mol, and the atomic mass of bromine is approximately 79.90 g/mol.
Molar mass of CBr₄ = (1 × 12.01 g/mol) + (4 × 79.90 g/mol) = 331.74 g/mol
To convert the number of molecules to moles, we divide the given number of molecules by Avogadro's number (6.022 × 10²³ molecules/mol):
Moles of CBr₄ = (8.35 × 10²² molecules) / (6.022 × 10²³ molecules/mol) = 0.138 mol
Finally, to find the mass in grams, we multiply the number of moles by the molar mass:
Mass of CBr₄ = (0.138 mol) × (331.74 g/mol) = 45.80 g
Therefore, the mass in grams of 8.35 × 10²² molecules of CBr₄ is approximately 45.80 grams.
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How many grams (of mass mm) of glucose are in 195 mL of a 5.50% (m/v) glucose solution?Express your answer with the appropriate units
The mass (mm) of glucose in 195 mL of a 5.50% (m/v) glucose solution is 10.725 grams (g).
The mass (mm) of glucose in 195 mL of a 5.50% (m/v) glucose solution is 10.725 grams (g).Explanation:The formula used to calculate the mass (mm) of glucose in 195 mL of a 5.50% (m/v) glucose solution is given as follows:Firstly, we need to know the formula of the percentage by mass/volume:%(m/v) = (mass of solute / volume of solution) × 100where,(mass of solute / volume of solution) is the concentration of the solution (C) and can be represented as:C = (mass of solute / volume of solution)Now, using the above formula we can find out the mass (mm) of glucose in 195 mL of a 5.50% (m/v) glucose solution:Given,Volume of solution (V) = 195 mLConcentration of solution (C) = 5.50% (m/v)The concentration of solution (C) is 5.50% (m/v)So, it means in 100 mL of the glucose solution, 5.50 g of glucose is present. Hence, in 195 mL of glucose solution, the mass of glucose is given as:Mass of glucose = (mass/volume) × volumeMass of glucose = (5.50 g/100 mL) × 195 mL= 0.055 g/mL × 195 mL= 10.725 gTherefore, the mass (mm) of glucose in 195 mL of a 5.50% (m/v) glucose solution is 10.725 grams (g).
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To remove a tight-fitting lid from a jar, Megan runs the lid under hot water.
What happens to the Jar lid when its temperature increases?
A.The temperature increases, and the lid expands
B.The temperature decreases, and the lid contracts.
C.The potential energy increases, and the lid contracts.
D.The potential energy decreases, and the lid expands
Answer:
A
Explanation:
For practice, the molecular formula for ascorbic acid is C6H8O6, and you used 6 g in this experiment, calculate the molarity of the ascorbic acid. Now calculate the concentration in moles per drop (assume 1 mL = 20 drops).
I do not want the answer. I want to know how to get the answer please. Just explain to me how to figure this out.
Answer:
0.68 M
Explanation:
The concentration of a solute refers to Molarity= [tex]\frac{moles}{Liter}[/tex]
We need to find moles, we can do that by using stoichiometry to convert 6g > moles using molar mass.
Molar mass of C6H8O6= 176 g/mol; this is our key to converting the amount we have to moles.
6g C6H8O6 x [tex]\frac{1 mol}{176 g}[/tex]C6H8O6 = 0.034 mol C6H8O6
Now we need our volume per drop.
We know 1000 mL=1 L
We have 20 drops per 1 mL, so 1 mL / 20 = 0.05 mL per drop. But we need Liters, so we'll convert it to Liters.
0.05 mL x [tex]\frac{1 L}{1000mL}[/tex] = 0.00005L
Now we can solve for Molarity.
Molarity of C6H8O6 = [tex]\frac{0.034 mols}{0.00005 L}[/tex] = 0.068M of C6H8O6
help? for me? yes? please?
Answer:
2 Mg + O2 -> 2 MgO
is a synthesis reaction.
A force caused by objects moving in opposite directions called
Answer: Balanced forces
Explanation: Equal forces acting in opposite directions are called balanced forces. Balanced forces acting on an object will not change the object's motion.
fire heats up a pot → pot cooks food → food fuels body → body releases heat during activity where is energy created in this energy chain?
In the given energy chain, the energy is created in the fire when it heats up the pot. When the pot heats up, it cooks the food that fuels the body. As the food fuels the body, the body releases heat during activity. Therefore, the energy is created in the initial stage of the chain, i.e., when the fire heats up the pot.
The given energy chain is as follows:
Fire heats up a pot → Pot cooks food → Food fuels body → Body releases heat during activity
In the given energy chain, the energy is created in the first step when the fire heats up the pot. As the fire heats up the pot, the pot becomes hot and cooks the food that is in it. This cooking of food releases energy, which can be used by the body.The energy is then transferred to the body as it consumes the cooked food. The food fuels the body and provides it with the necessary energy to carry out activities. As the body uses this energy, it also releases heat during activity, which can be measured in the form of sweat or an increase in body temperature.Therefore, the energy is created in the fire heats up the pot step of the energy chain.
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