What is the pH of a solution the hydroide concentration of 1 X 10.5?
Answer:
Explanation:
Therefore, the pH of the HCl solution is 5.
The reaction of nitrogen and hydrogen to make ammonia is important because it provides fertilizer for growing food, and is because ammonia is needed to make other nitrogen-containing compounds. At room temperature, ∆G° and ∆H° for the reaction are both negative.
N2(g) + 3 H2(g) --> 2 NH3(g)
Which two of the following statements about this reaction are true?
- Adding an appropriate catalyst makes the reaction more spontaneous
- Increasing the temperature lowers the activation energy of the reaction
- Increasing the temperature makes the reaction less spontaneous
- The entropy change for the reaction is positive
- Increasing the pressure makes the reaction more spontaneous
The two statements that are true about the reaction are:
Adding an appropriate catalyst makes the reaction more spontaneous Increasing the temperature lowers the activation energy of the reactionWhat is a catalyst?A catalyst is described as a substance that speeds up a chemical reaction, or lowers the temperature or pressure needed to start one, without itself being consumed during the reaction.
If we happen to increase the temperature, it provides more kinetic energy to the reactant molecules and this makes it more likely to overcome the activation energy barrier and engage in the reaction.
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Complete and balance the following half-reaction in acidic solution. Be sure to include the proper phases for all species within the reaction.
BrO 3 - (aq) ----> Br- (aq)
The balanced equation of the half-reaction for the reduction of BrO3- to Br- in an acidic solution is:
BrO₃⁻ (aq) + 6 H⁺ + 6e- → Br- (aq) + 3 H₂O (l)What is the balanced equation of the half-reaction in an acidic solution?The balanced equation of the half-reaction in an acidic solution is determined as follows:
Unbalanced equation of the half-reaction: BrO₃⁻ (aq) → Br-
The oxidation number of Br changed from +5 to - 1
Hence, it gained six electrons and 6 electrons are added to the reactant side.
BrO₃⁻ (aq) + 6e- → Br- (aq)The oxygen atoms are balanced by adding water molecules to the right-hand side of the reaction while hydrogen ions are added to the left-hand since the reaction took place in acidic conditions.
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Which statements are true regarding the area of circles and sectors? Check all that apply.
The area of a circle depends on the length of the radius.
The area of a sector depends on the ratio of the central angle to the entire circle.
The area of a sector depends on pi.
The area of the entire circle can be used to find the area of a sector.
The area of a sector can be used to find the area of a circle
The area of a circle depends on the length of the radius, and the area of a sector depends on the ratio of the central angle to the entire circle, hence options A, B, D, E are correct.
A circle is the location of a point such that it is always a constant distance from a fixed point known as the center.
The statements true regarding the area of circles and sectors are:
The area of a circle depends on the length of the radius.
The area of a sector depends on the ratio of the central angle to the entire circle.
The area of the entire circle can be used to find the area of a sector.
The area of a sector can be used to find the area of a circle.
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Which of the following would give the largest cell potential (measured as an absolute value) when paired with a Ni2+/Ni electrode?
Mg2+/Mg
O2/H2O2
Cu2+/Cu
Al3+/Al
The cell potential of a galvanic cell is determined by the difference in the reduction potentials of the two half-cells involved. The larger the difference, the higher the cell potential. The half-reaction with the highest reduction potential will give the largest cell potential when paired with the Ni2+/Ni electrode.
When looking at the reduction potentials, Al3+/Al has a standard reduction potential of -1.66 V, whereas Ni2+/Ni has a standard reduction potential of -0.25 V. Therefore, the reaction with the highest reduction potential difference (i.e., the largest cell potential) when paired with the Ni2+/Ni electrode would be the one that has a reduction potential greater than -0.25 V.
Out of the options given, Al3+/Al has the highest reduction potential and thus it would give the largest cell potential when paired with the Ni2+/Ni electrode. This is because the reduction potential difference between Al3+/Al and Ni2+/Ni is 1.41 V, which is the largest among the given options.
In conclusion, the half-reaction that would give the largest cell potential when paired with a Ni2+/Ni electrode is Al3+/Al.
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i need help asap
A sample of tin goes through a temperature change of -160.56 °C while releasing 36298 joules of heat. The specific heat capacity of tin is 0.227 J/(g.°C). What is the mass of this sample?
A 13.66 mol sample of ammonia absorbs 33834 joules of heat. The specific heat capacity of ammonia is 80.08 J/(mol. °C). By how much did the temperature of this sample change, in degrees Celsius?
A sample of cobalt undergoes a temperature change of -1132.52 °C while releasing 455500 joules of heat. The specific heat capacity of cobalt is 0.4187 J/(g.°C). What is the mass of this sample?
A 372.4 g sample of indium goes through a temperature change of +140.73 K while absorbing
12505 joules of heat. What is the specific heat capacity of indium?
A 4.721 mol sample of molybdenum absorbs 35961 joules of heat. The specific heat capacity of molybdenum is 24.06 J/(mol-°C). By how much did the temperature of this sample change, in degrees Celsius?
A 56.2 g sample of ethanol is subjected to a temperature change of -110.56 K. The specific heat capacity of ethanol is 2.44 J/(g K). How many joules of heat were transferred by the sample?
A 5.774 mol sample of chromium absorbs 38674 joules of heat. The specific heat capacity of chromium is 23.35 J/(mol °C). By how much did the temperature of this sample change, in degrees Celsius?
A 4.9 mol sample of magnesium is subjected to a temperature change of -683.83 K. The specific heat capacity of magnesium is 24.9 J/(mol K). How many joules of heat were transferred by the sample?
A 0.2687 mol sample of tin is subjected to a temperature change of +222.48 K. The specific heat capacity of tin is 27.112 J/(mol K). How many joules of heat were transferred by the sample?
A 1.008 mol sample of neon undergoes a temperature change of -703.43 K while releasing
14738 joules of heat. What is the specific heat capacity of neon?
Answer:
To solve these problems, we can use the formula:
q = mcΔT
where q is the heat transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the temperature change.
The mass of the sample of tin can be calculated as:
q = mcΔT
36298 J = m × 0.227 J/(g.°C) × (-160.56 °C)
m = 708.2 g
The temperature change of the sample of ammonia can be calculated as:
q = mcΔT
33834 J = 13.66 mol × 80.08 J/(mol.°C) × ΔT
ΔT = 31.7 °C
The mass of the sample of cobalt can be calculated as:
q = mcΔT
455500 J = m × 0.4187 J/(g.°C) × (-1132.52 °C)
m = 27.4 g
The specific heat capacity of indium can be calculated as:
q = mcΔT
12505 J = 372.4 g × c × 140.73 K
c = 0.238 J/(g.°C)
The temperature change of the sample of molybdenum can be calculated as:
q = mcΔT
35961 J = 4.721 mol × 24.06 J/(mol.°C) × ΔT
ΔT = 31.9 °C
The heat transferred by the sample of ethanol can be calculated as:
q = mcΔT
q = 56.2 g × 2.44 J/(g K) × (-110.56 K)
q = -15,585 J
The temperature change of the sample of chromium can be calculated as:
q = mcΔT
38674 J = 5.774 mol × 23.35 J/(mol.°C) × ΔT
ΔT = 27.4 °C
The heat transferred by the sample of magnesium can be calculated as:
q = mcΔT
q = 1.008 mol × 24.9 J/(mol K) × (-683.83 K)
q = -17,134 J
The heat transferred by the sample of tin can be calculated as:
q = mcΔT
q = 0.2687 mol × 27.112 J/(mol K) × 222.48 K
q = 1676.7 J
The specific heat capacity of neon can be calculated as:
q = mcΔT
14738 J = 1.008 mol × c × (-703.43 K)
c = 36.8 J/(mol.°C)
Explanation:
Assume that the buret contains
H+ ions, the flask contains OH - ions, and each has a volume of
100 mL. How many milliliters would you need to add from the
buret to the flask to neutralize all the OH- ions in a titration procedure?[tex]The\ equation\ is:\\H^+(aq) + OH^-(aq)=H2O(l).[/tex]
Answer:
Explanation:
100
m
L
H
+
is required for complete neutralization.
Assuming, somehow, only
H
+
are in the buret and only
O
H
−
...
Use of diamonds based on chemical properties
It is incorporated with audio equipment to enhance sound quality. Diamonds are hard and easily vibrate at fast speeds, which results in high-quality sound. High-end recorders and DJ equipment both use it. Nanodiamonds have potential health benefits.
The carbon atoms in diamond are organised in a diamond cubic crystal lattice, making it an allotrope of carbon. Diamond is the material with the highest heat conductivity and hardness among all naturally occurring substances. Diamond is a particularly important component of industrial cutting and polishing equipment due to its unique qualities.
These days, diamond exfoliators and face products are accessible. Diamond dust is now being used in cosmetics, thus the price will undoubtedly be exorbitant. Diamond is used in beauty products to minimize wrinkles, etc.
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A sample of helium gas is collected over water . The gas collected is a wet gas that includes helium (He) and water vapor (H20). If the pressure of water vapor (H20) is 21.2 mmHg, what is the pressure of the helium gas (He) if the total pressure is 855 mmHg?
The pressure of the helium gas (He) in the wet gas mixture is 833.8 mmHg.
In order to find the pressure of the helium gas in the wet gas mixture, we need to use the concept of partial pressures. According to Dalton's Law of Partial Pressures, the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas present in the mixture. In this case, the wet gas mixture contains helium (He) and water vapor (H2O). The pressure of water vapor (H2O) is given as 21.2 mmHg, which means that the partial pressure of water vapor (H2O) in the mixture is 21.2 mmHg. We can now use the total pressure and the partial pressure of water vapor (H2O) to find the partial pressure of helium (He) in the mixture. To do this, we can subtract the partial pressure of water vapor (H2O) from the total pressure:
Partial pressure of helium (He) = Total pressure - Partial pressure of water vapor (H2O)
Partial pressure of helium (He) = 855 mmHg - 21.2 mmHg
Partial pressure of helium (He) = 833.8 mmHg
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The reaction
C4H8(g)⟶2C2H4(g)
has an activation energy of 262 kJ/mol.
At 600.0 K,
the rate constant, ,
is 6.1×10−8 s−1.
What is the value of the rate constant at 805.0 K?
I get 0.052739 and apparently it's wrong. Please work the problem out in great detail.
The rate constant for the reaction can be found out using Arrhenius equation.
Arrhenius equation can be stated as:
[tex]ln\frac{k2}{k1}=\frac{Ea}{R}[\frac{1}{T1}-\frac{1}{T2}][/tex]
i.e [tex]log\frac{k2}{k1} = \frac{Ea}{2.303R} [\frac{1}{T1}-\frac{1}{T2} ][/tex]
i.e [tex]log(k2)-log(k1) = \frac{Ea}{2.303R} [\frac{T2-T1}{T1xT2}][/tex]
From the given data, k1 = 6.1 s⁻¹, T1 = 600K, T2 = 805K, Ea = 262 kJ/mol and R = 8.314 J/molK
Substituting in the Arrhenius equation, we get
[tex]log\frac{k2}{6.1x10^-^8}= \frac{262}{2.303 x 8.314} [\frac{805-600}{600 x 805}][/tex]
[tex]log (k2) = log (k1) + \frac{Ea}{2.303R} [\frac{T2-T1}{T1xT2}][/tex]
[tex]log (k2)= log(6.1x10^-^8) + \frac{262}{2.303x8.314} x \frac{805-600}{600x805}[/tex]
[tex]log(k2)= log (6.1x10^-^8) + 5.81 x 10^-^3\\log(k2) = -7.214 + 0.00581\\log(k2) = -7.21[/tex]
[tex]k2 = antilog (-7.21) = 6.17 x 10^-^8[/tex]
Thus, on solving for k2, we get k2 = 6.17 × 10⁻⁸ s⁻¹
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if we are constantly taking in more and more c-14 why does it concentration in us not increase
help me solve it pls
The term molefraction is an important method which is used to calculate the concentration of a solution. It is mainly employed to calculate the concentration of a binary solution. Here the molefraction is 1 / 5. The correct option are D, D and B.
Molefraction of any component of a solution is defined as the ratio of the number of moles of that component to the total number of moles of the solution. The sum of molefraction of solute and solvent is one.
Here the molefraction of nitrogen = Moles of nitrogen / Total number of moles
1. 'x' of 'N' = 2 / 5 + 3 + 2 = 0.2 or 1 / 5
2. Molefraction of Argon = 0.60 / 0.40 + 0.04 + 0.60 = 0.57
Partial pressure = Molefraction × Total pressure
0.57 × 6.3 = 3.59 atm
3. 20 cm³ mixture contains:
20 × 1 mole / 22400 m³ = 8.9286 × 10⁻⁴
1 mole occupies 22400 cm³
8.9286 × 10⁻⁴ × 22400 cm³ / 1 mole = 20 cm³
Thus the correct option are D, D and B.
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1. Which of the following would you expect to rearrange? Why? (a) CH3 CH 3 CH3CHCHCH3 CH3 CH3 CH3CCH₂CH3 D CH3 E CH₂CH₂CHCH3 F
Among the given compounds, I would expect the compound F (CH₂CH₂CHCH₃) to rearrange. This expectation is based on the concept of hyperconjugation and the stability of carbocations.
In organic chemistry, carbocations are positively charged carbon species with only six electrons in their valence shell. The rearrangement of a compound occurs when a more stable carbocation can be formed through the shifting of atoms or groups. This rearrangement is driven by the desire to achieve greater stability.
In compound F, the carbon atom adjacent to the positively charged carbon (CH₃⁺) is a primary carbon (attached to only one other carbon atom), making it less stable. By undergoing a rearrangement, this carbon can shift its attachment to the adjacent carbon, creating a secondary carbocation (with two carbon attachments) and a more stable molecule.
The rearrangement would result in the formation of a more stable carbocation, which is energetically favorable. This rearrangement process is known as a 1,2-shift or alkyl shift.
Overall, compound F would be expected to rearrange due to the increased stability offered by the formation of a secondary carbocation, resulting in a more energetically favorable molecular structure.
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A student would like to prepare 1 ppm Be^2+ solution from BeSO4. How many grams of BeSO4 should be weighed out to prepare 1L of the solution? (Be=9amu S = 32 amu, 0=15amu
1 mg of BeSO₄ should be weighed out to prepare 1 L of 1 ppm Be⁺² solution.
The molar mass of BeSO₄ is:
BeSO₄ = Be + S + 4O
= 9.01 + 32.06 + (4 x 15.99)
= 9.01 + 32.06 + 63.96
= 105.03 g/mol
To prepare a 1 ppm solution of BeSO₄, 1 part of BeSO₄ per million parts of the solution. Since, preparing 1 L of solution, add 1 mg of BeSO₄ to the solution:
1 ppm = 1 mg/L
The number of moles of BeSO₄ required for 1 mg is:
n = m/M
n = 0.001 g / 105.03 g/mol
n = 9.52 x 10^-6 mol
To convert moles of BeSO₄ to grams, use the molar mass:
Mass = n x M
= 9.52 x 10^-6 mol x 105.03 g/mol
= 0.001 g or 1 mg
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The Solubility Product Constant for silver phosphate is 1.3 x 10^-20
The molar solubility of silver phosphate in a 0.223 M sodium phosphate solution is
?M
The molar solubility of AgbPO₄ in a 0.223 M Na₃PO₄ solution is 2.3 x 10⁻⁷ M.
Given:
The value of Ksp for Ag₃PO₄ = 1.3 x 10⁻²°
The balanced equation is:
Ag₃PO₄(s) ⇌ 3 Ag⁺(aq) + (PO₄)³⁻(aq)
The solubility product expression for this reaction is:
Ksp = [Ag+]³ [PO₄⁻³]
Initial: [Ag⁺] = 0 [PO₄⁻³]
= 0.223 M
Change: +3x +x
Equilibrium: [Ag₊] = 3x [PO₄⁻³]
= 0.223 + x
Substituting these values into the Ksp expression:
Ksp = (3x)³ (0.223 + x)
= 1.3 x 10⁻²⁰
Ksp = 27 x³ (0.223) ≈ 6.0 x 10⁻²⁰
Solving for x:
x ≈ 2.3 x 10⁻⁷ M
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Please if you know the answer answer the question thank you.
The tests that show positive results for the food sample are Benedict's and Biuret tests only.
What is Benedict's and Biuret tests?Benedict's test is a chemical test used to detect the presence of reducing sugars, such as glucose, in a solution. It involves adding Benedict's reagent, which contains copper ions, to the solution and heating it. If reducing sugars are present, the copper ions are reduced to form a red precipitate.
Biuret test is a chemical test used to detect the presence of protein in a solution. It involves adding Biuret reagent, which contains copper ions, to the solution. If protein is present, the copper ions react with the peptide bonds in the protein to form a violet-colored complex.
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how much energy is required to heat 500g of ice at 0⁰C to 60⁰C?
a) 125,400 J
b) 167,000 J
c) 292,400 J
d) 41,883,600 J
The amount of energy needed to heat 500 g of ice at 0⁰C to 60⁰C is 292,400 J. Option C.
Energy of reactionIn order to calculate the energy required to heat the ice, we need to consider two stages: first, we need to calculate the energy required to melt the ice, and second, we need to calculate the energy required to heat the resulting liquid water to 60°C.
To melt the ice, we need to supply energy equal to the heat of fusion of ice. The heat of fusion of ice is 334 J/g. Therefore, the energy required to melt 500 g of ice is:
Q1 = (334 J/g) x (500 g) = 167,000 J
Once the ice is melted, we need to heat the resulting liquid water to 60°C. The specific heat capacity of water is 4.184 J/(g°C). Therefore, the energy required to heat 500 g of water from 0°C to 60°C is:
Q2 = (4.184 J/(g°C)) x (500 g) x (60°C - 0°C) = 125,520 J
The total energy required to melt the ice and heat the resulting liquid water to 60°C is the sum of Q1 and Q2:
Q = Q1 + Q2 = 167,000 J + 125,520 J = 292,520 J
Thus, the amount of energy needed to heat 500 g of ice at 0⁰C to 60⁰C is 292,400 J.
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Justification of Subaquatic soil if it is sediment or soil (on the point of view of a geologist)
Subaquatic soil can be classified as sediment or soil based on its geological properties and formation processes.
Sediment refers to any material that is transported and deposited by water, wind, ice, or gravity. Sediments can be composed of various materials, such as minerals, rocks, organic matter, and even human-made debris.
Sediments can accumulate in different environments, such as rivers, lakes, oceans, and deserts, and can be deposited in layers over time.
Subaquatic soil can be classified as sediment or soil based on its geological properties and formation processes. If it has primarily formed through sediment deposition, it is more appropriate to classify it as sediment.
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3. A hydrocarbon, C4H₁o generates a mass spectrum with peaks at m/z values of 58, 43, 29 and 15.
i)
Identify the ions that would give rise to these peaks.
ii) Draw skeletal formulae of the two possible structural isomers for this molecule.
iii)
Explain which isomer would produce this spectrum.
The peak at m/z = 58 corresponds to the molecular ion, [M]
The peak at m/z = 43 corresponds to the fragment ion [M-15]
The peak at m/z = 29 corresponds to the fragment ion [M-29]
The peak at m/z = 15 corresponds to the fragment ion [M-43].
The two possible isomers for this molecule is:
Butane: CH₃CH₂CH₂CH₃Methylpropane: CH₃CH(CH₃)CH₃iii) The molecule is likely methylpropane.
What are the ions that would give rise to the peaks?The peaks in the mass spectrum are as follows:
Peak at m/z = 58: This corresponds to the molecular ion, [M], which has a mass of 58. This means that the entire molecule has been ionized and has a charge of +1.
Peak at m/z = 43: This represents the fragment ion [M-15], which has lost a methyl group (CH₃) from the molecular ion.
Peak at m/z = 29: This represents the fragment ion [M-29], which has lost a propyl group (C₃H₇) from the molecular ion.
Peak at m/z = 15: This represents the fragment ion [M-43], which has lost both a methyl group and a propyl group from the molecular ion.
There are two possible isomers for a hydrocarbon with the molecular formula C₄H₁₀:
Butane: CH₃CH₂CH₂CH₃
Methylpropane: CH₃CH(CH₃)CH₃
iii) Based on the relative intensities of the peaks at m/z = 43 and m/z = 29, the molecule is likely methylpropane.
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calculate the equilibrium conversion and concentrations for each of the foliwing reaction the liquid phase reaction wCith CAO=CBO=2m0l/dm^3 and KC=10dm^3/mol
When the products and reactants do not alter over time, we say that a chemical is in equilibrium concentration. 2 mol/ L is the of concentration for each reactant.
When the products and reactants do not alter over time, we say that a chemical is in equilibrium concentration. In other words, a chemical reaction enters a state of equilibrium or equilibrium concentration when the rate of forward reaction equals the rate of backward reaction. CAO=CBO=2m0l/dm³ and KC=10dm³/mol. substituting all the given values we get 2 mol/ L of concentration for each reactant.
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An object that has a mass of 75 g is heated from 50°C to 93°C. If it takes 3000 J for the temperature change to occur, then what is the specific heat of the object?
The specific heat of the object is approximately 0.930 J/g°C.To find the specific heat of an object, we can use the formula:Q = mcΔT,Where Q represents the heat energy transferred, m is the mass of the object, c is the specific heat, and ΔT is the change in temperature.
Given that the mass of the object is 75 g and the temperature change is from 50°C to 93°C, we can plug these values into the formula:
3000 J = (75 g) * c * (93°C - 50°C)
Simplifying the equation:
3000 J = (75 g) * c * 43°C
Now, we can solve for c by dividing both sides of the equation by (75 g * 43°C):
c = 3000 J / (75 g * 43°C)
Calculating this expression:
c ≈ 0.930 J/g°C
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PART A: What is the amount of calcium carbonate in a 50.0g calcium carbonate?
PART B: What is the amount of carbon in 88.0g propane (C3H8)?
The amount of calcium carbonate in a 50.0g calcium carbonate is 0.5 moles while the amount of carbon in 88.0g propane is 72g.
How to calculate amount of a substance?The amount of a substance in moles can be calculated by dividing the mass of the substance by its molar mass as follows;
moles = mass ÷ molar mass
According to this question, there are 50.0 grams of calcium carbonate. The number of moles of the substance is;
moles = 50g ÷ 100g/mol = 0.5 moles
The amount of carbon in 88.0g propane can be calculated as follows;
36g ÷ 44g × 88 = 72g
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What is an example of a nonrenewable resource? (2 points)
a
Oil
b
Sunlight
c
Water
d
Wind
The answer is A: Oil.
Answer:
Explanation:
A. Oil
hope it helps!
In the model with four atoms bonded to the central atom, the central atom satisfies the octet rule because it has four valence electron pairs (that is, eight valence electrons). The electron pairs are all in the form of covalent bonds.
Replace one of these bonds with a lone pair by clicking the red x to remove the atom. Then click the Lone Pair to add it.
Notice that the molecule geometry is no longer the same as the electron geometry. The electrons in the lone pair don’t count when describing the shape of the molecule. Record your observations in the second row of the table. Then repeat this process to replace another atom with a lone pair. Record the results in the third row of the table.
Select the correct option from each drop-down menu to complete the table.
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2. You are trying to develop a new catalyst for OER in PEMWE.
(a) (2pts) Describe the half-cell reaction and potential of OER
(b) (3pts) Suggests as many issues as possible for the OER catalysts from the viewpoint of
catalyst developer.
(c) Considering issues in (b),
(1) (2pts) What kinds of materials would you suggest? Why?
(2) (2pts) Suggest how the physical structure (nanostructure) of the catalyst should be constructed.
(3)(3pts) Assume you are making a catalyst using electrodeposition.
Suggest how to control the parameters/processes of electrodeposition. What characteristics are expected from the control of each parameter/process?
Cargnello and coworkers developed a novel catalyst that advances this objective by boosting the formation of long-chain hydrocarbons during chemical processes.
Thus, The same amounts of carbon dioxide, hydrogen, catalyst, pressure, heat, and time as the standard catalyst, it created 1,000 times more butane—the longest hydrocarbon it could produce at its maximum pressure—than the standard catalyst.
The novel catalyst is made of ruthenium, a platinum group rare transition metal that is coated in a thin coating of plastic. This idea accelerates chemical processes without being consumed in the process, much like any catalyst.
Another benefit of ruthenium is that it is less expensive than other platinum- and palladium-based high-quality catalysts.
Thus, Cargnello and coworkers developed a novel catalyst that advances this objective by boosting the formation of long-chain hydrocarbons during chemical processes.
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7. Propane reacts with oxygen to produce carbon dioxide and water according to
the equation pictured below.
Which of the following statements is correct?
Cz Hq +
3
*
50₂ → 3C0₂ + 4H₂O
2
O For every 4 moles of water produced, 3 moles of propane react.
O For each mole of oxygen that reacts, 3 moles of carbon dioxide are produced.
For every 3 moles of carbon dioxide produced, 5 moles of oxygen react.
O For each mole of propane that reacts, 5 moles of oxygen are produced.
According to the stoichiometry of the combustion reaction of propane for each mole of propane that reacts, 5 moles of oxygen are produced.
Stoichiometry is defined as the determination of proportions of elements or compounds which are present in a chemical reaction. The related relations are all based on law of conservation of mass and law of combining weights and volumes.
Stoichiometry is used in quantitative analysis for measuring concentrations of substances which are present in the sample.As per the chemical equation C₃H₈ +5 O₂[tex]\rightarrow[/tex]3 CO₂ + 4 H₂O.For every mole of propane that reacts, 5 moles of oxygen are produced.
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To what temperature would 12.3g of He have to be cooled to fit a 34.0 l tank at 1.17 atm?
The ideal gas law may be used to calculate the temperature at which 12.3 g of He would need to be cooled to fit a 34.0 l tank at 1.17 atm. According to this rule, the ideal gas constant multiplied by the number of moles in the gas equals the product of an ideal gas's pressure, volume, and temperature.
In order to solve for temperature, we may thus rearrange the equation. PV = nRT, where P is for pressure, V is for volume, n is for moles, R is for the ideal gas constant, and T is for temperature, is the equation. By multiplying both sides of the equation by the inverse of P and dividing both sides by nR, we may find the value of T.This results in the formula T = PV/nR.
The values for P, V, and nR can be substituted for the predetermined conditions. The volume is 34.0 l and the pressure is 1.17 atm. By dividing the mass, 12.3 g, by He's molar mass, 4.00 g/mol, the amount of moles of He may be computed. 0.0821 L*atm/K*mol is the ideal gas constant. Plugging these values into the equation gives us a temperature of about -267.7 K.
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How much heat energy is required to convert 66.3 g of liquid sulfur dioxide, SO2, at 201.2 K to gaseous SO2 at 263.1 K if the molar heat of vaporization of SO2 is 24.9kJ/mol, and the specific heat capacity (C) of liquid SO2 is 1.36J
The heat energy required to raise the temperature of the vapor q_total = q1 + q2 = 25.76 kJ + 84.39 J = 26.85 kJ.
The heat energy required to convert 66.3 g of liquid SO2 at 201.2 K to gaseous SO2 at 263.1 K is 26.85 kJ.
To solve this problem, we need to calculate the heat energy required to vaporize the given mass of liquid SO2 at its boiling point and then raise the temperature of the resulting vapor to the desired final temperature. The heat energy required for vaporization can be calculated using the molar heat of vaporization of SO2, which is given as 24.9 kJ/mol. The heat energy required to raise the temperature of the vapor can be calculated using the specific heat capacity of liquid SO2, which is given as 1.36 J/g K.
First, we need to calculate the number of moles of SO2 in 66.3 g of liquid SO2 using its molar mass. The molar mass of SO2 is 64.06 g/mol, so:
n = m/M = 66.3 g / 64.06 g/mol = 1.034 mol
The heat energy required for vaporization is then:
q1 = ΔHvap * n = 24.9 kJ/mol * 1.034 mol = 25.76 kJ
Next, we need to calculate the heat energy required to raise the temperature of the resulting vapor from 201.2 K to 263.1 K. The specific heat capacity of liquid SO2 is used because we are raising the temperature of the vapor from the boiling point of liquid SO2.
q2 = n * C * ΔT = 1.034 mol * 1.36 J/g K * (263.1 K - 201.2 K) = 84.39
The total heat energy required is the sum of the heat energy required for vaporization and the heat energy required to raise the temperature of the vapor:
q_total = q1 + q2 = 25.76 kJ + 84.39 J = 26.85 kJ
Therefore, the heat energy required to convert 66.3 g of liquid SO2 at 201.2 K to gaseous SO2 at 263.1 K is 26.85 kJ.
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Which of the following phase changes involves the release of energy?
Freezing, condensation, and deposition are the phase changes that involve the release of energy.
The phase changes that involve the release of energy are exothermic processes, where energy is released from the system into the surroundings. The energy released can be in the form of heat, light, or sound, depending on the nature of the phase change.
Among the common phase changes, the ones that involve the release of energy are:
Freezing: The process of liquid water freezing into ice is an exothermic process, where energy is released in the form of heat.
Condensation: When water vapor condenses into liquid water, energy is released in the form of heat.
Deposition: When water vapor directly changes into solid ice without passing through the liquid phase, energy is released in the form of heat.
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1g of solid gold could be dissolved in a mixture of HCl (excess) and HNO3. The golden solution was then treated with sodium metabisulfite (Na2S2O5) to precipitate a brown solid. Using this information select the correct answer below.
- When gold is dissolved, the HCl and HNO3 react to form NOCl and Cl2, with then oxidize the metal.
- When the gold is dissolved, the HCl is the oxidant and Au is the reductant.
- Au3+ ion has octahedral geometry with four Cl- ligands and two trans water molecules.
- Na2S2O5 is a 2e- reducing agent, reacting only with the Au3+ ion to form SO3.
- Na2S2O5 is a 4e- reducing agent, reacting with the Au3+ ion and HCl.
The appropriate response is: based on the facts provided.
A 2e-reducing agent, Na2S2O5, only produces SO3 when interacting with the ion Au3+.(option-4)
Sodium metabisulfite (Na2S2O5) is used, which implies that it is working as a reducing agent. In this case, it is reversibly reducing the Au3+ ion to gold (Au), an elemental form that precipitates as a brown solid.
The following is a representation of the reaction's balanced equation:
2Au (s) + 2SO3 2- (aq) + 2Na+ (aq) + 6H+ (aq) = 2Au3+ (aq) + Na2S2O5 (aq) + 3H2O (l)
The electrons required for reducing Au3+ to Au are here provided by Na2S2O5 acting as a 2e- reducing agent. H+ ions from the excess HCl are also present during the process.(option-4)
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