Multiplying by Avogadro's number we find that 55.6 moles of water contains 3.34 × 1025 molecules.
Answer:
A molecular weight often is simply referred to as a mole. Thus, 1 L of water contains 55.6 moles of water. Multiplying by Avogadro's number we find that 55.6 moles of water contains [tex]3.34 * 10^2^5[/tex] molecules.
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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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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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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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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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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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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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How many moles are found in 25 grams of hydrogen chloride, HCl?
Here are 0.686 moles of hydrogen chloride in 25 grams of HCl.
Number of moles = Mass of substance (in grams) / Molar mass of substance
The molar mass of HCl can be calculated by adding the atomic masses of hydrogen (1.008 g/mol) and chlorine (35.45 g/mol), which gives a molar mass of 36.458 g/mol.
Now we can plug in the values:
Number of moles = 25 g / 36.458 g/mol
Number of moles = 0.686 moles
It's important to note that the molar mass of a substance is the mass in grams of one mole of that substance. This means that if we know the mass of a substance and its molar mass, we can find the number of moles present in that mass. This is a useful calculation in chemistry as it allows us to make accurate measurements and carry out calculations involving the reactions and properties of different substances.
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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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if we are constantly taking in more and more c-14 why does it concentration in us not increase
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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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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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.
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
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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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When Ba metal is added to an aqueous solution containing dissolved LiCl and MgCl2 what should occur based on the standard reduction potentials?
- Ba metal will be oxidized to Ba+2 and Mg2+ ions will be reduced to Mg metal.
- Ba metal will be oxidized to Ba2+ and H2, -OH, Li and Mg will form.
- Ba metal will be oxidized to Ba2+ and H2 and -OH will form.
- Ba metal will be oxidized to Ba2+ and the Li+ and Mg2+ ions will be reduced to Li and Mg metal.
- No reaction should occur.
The standard reduction potentials is Ba metal will be oxidized to Ba2+ and the Li+ and Mg2+ ions will be reduced to Li and Mg metal.
When Ba metal is added to an aqueous solution containing dissolved LiCl and MgCl2, Ba metal will be oxidized to Ba2+ ions based on the standard reduction potentials. However, the species that will undergo reduction depends on their respective reduction potentials.
According to the standard reduction potentials, Li+ has a more positive reduction potential than Mg2+, which means Li+ has a greater tendency to undergo reduction compared to Mg2+. Therefore, Li+ ions will be reduced to Li metal while Mg2+ ions will remain in solution.
The overall reaction can be represented as follows:
Ba(s) + 2Li+(aq) → Ba2+(aq) + 2Li(s)
Therefore, the correct answer is Ba metal will be oxidized to Ba2+ and the Li+ and Mg2+ ions will be reduced to Li and Mg metal. Mg2+ ions will not be reduced to Mg metal is incorrect. The formation of H2 gas and -OH ions, which are not supported by the standard reduction potentials is incorrect. -OH ions are not formed when Li+ ions undergo reduction is incorrect. A reaction does occur based on the standard reduction potentials is incorrect.
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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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a cool water sample absorbed 3135 j of energy from hot metal. the temperature of the 63 g piece of metal changed from c to 20 c what is the specific heat of the metal
Answer: 156.75 J/C
Explanation:
Q=CT (C is specific heat and T is change in temperature)
Disclaimer: I am assuming the initial temperature is 0 degrees Celsius because the question has been worded improperly. This assumption is not made because it is true, but because there is not enough information in the current question to solve for the specific heat. Also, the mass of the sample is given, which is unnecessary for solving the specific heat but necessary to solve for the specific heat capacity. Either the question has not been worded properly or the mass given is just to trick students.
Rearrange to isolate C: C = Q/T
Solve for C: C = 3135/(20-0) = 3135/20 = 156.75 J/C
Answer: 0.137
Explanation:
acellus
complete the given table by mentioning the quantum numbers for each orbits
Quantum number orbital
2p 3d
azimuthal quantum number ? ?
magnetic quantum number ? ?
What are the quantum numbers?
The orbital's orientation in space is described by the magnetic quantum number (m). Any number between -l and +l may represent the value of m.
The electron's orbital form is determined by a quantum number called the azimuthal quantum number. Any integer between 0 and n-1 can be used to represent the value of l, and as it rises, the orbital's form becomes more complex.
The quantum numbers that are involved have been shown above.
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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:
please help, dont understand
The required amount in moles are 3 moles of H₂O to produce 164 g of H₃PO₃.
How to find amount?To solve the problem, use the balanced chemical equation to relate the amount of H₃PO₃ formed to the amount of H₂O used. From the balanced chemical equation:
P2O₃ + 3H₂O → 2H₃PO₃
3 moles of H₂O are required to produce 2 moles of H₃PO₃. This can be written as:
2 moles H₃PO₃ / 3 moles H₂O
To find the number of moles of H₂O required to produce 164 g of H₃PO₃, use the molar mass of H₃PO₃:
1 mole H₃PO₃ = 82 g
So, 164 g of H₃PO₃ is equal to:
164 g H₃PO₃ / 82 g/mol = 2 moles H₃PO₃
Using the ratio above, calculate the number of moles of H₂O required:
2 moles H₃PO₃ × (3 moles H2O / 2 moles H₃PO₃) = 3 moles H₂O
Therefore, 3 moles of H₂O are required to produce 164 g of H₃PO₃.
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A radioactive sample contains 3.00 g of an isotope with a half-life of 3.8 days.
How much of the isotope in grams will remain after 19.8 days?
Answer:So, about 0.093 g of the isotope will remain after 19.8 days.
Explanation:
The first step is to find the number of half-lives that have passed during 19.8 days:
Number of half-lives = time elapsed / half-life
Number of half-lives = 19.8 days / 3.8 days per half-life
Number of half-lives ≈ 5.21
This means that the initial amount of the isotope has been halved 5.21 times. The remaining fraction of the original amount can be calculated using the following formula:
Remaining fraction = (1/2)^(number of half-lives)
Substituting the values, we get:
Remaining fraction = (1/2)^5.21
Remaining fraction ≈ 0.031
Therefore, the amount of the isotope remaining after 19.8 days is:
Remaining amount = Remaining fraction x Initial amount
Remaining amount = 0.031 x 3.00 g
Remaining amount ≈ 0.093 g
So, about 0.093 g of the isotope will remain after 19.8 days.
The isotope in grams will remain after 19.8 days would be 0.081 grams.
The formula to calculate the left mass of a radioactive element can be deduced as -
[tex] \qquad\star\longrightarrow \underline{\boxed{\sf{m =m_{o} \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{t}{T½}} }}} \\[/tex]
Where-
[tex]\sf m_{o} [/tex]is the initial mass of a radioactive elementT½ is the half life timet is the time periodm = Left mass of a radioactive element.According to the given specific parameters -
Initial mass,[tex]\sf m_{o} [/tex] = 3 gHalf life time, T½= 3.8 days Time period, t =19.8 daysNow that we have all the required values, so we can plug them into the formula and solve for the left mass of a radioactive element-
[tex] \qquad \longrightarrow \sf \underline{m =m_{o} \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{t}{T½} }} \\[/tex]
[tex] \qquad\longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{19.8}{3.8} } \\[/tex]
[tex]\qquad \longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{\cancel{19.8}}{\cancel{3.8}} } \\[/tex]
[tex] \qquad\longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ 5.21052..... } \\[/tex]
[tex] \qquad\longrightarrow \sf m =3 \times 0.02700... \\[/tex]
[tex] \qquad\longrightarrow \sf m =0.081020....\;g \\[/tex]
[tex] \qquad\longrightarrow \sf \underline{m =\boxed{\sf{0.081\;g}}} \\[/tex]
Henceforth,about 0.081 g of the isotope in grams will remain after 19.8 days.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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73.5 g of aluminum is heated in boiling water to a temperature of 98.7 degrees Celsius. The aluminum is then placed in a calorimeter containing 1500 g of water at a temperature of 25.4 degrees Celsius. The temperature of the water in the calorimeter increase to a final temperature of 28.2 degrees Celsius. What is the specific heat of the aluminum?
The specific heat of aluminum is 0.92 J/g°C.
Use the principle of conservation of energy.
Q aluminum = -Qwater-calorimeter
(m aluminum)(c aluminum)(ΔT aluminum) = -(m water + m calorimeter)(c water)(ΔT water)
where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Calculate the heat lost by the aluminum.
Q aluminum = (m aluminum)(c aluminum)(ΔT aluminum)
where ΔT aluminum is the change in temperature of the aluminum when it was heated in boiling water:
ΔT aluminum = 98.7°C - 100°C
= -1.3°C
Q aluminum = (73.5 g)(c aluminum)(-1.3°C)
Q water-calorimeter = -(m water + m calorimeter)(c water)(ΔT water)
where ΔT water is the change in temperature of the water in the calorimeter:
ΔT_water = 28.2°C - 25.4°C
= 2.8°C
Q water-calorimeter = -(1500 g + m calorimeter)(4.18 J/g°C)(2.8°C)
Q water = -(1500 g)(4.18 J/g°C)(2.8°C)
Substitute the values and solve for c aluminum:
(73.5 g)(c aluminum)(-1.3°C) = -(1500 g)(4.18 J/g°C)(2.8°C)
c aluminum = -(1500 g)(4.18 J/g°C)(2.8°C) / (73.5 g)(-1.3°C)
c aluminum = 0.92 J/g°C
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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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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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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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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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I need help with 5a and 5 b
The mass of the number of the reacting agent in the reaction producing Cr₂S₃ and H₂O are;
a. The mass of the H₂S is about 126.2 grams
b. Mass of Cr₂S₃ produced is about 242.242 grams
What is a reacting agent?A reacting agent in a chemical reaction are the elements, compounds or molecules on the reaction side of a chemical reaction.
Whereby the chemical reaction is; Cr₂O₃ + 3·H₂S → Cr₂S₃ + 3·H₂O
We get;
5 a. The molar mass of H₂O is; 18.01528 g/mol
The number of moles of H₂O in 66.6 grams of H₂O is; 66.6/18.01528 ≈ 3.7 moles
The number of moles H₂S required to produce 3 moles of H₂O = 3 moles
Therefore, the number of moles H₂S required to produce 3.7 moles of H₂O = 3.7 moles
The molar mass of H₂S =- 34.1 g/mol
The mass of H₂S required = 3.7 moles × 34.1 g/mol ≈ 126.2 grams5 b. The molar mass of Cr₂S₃ = 200.2 g/mol
The molar mass H₂S = 34.1 g/mol
The mass of the H₂S = 123.7 grams
The number of moles of H₂S is; 123.7 grams/(34.1 g/mol) ≈ 3.63 moles
The stoichiometry of the reaction indicates that the mole ratio of the number of moles of Cr₂S₃ to the number of moles of the H₂S is; 1 : 3
Therefore, the number of moles of the Cr₂S₃ in the reaction is; 3.63 moles/3 ≈ 1.21 moles
The mass of the Cr₂S₃ in the reaction is therefore; 200.2 g/mol × 1.21 moles = 242.242 gramsLearn more on the stoichiometry of chemical reaction here: https://brainly.com/question/29571173
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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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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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