It is polar or non polar is discussed below:
a) Beryllium chloride (BeCl2) is a linear molecule with two chlorine atoms on either side of the beryllium atom. Since the electronegativity of both chlorine and beryllium is similar, the bond between them is nonpolar. Therefore, BeCl2 is a nonpolar molecule. b) Hydrogen sulfide (H2S) is a bent molecule with the two hydrogen atoms and a sulfur atom. The sulfur atom has a higher electronegativity than the hydrogen atom, which leads to a polar covalent bond. Due to the bent shape of the molecule, the polar bonds do not cancel each other out, resulting in an overall polar molecule. c) Sulfur trioxide (SO3) is a trigonal planar molecule with three oxygen atoms surrounding a central sulfur atom. The electronegativity of oxygen is higher than that of sulfur, which creates polar covalent bonds. However, due to the symmetry of the molecule, the polar bonds cancel each other out, resulting in a nonpolar molecule. d) Water (H2O) is a bent molecule with two hydrogen atoms and one oxygen atom. The electronegativity of oxygen is higher than that of hydrogen, resulting in polar covalent bonds. Due to the bent shape of the molecule, the polar bonds do not cancel each other out, resulting in an overall polar molecule.
e) Trichloromethane (CHCl3) is a tetrahedral molecule with one carbon atom and three chlorine atoms. The electronegativity of chlorine is higher than that of carbon, resulting in polar covalent bonds. However, due to the tetrahedral shape of the molecule, the polar bonds do not cancel each other out, resulting in an overall polar molecule.
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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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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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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
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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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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What are paired and unpaired electrons
Answer:
Paired electrons are the electrons in an atom that occur in an orbital as pairs.
⇒paired electrons always occur as a couple of electrons
unpaired electrons are the electrons in an atom that occur in an orbital alone.
⇒unpaired electrons occur as single electrons in the orbital.
Answer:
Paired electrons are the electrons in an atom that occur in an orbital as pairs whereas unpaired electrons are the electrons in an atom that occur in an orbital alone. Therefore, paired electrons always occur as a couple of electrons while unpaired electrons occur as single electrons in the orbital.
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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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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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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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if we are constantly taking in more and more c-14 why does it concentration in us not increase
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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Which molecule is butene?
H H H H H
• A. H-0-C-C-C-C-H
H HHH Н
CH,
/
О в.
C=C
Is
H H H H
• D. H-CEC-C-C-H
H
The molecule that represents butene is option C: H2C=CHCH2CH3.
Butene is an alkene with four carbon atoms and a double bond between the second and third carbon atoms. In the given structure, the carbon atoms are connected in a linear chain with hydrogen atoms attached to them. The double bond between the second and third carbon atoms is denoted by the "=" symbol.
To identify butene, we can count the number of carbon atoms in the molecule. Butene has four carbon atoms, and option C satisfies this requirement. Additionally, the presence of a double bond between the second and third carbon atoms is another characteristic feature of butene, which is represented by the "=" symbol in option C.
Option A, H3C-O-C-C-C-H3, represents an ether molecule, not butene. Option B, HC≡CH, represents acetylene, a different hydrocarbon. Option D, H3C-EC-C-C-H3, does not correctly represent a recognizable organic molecule. option C, H2C=CHCH2CH3, is the structure that represents butene accurately. Therefore, Option C is correct.
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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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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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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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Balance the entire chemical
reaction using an atom inventory.
What is the correct whole
number coefficient for propane,
C3H8?
[?]C3H8+ [ 0₂
]CO2+[ ]H2O
The balanced chemical equation for the combustion of propane with oxygen is: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
To balance the equation, first balance the carbon atoms on both sides of the equation. There are three carbon atoms in the propane molecule and three in the carbon dioxide molecule, so balance the carbon atoms by putting a coefficient of 3 in front of the CO₂ molecule.
C3H8 + 5O2 → 3CO₂
Next, balance the hydrogen atoms. There are eight hydrogen atoms in the propane molecule and four in the water molecule, so balance the hydrogen atoms by putting a coefficient of 4 in front of the H₂O molecule.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Finally, balance the oxygen atoms. There are five oxygen atoms on the left side and 10 on the right side, so balance the oxygen atoms by putting a coefficient of 5 in front of the O₂ molecule.
Therefore, the correct whole number coefficient for propane, C3H8, is 1.
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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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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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Complete and balance the following half-reaction in acidic solution. Be sure to include the proper phases for all species within the reaction.
SO3 2- (aq) ---> SO4 2- (aq)
The balanced half reaction is obtained as [tex]SO_{3} ^2- (aq) + H_{2} O (l) --- > SO_{4}^2- (aq) + 2H^+ (aq).[/tex]
What does it mean to balance a redox reaction?We know that for the reaction to be seen as balanced we would haver to look at the masses and the charges and now we are going to have that, two protons are added to the product side to balance the charge. To balance the amount of hydrogen atoms on the reactant side, water is also supplied.
Then when we look at the balanced reaction for an acid medium as have been required by the question then we are going to have;
[tex]SO_{3} ^2- (aq) + H_{2} O (l) --- > SO_{4}^2- (aq) + 2H^+ (aq).[/tex]
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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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true or false?
If the Sun's surface became much hotter (while the Sun's size remained the same), the Sun would emit more ultraviolet light but less visible light than it currently emits.
Explain your reasoning.
If the Sun's surface became much hotter (while the Sun's size remained the same), the Sun would emit more ultraviolet light but less visible light than it currently emits. The statement is True.
Solar radiation is radiant (electromagnetic) energy from the sun. It provides light and heat for the Earth and energy for photosynthesis. This radiant energy is necessary for the metabolism of the environment and its inhabitants 1. The three relevant bands, or ranges, along the solar radiation spectrum are ultraviolet, visible (PAR), and infrared.
Ultraviolet radiation makes up just over 8% of the total solar radiation.
When heat increases, so does the frequency and energy of the wavelengths. Because of this, some visible light would be converted to ultraviolet light.
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Which of the two semiconductors shown in the illustration above is an n-type?
Which is a p-type? How are the two different?
In between conductors, which are typically metals, and not-conductors or insulators, such as ceramics, exist materials known as semiconductors. Semiconductors can be pure elements like germanium or silicon or compounds like gallium arsenide.
In the given pictures, 'As' is a N-type semiconductor whereas 'Ga' is a P-type semiconductor.
When pentavalent impurities (P, As, Sb, and Bi) are added to a pure semiconductor (germanium or silicon), four of the five valence electrons form a bond with the four electrons of the pure semiconductor.
The dopant's fifth electron is liberated and used for conduction in the lattice are called N-type semiconductors.
When a trivalent impurity (B, Al, In, or Ga) is added into a pure semiconductor, three of the semiconductor's four valence electrons form a bond with the impurity's three valence electrons.
In the impurity, this results in an electron (hole) being missing called P -type semiconductors.
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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.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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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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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:
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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After many generations, which trait will be most common? Why? amplify
Answer:
Over time, as generations of individuals with the trait continue to reproduce, the advantageous trait becomes increasingly common in a population, making the population different than an ancestral one.
Explanation:
have a nice day.
Answer:
Explanation:
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Write a balanced chemical equation for each of the following.
Solid lead (II) sulfide reacts with aqueous hydrochloric acid to form solid lead (II) chloride and dihydrogen sulfide gas.
Express your answer as a chemical equation. Identify all of the phases in your answer.
According to a balanced chemical equation, solid lead (II) sulphide reacts with aqueous hydrochloric acid to produce solid lead (II) chloride and dihydrogen sulphide gas.
PbCl2 + H2S (s) PbS + 2HCl PbS stands for solid lead (II) sulphide, 2HCl for aqueous hydrochloric acid, PbCl2 for solid lead (II) chloride, and H2S (s) for dihydrogen sulphide gas in this equation. Parentheses represent the stages of the reactants and products. While the products are solid and gaseous, the reactants are in the solid and aqueous phases.
Since each element has an equal amount of atoms on both sides of the equation in the sulfide and in the other one as hydrochloric well, the equation is balanced. There is one atom of lead on the left.
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Write the products and balance the chemical equation that results from the single replacement reaction of iron and silver nitrate.
The single replacement reaction of iron and silver nitrate can be represented by the chemical equation Fe + [tex]2AgNO_{3}[/tex] → [tex]Fe(NO_{3} )_{2}[/tex] + 2Ag
In this reaction, a single element, iron (Fe), replaces another element, silver (Ag), in the compound silver nitrate [tex](AgNO_{3} )[/tex], resulting in the formation of iron (II) nitrate [tex](Fe(NO_{3} )_{2} )[/tex] and elemental silver (Ag). The balanced equation shows that one iron atom reacts with two silver nitrate molecules to produce one molecule of iron (II) nitrate and two silver atoms.
To balance the equation, we need to ensure that the same number of each type of atom is present on both sides of the equation. We begin by counting the number of atoms of each element present in the reactants and products. The left side of the equation has one Fe atom, two Ag atoms, two N atoms, and six O atoms. The right side has one Fe atom, two Ag atoms, two N atoms, and six O atoms. Therefore, the equation is already balanced.
Overall, this reaction is an example of a single replacement reaction in which a more reactive element (iron) replaces a less reactive element (silver) in a compound. The resulting products are an ionic compound (iron (II) nitrate) and a pure element (silver). This type of reaction can be used to extract metals from their compounds or to convert one metal into another.
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