A sample of helium gas with a volume of 27.0 mL at 759 mm Hg is compressed at a constant temperature until its volume is 11.4 mL. What will be the new pressure in the sample?
Answer:
1797.6 mmHgExplanation:
The new pressure can be found by using the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we're finding the new pressure
[tex]P_2 = \frac{P_1V_1}{V_2} \\[/tex]
We have
[tex]P_2 = \frac{27 \times 759}{11.4} = \frac{20493}{11.4} \\ = 1797.631....[/tex]
We have the final answer as
1797.6 mmHgHope this helps you
Aluminium is a metal give reason
Answer:
Aluminium is ordinarily classified as a metal. It is lustrous, malleable and ductile, and has high electrical and thermal conductivity. Like most metals, it has a close-packed crystalline structure and forms a cation in an aqueous solution.
state what happens to the boiling point and freezing point of the solution when the solution is diluted with an additional 100. grams of h2o( ).
Diluting the solution with an additional 100 grams of water will increase the boiling point and decrease the freezing point of the solution.
When a solution is diluted with an additional 100 grams of water, the boiling point and freezing point of the solution will both be affected.
Boiling Point:
The addition of more water to the solution increases the total volume of the solution. As a result, the concentration of solute particles in the solution decreases. According to Raoult's law, the vapor pressure of the solvent above the solution is directly proportional to the mole fraction of the solvent in the solution.
With a decrease in the concentration of solute particles, the mole fraction of the solvent increases, leading to a decrease in the vapor pressure of the solution. As a consequence, the boiling point of the solution increases.
Therefore, when the solution is diluted with additional water, the boiling point of the solution will be higher compared to the original, more concentrated solution.
Freezing Point:
Similar to the boiling point, the addition of more water increases the total volume of the solution and decreases the concentration of solute particles. According to the colligative properties of solutions, such as the freezing point depression, a decrease in the concentration of solute particles leads to a decrease in the freezing point of the solution.
Hence, when the solution is diluted with additional water, the freezing point of the solution will be lower compared to the original, more concentrated solution.
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Which of the following general reactions appropriately models a combination reaction?
A. A+B⟶+B
B. C⟶A+B
C. A+B⟶C
D. A+B⟶C+D
The appropriate model for a combination reaction is A+B⟶C (option C)
What is combination reaction?A combination reaction represents a chemical transformation where multiple substances unite to generate a fresh entity. It follows a general format of A + B → C, with A and B acting as the reactants that undergo a fusion to yield the product C.
Combination reactions commonly exhibit exothermic characteristics, denoting the release of heat. This phenomenon arises due to the increased stability of the resultant products compared to the initial reactants.
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What should you use to compare one substance with another substance in a reaction?
Answer:
metal or other acids
Explanation: hope u get it right
Given the diagram to the right what is true about the missing volume?
Answer:
C
Explanation:
Boyle's law states that "the volume of a given mass of gas is inversely proportional to its pressure at constant temperature."
Inverse proportionality means that as one quantity is increasing, the other quantity is decreasing and vice versa.
Hence, as the pressure was increased, the volume decreases accordingly in obedience to Boyle's law.
Answer:
What is TRUE about the missing volume is option C.
C. The volume will decrease due to inverse relationship of V & P
Explanation:
The given parameters of the gas are;
The initial pressure of the gas, P₁ = 2 atm
The initial volume of the gas, V₁ = 1.5 L
The final pressure of the gas, P₂ = 2 atm
Boyle's law states that, at constant temperature, the volume, 'V', of a given mass is inversely proportional to its pressure, 'P';
Mathematically, Boyle's law can be expressed as follows;
P ∝ 1/V
From which we have;
P·V = Constant
∴ P₁·V₁ = P₂·V₂
For the given gas, we get;
2 atm × 1.5 L = 6 atm × V₂
∴ V₂ = 2 atm × 1.5 L/(6 atm) = 0.5 L
Therefore, the volume decreases from 1.5 L to 0.5 L.
Your friend Niko predicts that iron and silver nitrate will form when silver is placed into an iron (III) nitrate solution. Is this correct? Why or why not?
A
It is correct because the products are an element and a compound.
B
It is correct because iron is more reactive than silver.
C
It is incorrect because this reaction should form a single product.
D
It is incorrect because silver is less reactive than iron.
Answer: B
Explanation:
B. It is correct because iron is more reactive than silver [1]. When iron (III) nitrate solution comes in contact with silver, the iron ions and silver ions will exchange. The more reactive metal will displace the less reactive metal from its compound. In this case, iron is more reactive than silver, so iron ions will displace silver ions from silver nitrate to form iron nitrate and silver. This reaction will produce iron nitrate and silver as products [2][3].
The prediction made by Niko, stating that iron and silver nitrate will form when silver is placed into an iron (III) nitrate solution, is incorrect. This is because silver is less reactive than iron, and the reaction should not produce iron and silver nitrate as products.
The prediction made by Niko is incorrect. When evaluating the reactivity of metals, it is important to consider the reactivity series. The reactivity series ranks metals based on their tendency to undergo redox reactions. In the reactivity series, iron is typically more reactive than silver.
Iron (III) nitrate is a compound consisting of iron ions and nitrate ions . Silver nitrate, on the other hand, consists of silver ions and nitrate ions . When silver is placed into an iron (III) nitrate solution, a single replacement reaction can occur. However, since silver is less reactive than iron according to the reactivity series, it cannot displace iron from its compound.
Instead, if a reaction occurs, it would involve the displacement of silver by iron, resulting in the formation of iron nitrate and the deposition of silver metal. The correct prediction would be that iron displaces silver, not the formation of iron and silver nitrate as suggested by Niko.
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Considering the following precipitation reaction:
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
What is the correct net ionic equation?
A) 2(NO3)- (aq) + 2K+(aq) → 2KNO3
B) Pb2+(aq) + 2(NO3)- (aq) + 2K+ (aq) + 2I- (aq) → PbI2(s) + 2K+ (aq) + 2(NO3)- (aq)
C) Pb2+ (aq) + I2-(aq) → PbI2(s)
D) Pb2+ (aq) + 2I- (aq) → PbI2(s)
The correct net ionic equation for the given precipitation reaction is; Pb²⁺ (aq) + 2I⁻ (aq) → PbI₂(s). Option D is correct.
In the reaction, Pb²⁺ cations from lead nitrate (Pb(NO₃)₂) react with I- anions from potassium iodide (KI) to form solid PbI₂ (lead iodide). This is a precipitation reaction where the insoluble salt PbI₂ is formed.
Option A (2(NO₃)⁻ (aq) + 2K⁺(aq) → 2KNO₃) is not the correct net ionic equation because it represents the dissociation and reformation of the soluble salts potassium nitrate (KNO₃) and lead nitrate (Pb(NO₃)₂), but it does not include the formation of the insoluble PbI₂ precipitate.
Option B (Pb²⁺(aq) + 2(NO₃)⁻ (aq) + 2K⁺ (aq) + 2I⁻ (aq) → PbI₂(s) + 2K⁺ (aq) + 2(NO₃)⁻ (aq)) is not the correct net ionic equation because it includes the spectator ions (K⁺ and NO₃⁻) that do not participate in the actual formation of the precipitate.
Option C (Pb²⁺ (aq) + I₂⁻(aq) → PbI₂(s)) is not the correct net ionic equation because it represents a direct combination of Pb²⁺ and I⁻ ions to form PbI₂, but in the given reaction, the iodide ions come from potassium iodide (KI), not from I₂.
Hence, D. is the correct option.
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It takes 242. kJ/mol to break a chlorine-chlorine single bond. Calculate the maximum wavelength of light for which a chlorine-chlorine single bond could be broken by absorbing a single photon. Be sure your answer has the correct number of significant digits.
The maximum wavelength of light required to break a chlorine-chlorine single bond is approximately 4.947 x 10⁻⁷ meters or 494.7 nm (nanometers), rounded to four significant digits.
We can use the relationship between energy, wavelength, and speed of light to determine the maximum wavelength of light necessary to break a chlorine-chlorine single bond.
The equation which gives the energy of a photon is:
E = hc/λ
where:
E is the photon's energy.
Planck's constant, h, is (6.626 x 10⁻³⁴ Js)
2.998 x 10⁸ m/s is the speed of light, or c.
λ is the wavelength of light
We are aware that a chlorine-chlorine single bond must be broken with 242 kJ/mol of energy. This can be changed to joules per molecule as follows:
[tex]242 kJ/mol = \frac{242 \times 10^3 J}{(6.022 \times 10^{23} molecules/mol)} \approx 4.015 \times 10^{-19} J/molecule[/tex]
The equation can now be rearranged to find the maximum wavelength:
λ = hc/E
Putting in the values:
[tex]\lambda = \frac {(6.626 \times 10^{-34} Js)(2.998 \times 10^8 m/s)}{(4.015 \times 10^{-19} J/molecule)}[/tex]
Calculating the result:
[tex]\lambda \approx 4.947 \times 10^{-7} meters[/tex]
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How many moles are in 56 L of CO2?
Answer:
we have four moles in 56L of co2
How many moles are in 12 liters of Cl2?
Answer:
[tex]\boxed {\boxed {\sf 0.54 \ mol \ Cl_2}}[/tex]
Explanation:
A mole is any quantity of a substance that contains 6.02 × 10²³ particles. At standard temperature and pressure, or STP, 1 mole of as is equal to 22.4 liters. This is true for any gas, regardless of the specific kind.
Although it is not specified, we can assume this gas is at STP. Let's set up a ratio using this information: 22.4 L/mol
[tex]\frac {22.4 \ L \ Cl_2}{1 \ mol \ Cl_2}[/tex]
Multiply by the given number of liters: 12
[tex]12 \ L \ Cl_2 *\frac {22.4 \ L \ Cl_2}{1 \ mol \ Cl_2}[/tex]
Flip the ratio so the liters of chlorine cancel.
[tex]12 \ L \ Cl_2 * \frac {1 \ mol \ Cl_2}{22.4 \ L \ Cl_2}[/tex]
[tex]12 * \frac {1 \ mol \ Cl_2}{22.4 }[/tex]
[tex]\frac {12}{22.4 } \ mol \ Cl_2[/tex]
[tex]0.53571428571 \ mol \ Cl_2[/tex]
The original measurement of liters has 2 significant figures, so our answer must have the same.
For the number we found, that is the hundredth place.
0.53571428571The 5 in the thousandth place tells us to round the 3 up to a 4.
[tex]0.54 \ mol \ Cl_2[/tex]
12 liters of chlorine gas at STP is approximately 0.54 moles of chlorine gas.
For the reaction between nitrogen monoxide and chlorine to produce nitrosyl chloride, 2NO(g) + Cl2(g) à 2NOCl(g), it is found that tripling the initial concentration of both reactants increases the initial rate by a factor of 27. If only the initial concentration of chlorine is tripled, the initial rate increases by a factor of 3. What is the order of the reaction with respect to Cl2?
A) ½
B) 0
C) 3
D) 1
E) 2
Comparing these two scenarios, we can conclude that the order of the reaction with respect to Cl₂ is 1. Therefore, the correct answer is (D) 1.
To determine the order of the reaction with respect to Cl₂, we can use the information provided about the effect of concentration changes on the initial rate.
Let's analyze the given data:
When the initial concentrations of both reactants, NO and Cl₂, are tripled, the initial rate increases by a factor of 27. This indicates that the rate is proportional to the cube of the initial concentration of both reactants.
When only the initial concentration of Cl₂ is tripled, the initial rate increases by a factor of 3. This indicates that the rate is directly proportional to the initial concentration of Cl₂.
Comparing these two scenarios, we can conclude that the order of the reaction with respect to Cl₂ is 1.
Therefore, the correct answer is (D) 1.
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Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
(a) K2CO3
(b) CaCl2
(c) KH2PO4
(d) (NH4)2CO3
Can you show work that was used to find out of the compound was acidic, basic, neutral.
The aqueous solutions of the following salts are:
(a) [tex]K_2CO_3[/tex]: Basic
(b) [tex]CaCl_2[/tex]: Neutral
(c) [tex]KH_2PO_4[/tex]: Acidic
(d) [tex](NH_4)_2CO_3[/tex]: Acidic
(a) [tex]K_2CO_3[/tex]:
The cation, [tex]K^+[/tex] (potassium ion), is derived from a strong base (KOH), which does not hydrolyze in water. Therefore, it does not contribute to the acidity or basicity of the solution. The anion, [tex]CO_3^{2-[/tex] (carbonate ion), is derived from a weak acid [tex]H_2CO_3[/tex], which can hydrolyze in water.
The hydrolysis of [tex]CO_3^{2-[/tex] can be represented as follows:
[tex]CO_3^{2-} + H_2O[/tex] ⇌ [tex]HCO_3^- + OH^-[/tex]
Since the hydrolysis of [tex]CO_3^{2-[/tex] results in the formation of [tex]OH^-[/tex] ions, the solution will be basic.
(b) [tex]CaCl_2[/tex]:
Both the cation, [tex]Ca_2^+[/tex] (calcium ion), and the anion, [tex]Cl^-[/tex] (chloride ion), are derived from strong acids and bases (HCl and [tex]Ca(OH)_2[/tex]). As a result, they do not undergo hydrolysis in water. Therefore, the solution will be neutral.
(c) [tex]KH_2PO_4[/tex]:
The cation, [tex]K_+[/tex] (potassium ion), does not undergo hydrolysis in water. The anion, [tex]H_2PO_4^-[/tex] (dihydrogen phosphate ion), can hydrolyze in water.
The hydrolysis of [tex]H_2PO_4^-[/tex] can be represented as follows:
H2PO4- + H2O ⇌ [tex]HPO_4^{2-} + H_2PO_4^- + H_3O^+[/tex]
Since the hydrolysis of [tex]H_2PO_4^-[/tex] results in the formation of [tex]H_3O^+[/tex] ions, the solution will be acidic.
(d) [tex](NH_4)_2CO_3[/tex]:
Both the cation, [tex]NH_4^+[/tex] (ammonium ion), and the anion, [tex]CO_3^{2-[/tex] (carbonate ion), can hydrolyze in water.
The hydrolysis of [tex]NH_4^+[/tex] can be represented as follows:
[tex]NH_4^+ + H_2O[/tex] ⇌ [tex]NH_3 + H_3O^+[/tex]
Since the hydrolysis of [tex]NH_4^+[/tex] results in the formation of [tex]H_3O^+[/tex] ions, the solution will be acidic.
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using hess's law, calculate δh° for the process: co3o4 (s) 3 co (s) 2o2 (g) from the following information: co(s) o2(g) coo (s) δh° = −237.9 kj 3 coo (s) o2(g) co3o4 (s) δh° = −177.5 kj
The enthalpy change (ΔH°) for the given process, Co₃O₄ (s) → 3 Co (s) + 2 O₂ (g), is +891.2 kJ.
To calculate ΔH° for the given process using Hess's Law, we need to manipulate the provided equations and their enthalpy changes to obtain the reactions;
Reverse the first reaction; Coo (s) → Co (s) + 0.5 O₂ (g) with ΔH° = +237.9 kJ.
Multiply the first reaction by 3; 3 Coo (s) → 3 Co (s) + 1.5 O₂ (g) with ΔH° = 3 × (+237.9) kJ = +713.7 kJ.
Reverse the second reaction; Co₃O₄ (s) → 3 Coo (s) + 0.5 O₂ (g) with ΔH° = -(-177.5) kJ = +177.5 kJ.
Add the manipulated reactions together;
3 Coo (s) + 0.5 O2 (g) → 3 Co (s) + 1.5 O₂ (g) with ΔH° = +713.7 kJ
Co₃O₄ (s) → 3 Coo (s) + 0.5 O₂ (g) with ΔH° = +177.5 kJ
Cancel out the common species on both sides;
Co3O4 (s) → 3 Co (s) + O2 (g) with ΔH°
= +713.7 kJ + 177.5 kJ
= +891.2 kJ.
Therefore, the ΔH° is +891.2 kJ.
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Which of these is true about nuclear power?
A. Nuclear power is location specific
B. The neutrons released from the reactions are used to do work
C. Extracting energy from nuclear fuels is more expensive than
extracting energy from fossil fuels.
D. Nuclear fuel has a lower energy density than fossil fuels.
SUBN
extracting energy from nuclear fuels is more expensive than extracting energy from fossil fuels
a sealed vessel contains 15 kpa chlorine and 89 kpa fluorine in addition to some helium. the total pressure is 205 kpa. what is the partial pressure of heium in kpa
When more than one gas is present in a container, each gas exerts pressure, which is known as partial pressure. The partial pressure refers to the pressure of any gas inside the container. The partial pressure of the helium in kpa is 101.
Partial pressure is the pressure that one gas in a gas mixture will exert if it occupies the same volume on its own. In a mixture, every gas exerts a particular pressure. The partial pressures of the various gases in a mixture of an ideal gas add up to its total pressure.
The overall pressure exerted by a mixture of gases is equal to the sum of the partial pressures, according to Dalton's law of partial pressures.
P total= P₁ + P₂ + P₃ …
Here,
205 = pf + pCl + pHe
pHe = 205 - (pf + pCl)
pHe = 205 - (89+15) = 101 kpa
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an object is placed 25 cm in front of a lens of focal length 20 cm. 60 cm past the first lens is a second lens of focal length 25 cm. how far past the 25-cm lens does the final image form?
When an object is placed at a distance of 25 cm from the lens of focal length 20 cm, the image formed is virtual, erect and enlarged.
A virtual image is an image that appears to be behind the lens, and it is formed by light rays that do not actually pass through it but appear to diverge from it. The image is erect because it is the same size as the object, and it is enlarged because it is closer to the lens than the object.
The image formed by the first lens acts as the object for the second lens. The second lens has a focal length of 25 cm, and the final image is formed 60 cm past the second lens. Using the lens formula, the position of the image formed by the first lens is given by: 1/f = 1/v - 1/u Where f is the focal length of the lens, u is the object distance from the lens and v is the image distance from the lens.
Using the values given, we have: 1/20 = 1/v - 1/25So, 1/v = 1/20 + 1/25 = 9/1000v = 1000/9 = 111.11 cmThis means that the image formed by the first lens is 111.11 cm behind the first lens.
This image acts as the object for the second lens, and we can use the same formula to find the position of the final image. Using the same formula, we have: 1/25 = 1/v' - 1/111.11So, 1/v' = 1/25 + 1/111.11 = 4.84/1000v' = 1000/4.84 = 206.61 cm. Therefore, the final image is formed 206.61 cm past the first lens.
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2. An ideal machine would be a machine that did not have to work against the force of Type your ansiler here
Answer:
100% efficiency, this could only happen if there was not any friction Simple Machines Inclined plane, wedge, screw, lever, pulley and wheel and axel they are the most basic devices for making work easier
Explanation:
Microwave ovens heat food by exciting the quantum rotational frequencies of water and certain other molecules in the food sample. Most household microwave ovens emit radiation with a wavelength of 12.2 cm. a. What is the energy of a single photon of this radiation? b. Assuming all of the photon energy is converted into heat, how many photons of this radiation must be absorbed to warm 250. mL of water from 23.1 ºC to its boiling point?
Approximately 4.44 × [tex]10^{26[/tex] photons of this microwave radiation must be absorbed to warm 250 mL of water from 23.1 ºC to its boiling point .
a. To calculate the energy of a single photon of microwave radiation, we can use the equation E = hc/λ, where E is the energy of the photon, h is Planck's constant (6.626 × 10^-34 J·s), c is the speed of light (2.998 × 10^8 m/s), and λ is the wavelength of the radiation.
Converting the wavelength to meters, we have λ = 12.2 cm = 0.122 m.
Using the equation, E = (6.626 × 10^-34 J·s × 2.998 × 10^8 m/s) / 0.122 m = 1.632 × 10^-24 J.
Therefore, the energy of a single photon of this microwave radiation is 1.632 × 10^-24 J.
b. To calculate the number of photons required to warm 250 mL of water, we need to determine the heat energy required to raise the temperature from 23.1 ºC to its boiling point (100 ºC). The heat energy can be calculated using the formula Q = mcΔT, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water (4.184 J/g·ºC), and ΔT is the change in temperature.
First, we need to convert the mass of water to grams. Since 1 mL of water is approximately equal to 1 gram, 250 mL of water is equal to 250 grams.
Next, we calculate the heat energy:
Q = (250 g) × (4.184 J/g·ºC) × (100 ºC - 23.1 ºC) = 723,280 J.
To find the number of photons, we divide the total energy (723,280 J) by the energy of a single photon:
Number of photons = 723,280 J / (1.632 ×[tex]10^{-24[/tex] J) ≈ 4.44 × 10^26 photons.
Converting the wavelength to meters, we have λ = 12.2 cm = 0.122 m.
Using the equation, E = (6.626 × [tex]10^{-34[/tex] J·s × 2.998 × [tex]10^8[/tex] m/s) / 0.122 m = 1.632 × [tex]10^{-24[/tex] J.
Therefore, the energy of a single photon of this microwave radiation is 1.632 × [tex]10^{-24[/tex] J.
b. To calculate the number of photons required to warm 250 mL of water, we need to determine the heat energy required to raise the temperature from 23.1 ºC to its boiling point (100 ºC). The heat energy can be calculated using the formula Q = mcΔT, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water (4.184 J/g·ºC), and ΔT is the change in temperature.
First, we need to convert the mass of water to grams. Since 1 mL of water is approximately equal to 1 gram, 250 mL of water is equal to 250 grams.
Next, we calculate the heat energy:
Q = (250 g) × (4.184 J/g·ºC) × (100 ºC - 23.1 ºC) = 723,280 J.
To find the number of photons, we divide the total energy (723,280 J) by the energy of a single photon (1.632 × [tex]10^{-24[/tex] J):
Number of photons = 723,280 J / (1.632 × [tex]10^{-24[/tex] J) ≈ 4.44 × [tex]10^{26[/tex] photons.
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Which of the following properties is likely to decrease in a small, shallow tide pool at low tide? Multiple Choice Ο. oxygen Ο carbon dioxide Ο temperature Ο acidity Ο salinity
Oxygen is likely to decrease in a small, shallow tide pool at low tide.
When the tide goes out, the water in a tide pool becomes shallower and more exposed to the sun. This causes the water to heat up and the oxygen levels to decrease. The decrease in oxygen can be harmful to the organisms that live in the tide pool, as they need oxygen to survive.
The other properties are not likely to decrease in a small, shallow tide pool at low tide. Carbon dioxide levels may increase as the water heats up, but they are not likely to decrease.
Temperature is likely to increase, not decrease. Acidity is not likely to change significantly. Salinity is also not likely to change significantly.
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In the quantum mechanical description of a hydrogen atom, the electron is in a state that has orbital angular momentum squareroot 2h. What is the maximum possible ionization energy of this state of the atom? (a) 0.378 eV (b) 0.544 eV (c) 0.850 eV (d) 1.51 eV (e) 3.40 eV (f) none of the above answers
The ionization energy of an atom is the amount of energy required to completely remove an electron from its ground state. In the case of a hydrogen atom in a state with quantum number n, the ionization energy is given by the following equation: Ionization energy = -13.6 eV / n^2
For a hydrogen atom in a state with quantum number n=3, the ionization energy can be calculated as follows:
Ionization energy = -13.6 eV / 3^2 = -13.6 eV / 9 = -1.51 eV
The negative sign indicates that energy is required to remove the electron. Therefore, the largest possible ionization energy of the atom in this state is 1.51 eV.
In the quantum mechanical description, the ionization energy of a hydrogen atom is given by the formula:
Ionization Energy (IE) = -13.6 eV * (1/n²)
where n is the principal quantum number. In this case, n = 3.
IE = -13.6 eV * (1/3²) = -13.6 eV * (1/9) = -1.51 eV
Since the ionization energy is negative, the largest possible ionization energy is the least negative value. Therefore, the answer is (b) 1.51 eV.
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calculate the molecular mass of N2O3
the answer is 76.01g/mol
and the density:1.4g/cm³
PLEASE HELP FAST! NO LINKS
50 POINTS
Answer:
a
Explanation:
i am wrong haha
Answer:
are u out of time ?
Explanation:
which of the following laboratory procedures best illustrates the law of conservation of mass? (assume the product of the reaction includes the mass of any unused reactants.
The laboratory procedure best illustrates the law of conservation of mass is Option c "Burning 2.4 g of Mg in an open crucible to produce 2 g of MgO and unused reactants"
What is the law of conservation of mass?According to the law of conservation of mass, within an isolated system, mass remains constant and cannot be generated or annihilated, but rather undergoes transformations from one state to another.
For instance, by subjecting 32 grams of sulfur (S) and 56 grams of iron (Fe) to heat, they combine to form 88 grams of iron sulfide (FeS) alongside any unreacted starting materials.
Hence, the cumulative mass of the reactants, namely 32 grams of sulfur and 56 grams of iron, adds up to 88 grams, which aligns with the combined mass of the resultant product.
In this manner, it becomes evident that mass is neither eradicated nor brought into existence; rather, it seamlessly converts from one manifestation to another.
Consequently, the example involving the heating of 32 grams of sulfur and 56 grams of iron to yield 88 grams of iron sulfide, along with any remaining unreacted substances, aptly exemplifies the principle of the conservation of mass.
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Complete question:
Which of the following laboratory procedures best illustrates the law of conservation of mass?
a. Calculating the number of atoms in 11 g of Na
Using 250 g of impure Cu to obtain 200 g of pure Cu
b. Heating 32 g of S and 56 g of Fe to produce 88 g of FeS and unused reactants
c. Burning 2.4 g of Mg in an open crucible to produce 2 g of MgO and unused reactants
8. Extinction of a species could result from
*
1
evolution of a type of behavior that produces greater reproductive success
synthesis of a hormone that controls cellular communication
limited genetic variability in the species
fewer unfavorable mutations in the species
Answer:
Limited Genetic Variability
Explanation:
its 3
Which of these reagents would not react with stearic acid? H2, Ni SOCl2 CH3MgI NH3/H2O LAH
The reagent that would not react with stearic acid is Lithium Aluminium Hydride (LAH).
Stearic acid is a saturated fatty acid consisting of eighteen carbon atoms. In organic chemistry, stearic acid is a common reagent that reacts with other reagents to create a range of useful compounds. In this question, we will discuss which of the following reagents would not react with stearic acid.The four reagents given are: H2, Ni; SOCl2; CH3MgI; NH3/H2O; and LAH.LAH (Lithium Aluminium Hydride) is the reagent that would not react with stearic acid. Lithium Aluminium Hydride (LAH) is a strong reducing agent that reduces carboxylic acids into alcohols. However, stearic acid, being a saturated fatty acid, does not contain any double bonds that can be reduced by the strong reducing agent LAH. Therefore, it does not react with stearic acid.The other reagents mentioned such as H2, Ni, SOCl2, CH3MgI, NH3/H2O all react with stearic acid, and would give different products depending on the reaction conditions and reagents used. Hence, the reagent that would not react with stearic acid is Lithium Aluminium Hydride (LAH).
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Characteristic orange light produced by sodium in a fl ame is due to an intense emission called the sodium D line, which is actually a doublet, with wavelengths (measured in vacuum) of 589.157 88 and 589.755 37 nm. The index of refraction of air at a wavelength near 589 nm is 1.000 292 6. Calculate the frequency, wavelength, and wavenumber of each component of the D line, measured in air.
Answer:
a. i. 5.092 × 10²⁰ Hz ii. 588.98554 nm ii. 10667809.11 rad/m
b. i. 5.087 × 10²⁰ Hz ii. 589.58286 nm iii. 10657001.3 rad/m
Explanation:
refractive index, n = λ/λ' where λ = wavelength in vacuum = and λ' = wavelength in air
a. For λ = 589.15788 nm,
i. Frequency,
f = c/λ where c = speed of light in vacuum = 3 × 10⁸ m/s and λ = 589.15788 nm = 589.15788 × 10⁻⁹ m
So, f = 3 × 10⁸ m/s ÷ 589.15788 × 10⁻⁹ m
= 0.005092 × 10¹⁷ /s
= 5.092 × 10²⁰ /s
= 5.092 × 10²⁰ Hz
ii. Wavelength,
Since n = λ/λ' where λ = wavelength in vacuum = and λ' = wavelength in air
and n = 1.0002926
λ' = λ/n
= 589.15788 nm/1.0002926
= 588.98554 nm
iii. Wave number, kk = 2π/λ'
= 2π/588.98554 nm
= 0.01066780911 rad/nm
= 0.01066780911 rad/nm × 10⁹ nm/1m
= 10667809.11 rad/m
b. For λ = 589.755 37 nm.,
i. Frequency,
f = c/λ where c = speed of light in vacuum = 3 × 10⁸ m/s and λ = 589.755 37 nm. = 589.755 37 × 10⁻⁹ m
So, f = 3 × 10⁸ m/s ÷ 589.755 37 × 10⁻⁹ m
= 0.005087 × 10¹⁷ /s
= 5.087 × 10²⁰ /s
= 5.087 × 10²⁰ Hz
ii. Wavelength,
Since n = λ/λ' where λ = wavelength in vacuum = and λ' = wavelength in air
and n = 1.0002926
λ' = λ/n
= 589.755 37 nm./1.0002926
= 589.58286 nm
iii. Wave number, kk = 2π/λ'
= 2π/589.58286 nm
= 0.0106570013 rad/nm
= 0.0106570013 rad/nm × 10⁹ nm/1m
= 10657001.3 rad/m
consider the following reaction: 2 c2h6(g) 7 o2(g) → 4 co2(g) 6 h2o(g) at stp, what is the total volume of co2 formed when 6.0 liters of c2h6 are combusted?
When 6.0 liters of [tex]C_2H_6[/tex] are combusted, the total volume of [tex]CO_2[/tex] formed at STP is approximately 12.02 liters.
To determine the total volume of [tex]CO_2[/tex] formed when 6.0 liters of [tex]C_2H_6[/tex] are combusted, we need to use the balanced chemical equation and stoichiometry.
From the balanced chemical equation:
2 [tex]C_2H_6[/tex](g) + 7 O2(g) → 4 [tex]CO_2[/tex](g) + 6 H2O(g)
We can see that the molar ratio between [tex]C_2H_6[/tex] and [tex]CO_2[/tex] is 2:4, which simplifies to 1:2. This means that for every 2 moles of [tex]C_2H_6[/tex] combusted, 4 moles of [tex]CO_2[/tex] are produced.
To solve this problem, we need to convert the given volume of [tex]C_2H_6[/tex] to moles and then use the stoichiometry to determine the volume of [tex]CO_2[/tex] produced.
Step 1: Convert volume of [tex]C_2H_6[/tex] to moles:
Using the ideal gas law, PV = nRT, at STP (Standard Temperature and Pressure), one mole of any ideal gas occupies 22.4 liters. Therefore, 6.0 liters of [tex]C_2H_6[/tex] is equal to 6.0/22.4 = 0.268 moles of [tex]C_2H_6[/tex].
Step 2: Apply stoichiometry to find moles of [tex]CO_2[/tex]:
Since the molar ratio between [tex]C_2H_6[/tex] and [tex]CO_2[/tex] is 1:2, we multiply the moles of C2H6 by the stoichiometric coefficient ratio:
0.268 moles of [tex]C_2H_6[/tex] * (4 moles [tex]CO_2[/tex] / 2 moles [tex]C_2H_6[/tex]) = 0.536 moles of CO2.
Step 3: Convert moles of [tex]CO_2[/tex] to volume:
At STP, 1 mole of any ideal gas occupies 22.4 liters. Therefore, 0.536 moles of [tex]CO_2[/tex] is equal to 0.536 * 22.4 = 12.02 liters of [tex]CO_2[/tex].
Thus, when 6.0 liters of [tex]C_2H_6[/tex] are combusted, the total volume of [tex]CO_2[/tex] formed at STP is approximately 12.02 liters.
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5 The atoms of a certain element each contain 36 protons and 8 valence electrons. Which
statement correctly identifies this element and describes its chemical reactivity? (8.58)
A The element is krypton, Kr, and it is not very reactive.
B The element is chlorine, Cl, and it is highly reactive.
C The element is krypton, Kr, and it is highly reactive.
D The element is chlorine, Cl, and it is not very reactive.
Answer:
dfdfaxc b
Explanation:
this ka value is listed as 6.6 x 10−4. is hf classified as a strong or weak acid? briefly justify your answer.
HF is classified as a weak acid because it only partially dissociates into ions when dissolved in water, and its relatively low Ka value confirms its limited ionization and weaker acid properties.
Hydrofluoric acid (HF) is classified as a weak acid. The classification of acids as strong or weak depends on their ability to dissociate into ions when dissolved in water. Strong acids readily dissociate completely into ions, while weak acids only partially dissociate.
HF is a weak acid because it does not fully dissociate into H+ ions and F- ions when dissolved in water. Instead, only a fraction of HF molecules dissociate, resulting in a small concentration of H+ ions in solution.
This limited dissociation is due to the strength of the bond between hydrogen and fluorine in HF. The bond is relatively strong, requiring a significant amount of energy to break, and as a result, the dissociation of HF is incomplete.
The Ka value of HF, listed as 6.6 x 10^−4, further supports its classification as a weak acid. Ka is the acid dissociation constant and provides a measure of the extent of dissociation of an acid in solution. A lower Ka value indicates weaker acid strength and a smaller degree of dissociation.
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