When, 1 mole of methane at 25°C in a 244.6 mL bulb would exert a pressure of 2.79 atm assuming it behaves as a real gas and obeys the van der Waals equation.
First, let's calculate the pressure exerted by 1 mole of methane at 25°C assuming it behaves as an ideal gas;
We can use Ideal Gas Law to calculate the pressure;
PV = nRT
where P is pressure, V is volume, n is number of moles, R is gas constant, and T is the temperature in Kelvin.
Converting the volume of the bulb to liters and the temperature to Kelvin;
V = 244.6 mL = 0.2446 L
T = 25°C = 298 K
For 1 mole of methane;
n = 1 mole
The gas constant for the Ideal Gas Law is;
R = 0.0821 L·atm/(mol·K)
Substituting the values into Ideal Gas Law equation;
P = (nRT) / V
P = (1 mole) x (0.0821 L·atm/(mol·K)) x (298 K) / (0.2446 L)
P = 3.24 atm
Therefore, 1 mole of methane at 25°C in a 244.6 mL bulb would exert a pressure of 3.24 atm assuming it behaves as an ideal gas.
Now, let's calculate the pressure exerted by 1 mole of methane at 25°C assuming it behaves as a real gas and obeys the van der Waals equation;
The van der Waals equation is;
(P + a(n/V)²) (V - nb) = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is the temperature in Kelvin, a is a constant that takes into account the attractive forces between molecules, b is a constant that takes into account the volume of the molecules, and (n/V) is the molar density.
For methane, the values of the van der Waals constants are;
a = 2.253 atm L²/mol
b = 0.0428 L/mol
Substituting the values into the van der Waals equation and solving for P;
P = (nRT / (V - nb)) - (a(n/V)² / V²)
P = (1 mole) x (0.0821 L·atm/(mol·K)) x (298 K) / (0.2446 L - (0.0428 L/mol x 1 mole)) - (2.253 atm L²/mol² / (0.2446 L)²)
P = 2.79 atm
Therefore, the pressure is 2.79 atm.
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The first law the of thermodynamic also known as the "Law of Conservation of Mass" states that
A. heat changes occur during chemical and physical changes.
B. there are two types of energy, kinetic and potential
C. In any chemical or physical change, energy cannot be created or destroyed, only transformed in form.
D. energy is the capacity to do work or to supply heat
In any chemical or physical change, energy cannot be created or destroyed, only transformed in form.
option C.
What is the first law of thermodynamics?The first law of thermodynamics is known as the law of Conservation of Energy.
This law states that energy can neither be created nor destroyed but can be converted from one form to another.
So the first law of thermodynamics is not known as the "Law of Conservation of Mass", but rather as the "Law of Conservation of Energy".
The statement that best corresponds to the first law of thermodynamics is option C: "In any chemical or physical change, energy cannot be created or destroyed, only transformed in form."
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Lab: solubility
lap report essay! need help now!! please!!
Solubility is the ability of a substance to dissolve in a liquid to form a solution. It is an important physical property of substances that must be taken into account.
What is substance?Substance is a term used to refer to a material that has mass and occupies space. It is something that has physical properties that can be identified and measured. Substance can be either a solid, liquid, gas, or plasma. Examples of substances include solids such as iron, liquids like water, gases like oxygen, and plasma like fire.
Solubility is the ability of a substance to dissolve in a liquid to form a solution. It is an important physical property of substances that must be taken into account when studying topics such as solute-solvent interactions, chemical reactions, and phase changes. In this lab, we will be exploring the solubility of various substances, including sugar, salt, and baking soda, to determine how their solubility is affected by changes in temperature. To begin, we will measure out one gram of each substance into separate test tubes and dissolve them in 10mL of water. We will then place each test tube into a beaker of hot (90°C) and cold (0°C) water and observe the differences in solubility. We will use a thermometer to measure the temperatures of each beaker and record the results. Next, we will measure out two grams of each substance and repeat the same procedure as before. We will then measure out five grams of each substance and repeat the experiments. We will record our observations and results for each experiment.
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What is the percentage composition of each element in dinitrogen monoxide, n2o? (5 points) a 58.32% n; 41.68% o b 60.55% n; 39.45% o c 63.64% n; 36.36% o d 62.66% n; 37.34% o
The percentage composition of each element in dinitrogen monoxide is 63.64% N; 36.36% O.
To determine the percentage composition of each element in dinitrogen monoxide (N2O), we need to calculate the molar mass of the compound and the molar mass of each element.
Molar mass of N2O = (2 x molar mass of N) + molar mass of O
= (2 x 14.01 g/mol) + 16.00 g/mol
= 44.02 g/mol
The percentage composition of each element can be calculated as follows:
Percentage composition of N = (2 x molar mass of N) / molar mass of N2O x 100%
= (2 x 14.01 g/mol) / 44.02 g/mol x 100%
= 63.64%
Percentage composition of O = molar mass of O / molar mass of N2O x 100%
= 16.00 g/mol / 44.02 g/mol x 100%
= 36.36%
Therefore, the correct answer is option c: 63.64% N; 36.36% O.
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What is the resultant pressure if 3. 5 mol of
ideal gas at 273 K and 0. 96 atm in a closed
container of constant volume is heated to
619 K? Answer in units of atm
The resultant pressure after heating the ideal gas to 619 K is approximately 2.17 atm.
To find the resultant pressure of the ideal gas after being heated, we can use the Ideal Gas Law formula, which is:
PV = nRT
where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Since the volume is constant, we can compare the initial and final states of the gas using the following equation:
P1/T1 = P2/T2
Given the initial conditions: P1 = 0.96 atm, T1 = 273 K, and the final temperature T2 = 619 K. We need to find the final pressure P2.
0.96 atm / 273 K = P2 / 619 K
Now, solve for P2:
P2 = (0.96 atm * 619 K) / 273 K
P2 ≈ 2.17 atm
Therefore, the resultant pressure after heating the ideal gas to 619 K is approximately 2.17 atm.
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2C6H6(g) + 150₂(g) → 12CO₂(g) + 6H₂O(g)
What volume of water vapor, in liters,
forms at STP when 1 L of C6H6 reacts
with oxygen?
[?] LH₂O
Volume (L) H₂O
Enter
Answer:
3 L
Explanation:
You want to know the volume of water vapor produced at STP when 1 L of C₆H₆ reacts with oxygen.
ReactantsThe given balanced reaction equation tells us that 6 moles of water vapor are produced from each 2 moles of C₆H₆. At STP, the volume of water vapor will be 3 times the volume of C₆H₆.
3 liters of water vapor are produced by reacting 1 liter of C₆H₆ with oxygen.
Food web
wolf
rabbit
deer
plants
i
a student drew a basic food web of a forest ecosystem.
part a: describe what the arrows represent in the food web
part b: explain why the ecosystem supports fewer wolves than deer
Part a: The arrows in the food web represent the flow of energy and nutrients.
Part b: Ecosystem supports fewer wolves than deer because wolves are at a higher trophic level in food chain.
Part a: The movement of nutrients and energy from one organism to another is depicted by arrows in food chain. They specifically point to the direction of matter and energy transfer when one organism feeds another.
Part b: Due to wolves' higher trophic level in food chain, the ecology can only support a smaller population of them than deer. Due to energy loss from heat and metabolism, the amount of energy available at each level of the food chain diminishes as it progresses up the chain.
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Write the following chemical reactions and balance:
Potassium reacts with sodium oxide to produce potassium oxide and sodium
The chemical reaction is
2 K + Na2O -> K2O + 2 Na
The given chemical equation represents a reaction between potassium (K) and sodium oxide (Na2O). The products formed in this reaction are potassium oxide (K2O) and sodium (Na).
On the reactant side, we have two atoms of potassium and two atoms of sodium, while on the product side, we have two atoms of potassium and two atoms of sodium as well.
Therefore, the equation is already balanced with respect to the number of potassium and sodium atoms.
However, we need to balance the oxygen atoms. On the reactant side, we have one molecule of Na2O, which contains two atoms of oxygen. On the product side, we have one molecule of K2O, which also contains two atoms of oxygen. Thus, the equation is balanced.
Finally, we can write the balanced equation as:
2 K + Na2O → K2O + 2 Na
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What is the concentration of KBr in a solution prepared by mixing 0. 200 L of 0. 053 M KBr with
0. 550 L of 0. 078 M KBr?
To find the concentration of KBr in the solution, we can use the formula:
C1V1 + C2V2 = C3V3
where C1 and V1 are the concentration and volume of the first solution, C2 and V2 are the concentration and volume of the second solution, and C3 and V3 are the concentration and volume of the resulting mixed solution.
Plugging in the given values, we get:
(0.053 M x 0.200 L) + (0.078 M x 0.550 L) / (0.200 L + 0.550 L)
= (0.0106 mol + 0.0429 mol) / 0.750 L
= 0.0587 M
Therefore, the concentration of KBr in the final solution is 0.0587 M.
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An aircraft flying from a region of higher air pressure towards a region of lower air pressure will _____ altitude, and the aircraft’s pressure altimeter will read an altitude _____ than the plane’s true elevation, unless corrections are made to the altimeter
An aircraft flying from a region of higher air pressure towards a region of lower air pressure will experience a change in altitude, and the aircraft's pressure altimeter will read an altitude different than the plane's true elevation, unless corrections are made to the altimeter.
When an aircraft moves from an area of higher air pressure to an area of lower air pressure, the aircraft will generally gain altitude. This occurs because the pressure difference causes the air to become less dense, allowing the aircraft to rise more easily. As a result, the aircraft's wings will generate more lift, enabling it to climb higher.
However, the pressure altimeter, which measures an aircraft's altitude based on the surrounding air pressure, will not accurately reflect the plane's true elevation in this situation. The altimeter will typically read an altitude lower than the actual elevation of the aircraft.
This discrepancy occurs because the altimeter is calibrated for a standard pressure setting and will not account for variations in air pressure without adjustments.
To ensure accurate altitude readings, pilots must make corrections to the altimeter by setting the appropriate pressure setting for the area they are flying in, known as the "altimeter setting." This setting can be obtained from air traffic control or other aviation weather sources.
By inputting the correct altimeter setting, the pressure altimeter will provide a more accurate altitude reading, reflecting the plane's true elevation.
In summary, an aircraft flying from a region of higher air pressure towards a region of lower air pressure will gain altitude, and the aircraft's pressure altimeter will read an altitude lower than the plane's true elevation unless corrections are made to the altimeter.
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A truck tire is filled with 1. 85 moles of gas at a pressure of 2. 1 atm and a temperature of 300 K. What is the volume of the gas in the tire?
The volume of the gas in the truck tire is approximately 44.2 L.
The ideal gas law equation can be used to solve for the volume of the gas in the tire:
PV = nRT
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas.
Substituting the given values into the equation, we get:
(2.1 atm)(V) = (1.85 mol)(0.08206 L atm/mol K)(300 K)
Solving for V, we get:
V = (1.85 mol)(0.08206 L atm/mol K)(300 K)/(2.1 atm) ≈ 44.2 L
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2.
for the reaction c + 2h2 - ch4, how many grams of hydrogen are required
to produce 0.6 moles of methane, ch4 ?
cannu help em do the whole paper
1.21 grams of hydrogen are required to produce 0.6 moles of methane (CH₄) in the given reaction.
The given reaction is:
C + 2H₂ → CH₄
We can see that 2 moles of hydrogen (H₂) are required to produce 1 mole of methane (CH₄) according to the balanced chemical equation. Therefore, to produce 0.6 moles of methane, we will need 2 times as many moles of hydrogen, which is:
number of moles of hydrogen = 2 × number of moles of methane
number of moles of hydrogen = 2 × 0.6 moles
number of moles of hydrogen = 1.2 moles
To convert the number of moles of hydrogen to grams, we need to use the molar mass of hydrogen, which is approximately 1.008 g/mol. Thus, the mass of hydrogen required can be calculated as:
mass of hydrogen = number of moles of hydrogen × molar mass of hydrogen
mass of hydrogen = 1.2 moles × 1.008 g/mol
mass of hydrogen = 1.21 g
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The complete question is:
For the reaction C+2H₂ → CH₄, how many grams of hydrogen are required to produce 0.6 moles of methane, CH₄?
ENDOTHERMIC
During this chemical reaction energy is absorbed. In the chemistry lab, this would be indicated by a decrease in temperature or if the reaction took place in a test tube, the test tube would feel colder to the touch. Reactions like this one absorb energy because
The reactants have less potential energy than the products
In chemistry, a chemical reaction can be classified as either endothermic or exothermic based on whether the reaction releases or absorbs energy, respectively. An endothermic reaction is one in which energy is absorbed from the surroundings, resulting in an increase in the internal energy of the system.
The term potential energy refers to the stored energy within a system due to the position or configuration of the particles that make up that system. In the case of a chemical reaction, potential energy is stored within the chemical bonds between atoms and molecules.
In an endothermic reaction, the reactants have less potential energy than the products. This is because energy is required to break the chemical bonds in the reactants, which absorbs energy from the surroundings. As a result, the products have higher potential energy than the reactants because they have absorbed energy from the surroundings during the reaction.
Examples of endothermic reactions include the process of melting ice, where energy is absorbed from the surroundings to break the bonds between water molecules, and the reaction between baking soda and vinegar, where energy is absorbed to break the bonds between the molecules of the reactants.
In summary, endothermic reactions are those that require energy to be absorbed from the surroundings. This results in the products of the reaction having more potential energy than the reactants, which have had their bonds broken and therefore have less potential energy.
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what reactions take place during the electrolysis of water? group of answer choices hydrogen is reduced; oxygen is oxidized oxygen is reduced; hydrogen is oxidized. oxygen gas is reduced; water is oxidized. water is reduced; oxygen gas is oxidized. both oxygen and hydrogen are oxidized and reduced.\
The electrolysis of water involves the passage of an electric current through water, which leads to the splitting of water molecules into their constituent elements, hydrogen and oxygen. The correct answer is (a)
This process occurs through two simultaneous half-reactions at the cathode and anode of the electrolysis cell.
At the cathode, hydrogen ions (H+) are reduced to hydrogen gas (H2) as they gain electrons from the electrode: [tex]2H+ + 2e-[/tex]→ [tex]H2[/tex]
At the anode, water molecules (H2O) are oxidized to oxygen gas (O2) and positively charged hydrogen ions (H+): [tex]2H2O[/tex]→ [tex]O2 + 4H+ + 4e-[/tex]
Therefore, the correct answer is (a) hydrogen is reduced; oxygen is oxidized. During the electrolysis of water, hydrogen is reduced at the cathode, while oxygen is oxidized at the anode.
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--The complete question is, what reactions take place during the electrolysis of water?
group of answer choices
a. hydrogen is reduced; oxygen is oxidized
b. oxygen is reduced; hydrogen is oxidized
c. . oxygen gas is reduced; water is oxidized.
d. water is reduced; oxygen gas is oxidized.
e. both oxygen and hydrogen are oxidized and reduced.--
Bombardment of boron-10 with a neutron produces a hydrogen-1 atom and another nuclide. what is this nuclide?
The nuclide produced when boron-10 is bombarded with a neutron is lithium-7 besides a hydrogen-1 atom.
When boron-10 is bombarded with a neutron, it undergoes a nuclear reaction called neutron capture, which produces lithium-7 and a highly excited compound nucleus.
The compound nucleus then emits an alpha particle and a gamma ray to reach a stable state. This reaction is commonly used in nuclear reactors to produce tritium, which is a fuel for fusion reactions.
Lithium-7 is a stable isotope of lithium and is commonly used in nuclear reactions as a neutron detector.
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Calculate the pressure exerted by 200. g of A r in a rigid 4.50 L container at 21.0 ˚ C . Assume ideal gas behavior. Note that R = 0.08206 L ⋅ atm K ⋅ mol .
The pressure exerted by 200 g of Ar in a rigid 4.50 L container at 21.0 ˚ C would be 19.6 atm.
Ideal gas problemTo calculate the pressure exerted by the Argon gas, we can use the ideal gas law:
PV = nRT
where
P is the pressureV is the volumen is the number of molesR is the ideal gas constantT is the temperature in Kelvin.First, we need to determine the number of moles of Argon gas present:
n = mass / molar massn = 200/39.95 = 5.004 molesNext, we convert the volume and temperature:
V = 4.50 L = 0.00450 [tex]m^3[/tex]T = 21.0 ˚C + 273.15 = 294.15 KNow we can substitute the values into the ideal gas law and solve for P:
P = nRT/VP = (5.004) x (0.08206) x (294.15) / (0.00450)P = 19.6 atmIn other words, the pressure exerted by 200 g of Argon gas in a 4.50 L container at 21.0 ˚C is 19.6 atm.
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Match each decimal number to its equivalent in scientific notation
Decimal number to its equivalent in scientific notation: 15 = 1.5 × 102, 0.015 = 1.5 × 10-2, 0.15 = 1.5 × 10-1, 150 = 1.5 × 101 and 1.5 = 1.5 × 100.
What is notation?Notation is a system of symbols used to represent a set of ideas or concepts. It is used to communicate complex musical, mathematical, and scientific concepts. Notation helps to make information easier to understand and is widely used in many fields. Notation can range from simple symbols such as musical notes, to complex formulas and equations used in mathematics and science. It allows for the efficient and organized communication of ideas and can be used to represent abstract concepts. Notation makes it easier to understand and learn complex topics, and is an important tool for communication.
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Complete Question:
Match each decimal number to its equivalent in scientific notation
15
1.5
0.015
0.15
150
1.5 × 10-2
1.5 × 101
1.5 × 102
1.5 × 100
1.5 × 10-1
What is the temperature of 5. 16g of helium gas at a pressure of 785 mmHg that occupies a 1. 00 L container?
The temperature of 5.16g of helium gas at a pressure of 785 mmHg that occupies a 1.00 L container is approximately 248 Kelvin.
The temperature of 5.16g of helium gas at a pressure of 785 mmHg that occupies a 1.00 L container can be calculated using the ideal gas law equation:
PV = nRT
where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.
Convert the pressure from mmHg to atm by dividing by 760 mmHg/atm:
785 mmHg ÷ 760 mmHg/atm = 1.033 atm
Calculate the number of moles of helium gas using its molecular weight:
molecular weight of helium = 4.00 g/mol
moles of helium = 5.16 g ÷ 4.00 g/mol = 1.29 mol
Now, we can rearrange the ideal gas law equation to solve for T:
T = PV ÷ nR
T = (1.033 atm)(1.00 L) ÷ (1.29 mol)(0.0821 L atm/mol K)
T = 248 K
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Determine the mass of ammonium chloride, NH4Cl, required to prepare 0. 250 L of a 0. 35 M solution of ammonium chloride
We need 4.68 grams of ammonium chloride to prepare 0.250 L of a 0.35 M solution.
To determine the mass of ammonium chloride needed to prepare a 0.35 M solution in 0.250 L of solution, we can use the formula:
moles of solute = concentration x volume
We can rearrange this formula to solve for the mass of solute needed:
mass of solute = moles of solute x molar mass of solute
First, we need to calculate the number of moles of ammonium chloride needed for this solution:
moles of NH4Cl = concentration x volume
moles of NH4Cl = 0.35 mol/L x 0.250 L
moles of NH4Cl = 0.0875 mol
Next, we need to calculate the molar mass of ammonium chloride:
Molar mass of NH4Cl = 14.01 g/mol (mass of N) + 4(1.01 g/mol) (mass of H) + 35.45 g/mol (mass of Cl)
Molar mass of NH4Cl = 53.49 g/mol
Finally, we can calculate the mass of ammonium chloride needed:
mass of NH4Cl = moles of NH4Cl x molar mass of NH4Cl
mass of NH4Cl = 0.0875 mol x 53.49 g/mol
mass of NH4Cl = 4.68 g
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A system starts with a multiplicity of 2000. two kj of heat are transferred into the system reversibly at 298 k. what is the multiplicity now
In statistical mechanics, multiplicity refers to the number of microstates corresponding to a given macrostate of a system. Microstates represent the different ways in which the system's particles can be arranged while still satisfying the constraints imposed by the macrostate (e.g., total energy, volume, etc.).
To determine the change in multiplicity due to the transfer of heat, we typically need to know more about the system's properties, such as the number of particles, the energy levels available to those particles, and any other relevant information about the system's configuration.
Without further information, it is not possible to calculate the precise change in multiplicity resulting from the transfer of two kilojoules of heat at 298 Kelvin. Multiplicity is a system-specific property that depends on the unique characteristics and constraints of the system under consideration.
If you can provide additional details about the system, its properties, or the specific problem you are working on, I'll be happy to assist you further in understanding or calculating the multiplicity.
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what does the size of kf indicate regarding the stability of transition metal complexes? group of answer choices the large values of kf indicate that transition metal complexes are often very stable. the tiny values of kf indicate that transition metal complexes are often very unstable. the kf values have nothing to do with stability.
The size of the Kf indicate regarding the stability of the transition metal complexes is the large values of the Kf indicate that transition metal complexes are often very stable.
The larger the value of the Kf of the complex ion, the more stable will be the transition metal complexes. Due to the how large formation constants are often is not uncommon to listed as the logarithms in the form of the log Kf.
The Kf values that are the very large in the magnitude for the complex ion formation that indicate that the reaction is heavily favors the products. The complex ions that are the poorly formed and this value will be the very small.
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Lab Report: Titration
HELP!!! I don’t understand this!! Anyone done this before??
A titration is a laboratory technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration.
In a typical titration, a burette is filled with the known solution (titrant) and is gradually added to the unknown solution (analyte) in a flask, until the reaction between the two solutions is complete.
A lab report on titration should include the following sections:
1. Introduction: Provide an overview of the purpose of the experiment and the concept of titration.
2. Materials and Methods: List the chemicals, glassware, and equipment used in the experiment, and describe the step-by-step procedure followed during the titration.
3. Results: Present your raw data, including initial and final burette readings and the volume of titrant used. Calculate the concentration of the unknown solution using the stoichiometry of the reaction and the known concentration of the titrant.
4. Discussion: Analyze your results and explain any discrepancies or sources of error that may have occurred during the experiment.
5. Conclusion: Summarize the main findings of the experiment and emphasize their significance.
Remember to always follow any specific guidelines provided by your instructor or institution.
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You have a sample of gas with a volume of 22. 4 L, a pressure of 1663 mmHg, and a temperature of 83 ºC. How many moles of gas are in the sample?
In your gas sample, there are approximately 1.21 moles of gas.
To determine the number of moles of gas in the sample, you can use the ideal gas law formula: PV = nRT. In this formula, P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.
1. Convert the pressure to atm: (1663 mmHg) * (1 atm/760 mmHg) = 2.19 atm.
2. Convert the temperature to Kelvin: (83°C) + 273.15 = 356.15 K.
3. Rearrange the formula to solve for n: n = PV/RT.
4. Plug in the values: n = (2.19 atm) * (22.4 L) / (0.0821 L atm/mol K) * (356.15 K).
5. Calculate the number of moles: n = 1.21 moles (rounded to two decimal places).
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1. A gas takes up a volume of 10 ml, has a pressure of 6 atm, and a temperature of 100 K. What is the new volume of the gas at stp?
2. The gas in an aerosol can is under a pressure of 8 atm at a temperature of 45 C. It is dangerous to dispose of an aerosol can by incineration. (V constant)What would the pressure in the aerosol can be at a temperature of 60 C ?
3. A sample of nitrogen occupies a volume of 600mL at 20 C. What volume will it occupy at STP?(P constant)
The new volume of the gas at STP is 16.36 ml, the pressure in the aerosol at the 60 degree temperature is 9.46 atm and the volume that it will occupy is 557.66 m.
1. We must apply the combined gas law equation to determine the new volume of the gas at STP,
P₁V₁/T₁ = P₂V₂/T₂.
At STP, the pressure is 1 atm and the temperature is 273 K.
Plugging in the values, we get:
6 atm * 10 ml / 100 K = 1 atm * V₂/273 K
V₂ = 16.36 ml (rounded to two decimal places)
2. To find the new pressure of the gas in the aerosol can at a temperature of 60 C, we can use the ideal gas law equation: PV = nRT, where n is the number of moles of gas and R is the gas constant,
P₁/T₁ = P₂/T₂.
Plugging in the values, we get:
8 atm/(45 + 273) K = P₂/(60 + 273) K
P₂ = 9.46 atm (rounded to two decimal places)
3. Using the relation, V₁/T₁ = V₂/T₂. At STP, the temperature is 273 K.
Plugging in the values, we get:
600 ml / (20 + 273) K = V2 / 273 K
V₂ = 557.66 ml (rounded to two decimal places)
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Identify whether the mixing of each pair of solutions results in a buffer Check all that apply. The mixing of Check all that apply. 100. 0 mL of 0. 10 M NH3 and 70. 0 mL of 0. 15 MNH4Cl 50. 0 mL of 0. 10 MHCl and 35. 0 mL of 0. 150 MNaOH 125. 0 mL of 0. 17 MNH3 and 160. 0 mL of 0. 20 MNaOH 155. 0 mL of 0. 10 MNH3 and 150. 0 mL of 0. 11 MNaOH 50. 0 mL of 0. 20 MHF and 20. 0 mL of 0. 20 MNaOH will result in a buffer
Only the first pair (100.0 mL of 0.10 M NH3 and 70.0 mL of 0.15 M NH4Cl) will result in a buffer solution.
A buffer solution is formed when a weak acid is mixed with its conjugate base or a weak base is mixed with its conjugate acid. Let's analyze each pair of solutions:
1. 100.0 mL of 0.10 M NH3 and 70.0 mL of 0.15 M NH4Cl: This mixture is a weak base (NH3) with its conjugate acid (NH4Cl). Therefore, it will result in a buffer.
2. 50.0 mL of 0.10 M HCl and 35.0 mL of 0.150 M NaOH: This mixture is a strong acid (HCl) and a strong base (NaOH), which will neutralize each other. It will not result in a buffer.
3. 125.0 mL of 0.17 M NH3 and 160.0 mL of 0.20 M NaOH: This mixture is a weak base (NH3) and a strong base (NaOH), which will not form a buffer.
4. 155.0 mL of 0.10 M NH3 and 150.0 mL of 0.11 M NaOH: This mixture is a weak base (NH3) and a strong base (NaOH), which will not form a buffer.
5. 50.0 mL of 0.20 M HF and 20.0 mL of 0.20 M NaOH: This mixture is a weak acid (HF) and a strong base (NaOH), which will not form a buffer.
In conclusion, only the first pair (100.0 mL of 0.10 M NH3 and 70.0 mL of 0.15 M NH4Cl) will result in a buffer solution.
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What is the mass of 6. 02 x 10^22 molecules of fluorine gas at stop
Answer:
3.7996 g
Explanation:
From the number of molecules we can find the number of moles of Fluorine gas (F2) and multiply by Fluorine Gas' molecular weight. Fluorine gas is F2,
F = 18.998g/mol.
F2 (g) = 18.998*2 =37.996g F2(g)/mol
1 mol = 6.02 x 10^23 molecules
[tex]\frac{6.02*10^{22} molecules}{6.02*10^{23}molecules / mole }\\\\ = 0.1 mole[/tex]
0.1 mol x 37.996g F2 (g) / mol
3.7996 g F2
If the boiling point of ethanol went up 6. 8 degrees, how many grams of PbCl4 were added to 2700 grams of ethanol? round to nearest tenth
Approximately 5272.2 grams of PbCl4 were added to 2700 grams of ethanol to increase the boiling point by 6.8 degrees.
To determine the grams of PbCl4 added to 2700 grams of ethanol, causing the boiling point to increase by 6.8 degrees, we will use the molality-based boiling point elevation formula, which is:
ΔTb = Kb * m
Here, ΔTb is the change in boiling point (6.8 degrees), Kb is the molal boiling point elevation constant of ethanol (1.22 °C kg/mol), and m is the molality (moles of solute per kg of solvent).
First, we need to find the molality (m) of the solution:
6.8 = 1.22 * m
m = 6.8 / 1.22 ≈ 5.57 mol/kg
Now, we can calculate the moles of PbCl4 added to the ethanol:
5.57 mol/kg * (2700 g / 1000 g/kg) ≈ 15.03 mol of PbCl4
Next, we need to find the molar mass of PbCl4:
Pb: 207.2 g/mol
Cl: 35.45 g/mol
Molar mass of PbCl4 = 207.2 + (4 * 35.45) ≈ 350.6 g/mol
Finally, we can calculate the grams of PbCl4 added to the ethanol:
15.03 mol * 350.6 g/mol ≈ 5272.2 g
Therefore, approximately 5272.2 grams of PbCl4 were added to 2700 grams of ethanol to increase the boiling point by 6.8 degrees.
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Question 2:heat (5 points) a. Describe the following heat equations and identify the indicated variables. (3 points) I. Q= mct; identify c. (1 point) ii. Q=ml vapor; identify l vapor (1 point) iii. Q= ml fusion; identify l fusion (1 point)
I. Q = mct; c is the specific heat capacity, II. Q = ml vapor; l vapor is the latent heat of vaporization,III. Q = ml fusion; l fusion is the latent heat of fusion.
What is vaporization?Vaporization is the process of a substance changing from its liquid form to its gaseous form. It occurs when the substance absorbs heat, causing its molecules to move faster and further apart, converting it from a liquid to a gas. Vaporization is a process that occurs when a liquid is heated to its boiling point and then cooled, causing the molecules to break apart and form a vapor. Vaporization can also occur when a solid is heated until it sublimates, or when the molecules of the solid are broken down into a gas. Vaporization is an important part of the water cycle, and it is also used in many industries, such as chemical production, pharmaceutical manufacturing, and food processing.
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Please help
3) a student claims that the reaction of hydrogen and oxygen to form
water is evidence supporting the claim that mass is conserved in a
chemical reaction. the chemical equation the student uses for the reaction
is h2 + o2 --> h2o. does this evidence support the claim? why or why not?*
a.) yes, it supports the claim because all the elements in the reactants appear in the
product.
b.) no, it does not support the claim because it is not a closed system.
c.) yes, it supports the claim because the reaction equation is balanced.
d.) no, it does not support the claim because the reaction equation is not balanced.
Yes, this evidence supports the claim that mass is conserved in a chemical reaction because the reaction equation is balanced.
This means that the same number of atoms of each element is present in the reactants as in the products. This is the fundamental principle of conservation of mass, which states that mass is neither created nor destroyed during a chemical reaction.
The conservation of mass can also be verified by calculating the total mass of the reactants and comparing it to the total mass of the products.
If the same amount of mass is present in both reactants and products, then the reaction equation is balanced and the conservation of mass is supported.
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A sample of neon gas exerts a pressure of 1. 18 atm when the temperature is
400 K. What pressure is produced when the temperature is raised to 600 K?
The pressure produced when the temperature is raised to 600 K is 1.77 atm.
The pressure of a gas is directly proportional to its temperature, according to the ideal gas law. This means that if the temperature of a gas increases, its pressure will increase proportionally, assuming that the volume and number of gas molecules remain constant.
In this problem, we are given the initial pressure of neon gas at 400 K, which is 1.18 atm. We need to find the pressure of the gas when the temperature is raised to 600 K.
To solve this problem, we can use the following formula:
P₂ = P₁ x (T₂/T₁)where P₁ is the initial pressure, T₁ is the initial temperature, P₂ is the final pressure, and T₂ is the final temperature.
Substituting the given values, we get:
P₂ = 1.18 atm x (600 K/400 K)P₂ = 1.77 atmTherefore, the pressure produced when the temperature is raised to 600 K is 1.77 atm. This means that the pressure of the neon gas increases by a factor of 1.5 when the temperature is increased from 400 K to 600 K.
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Of the types of waves listed, which come naturally from the decay of radioactive
isotopes and are used in medicine for diagnostic imaging?
The type of waves that come naturally from the decay of radioactive isotopes and are used in medicine for diagnostic imaging are gamma rays.
Gamma rays are a type of electromagnetic radiation with the highest energy and shortest wavelength in the electromagnetic spectrum. They are produced naturally by the decay of radioactive isotopes, such as uranium and radon, and are also emitted during nuclear reactions and explosions.
In medicine, gamma rays are used in a diagnostic imaging technique called gamma-ray spectroscopy, which detects and measures gamma rays emitted by radioactive isotopes in the body. This technique can be used to diagnose various conditions, such as cancer and heart disease, by identifying areas of the body with abnormal radioactive activity.
Gamma rays are also used in radiation therapy to treat cancer. In this treatment, high-energy gamma rays are directed at cancerous cells to damage and kill them. However, the high energy of gamma rays can also damage healthy cells, so careful targeting and dose management is necessary to minimize side effects.
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