Answers
Under the condition A. 273 K and 1.01 × [tex]10^{5}[/tex] Pa, one mole of methane gas, [tex]CH_{4}[/tex], occupy the smallest volume.
What is the smallest volume ?
The ideal gas law, PV = nRT, can be rearranged to solve for volume as V = nRT/P, where V is volume, n is the number of moles, R is the gas constant, T is temperature in Kelvin, and P is pressure.
To find the conditions under which one mole of methane gas occupies the smallest volume, we need to find the combination of temperature and pressure that results in the smallest volume.
Using the equation V = nRT/P, we can see that volume is inversely proportional to pressure, so we want the highest pressure possible. Therefore, we can eliminate choices A and C, which have lower pressures than B and D.
Next, we need to consider temperature. Using the same equation, we can see that volume is directly proportional to temperature, so we want the lowest temperature possible. Therefore, the correct choice is A, which has a temperature of 273 K, the lowest temperature of all the choices.
Therefore, the answer is A. 273 K and 1.01 × [tex]10^{5}[/tex] Pa.
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Complete question is: Under the 273 K and 1.01 × [tex]10^{5}[/tex] Pa condition, one mole of methane gas, [tex]CH_{4}[/tex], occupy the smallest volume.
Related Questions
Calculate the mass of water produced when 1.03 g
of butane reacts with excess oxygen.
Answers
1.03 g of butane will yield 1.59 g of water when it reacts with too much oxygen.
What is mass?The amount of matter in an item is measured by the fundamental physical quantity known as mass. It is a scalar quantity that is measured in grams (g) or kilograms (kg) (g). No matter where it is or what force is pushing against it, an object's mass always remains constant.
How do you determine it?For butane and oxygen to burn, the chemical equation is balanced as follows:
2C4H10+13 O2→ 8 CO2+ 10 H2O
According to the equation, 1 mole of butane (C4H10) and 13/2 moles of oxygen (O2) combine to form 5 moles of water (H2O).
To begin with, we must count the butane moles that are present:
Mass of butane divided by its molar mass yields moles of butane.
1.03 g/58.12 g/mol is the formula for butane.
A mole of butane weighs 0.0177 mol.
Then, we can calculate the quantity of water created using the butane-to-water mole ratio:
Moles of water = [tex]\frac{5 mol H2O}{(1 mol C4H10 ×0.0177 mol C4H10)}[/tex] = 0.0885 mol.
Eventually, we can determine how much water was generated:
moles of water equal 0.0885 mol when 5 mol H2O is divided by 1 mol C4H10 and multiplied by 0.0177 mol C4H10.
Lastly, we can determine the mass of created water:
Water's mass is equal to its moles times its molar mass= 0.0885 mol × 18.02 g/mol = 1.59 g.
As a result, 1.03 g of butane will yield 1.59 g of water when it reacts with to moles of water equal 0.0885 mol when 5 mol H2O is divided by 1 mol C4H10 and multiplied by 0.0177 mol C4H10.
Lastly, we can determine the mass of created water:
Water's mass is equal to its moles times its molar mass = 0.0885 mol × 18.02 g/mol = 1.59 g.
As a result, 1.03 g of butane will yield 1.59 g of water when it reacts with too much oxygen.
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The alcoholic, blue solution from Part I of your experiment is commonly used in weather-forecasting devices found in coastal areas of the USA. Based on your observations in the lab explain how this reaction can indicate coming rain
Answers
Answer:
The reaction referred to in this question is likely the reaction between hydrated copper(II) sulfate and anhydrous copper(II) sulfate, where the former is blue and the latter is white.
When the blue solution of hydrated copper(II) sulfate is exposed to moist air, it slowly turns white as water is absorbed, forming anhydrous copper(II) sulfate. This reaction is exothermic, meaning it releases heat, and is reversible. The reverse reaction occurs when anhydrous copper(II) sulfate is exposed to water vapor in the air, forming hydrated copper(II) sulfate and releasing heat.
In coastal areas, the humidity in the air tends to increase before a storm, which can trigger the reverse reaction between anhydrous copper(II) sulfate and water vapor. This releases heat, causing the weather-forecasting device to warm up, indicating that rain may be on the way.
Therefore, the observation of the blue solution turning white in the lab, which indicates the reversible reaction between hydrated copper(II) sulfate and anhydrous copper(II) sulfate, can indirectly indicate the presence of moisture in the air and the possibility of rain, similar to the process in weather-forecasting devices.
A 0.675 mole sample of oxygen gas has a pressure of 2.50 atm at 50. °C. What volume of gas is present
in the sample? Show the rearranged ideal gas law solving for volume. Cancel units in work.
Answers
Considering the ideal gas law, the volume of gas present in the sample is 7.15122 L.
Definition of Ideal Gas LawIdeal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar gas constant:
P×V = n×R×T
Volume of gas in this caseIn this case, you know:
P= 2.50 atmV= ?n= 0.675 molesR= 0.082 (atmL)/(molK)T= 50 C= 323 K (being 0 C= 273 K)Replacing in the ideal gas law:
2.50 atm×V = 0.675 moles×0.082 (atmL)/(molK)× 323 K
Solving:
V= (0.675 moles×0.082 (atmL)/(molK)× 323 K)÷ 2.50 atm
V= 7.15122 L
Finally, the volume of gas is 7.15122 L.
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A sample of sodium azide (NaN3), a compound used in automobile air bags, was thermally decomposed, and 15.3 mL nitrogen gas was collected over water at 25°C and 755 torr. Given the vapour pressure of water at 25°C is 23.6 torr, how many grams of nitrogen were collected?
Answers
A sample of sodium azide (NaN3), a compound used in automobile air bags, was thermally decomposed, and 15.3 mL nitrogen gas was collected over water at 25°C and 755 torr. 131.1g is the mass of nitrogen.
In physics, mass is a quantitative measurement of inertia, a basic characteristic of all matter. It essentially refers to a body of matter's resistance to changing its speed or location in response to the application for a force.
The change caused by a force being applied is smaller the more mass a body has. The kilogramme, which is defined approximately equal to 6.62607015 1034 joule second in terms of Planck's constant, is the unit of mass within the Internacional System of Units (SI).
P×V = n×R×T
755×15.3 = n×0.0821×300
11551.5=n×24.63
n= 46.9
mass = 46.9×28=131.1g
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A fjord is _____ .
a high mountain
a steep-sided glacial valley
an oceanic mountain range
a glacial plain
Answers
Answer:
a steep-sided glacial valley
Explanation:
A fjord is a long, narrow inlet with steep sides or cliffs, created by a glacier. It is a long, deep, narrow body of water that reaches far inland. Fjords are found mainly in Norway, Chile, New Zealand, Canada, Greenland, and Alaska. They are formed when a glacier cuts a U-shaped valley by ice segregation and abrasion of the surrounding bedrock. According to the standard model, glaciers formed in pre-glacial valleys with a gently sloping valley floor. The work of the glacier then left an over deepened U-shaped valley that ends abruptly at a valley or trough end. Such valleys are fjords when flooded by the ocean. Thresholds above sea level create freshwater lakes.
PLS HELP!!!!!
Convert the following measurements. Show all work, including units that cancel.
9.3 mol SO3 -> liters
Answers
Answer: 9.3 mol of SO3 at STP occupies 20.2 liters.
Explanation: To convert from moles of a gas to liters, we need to use the ideal gas law:
PV = nRT
where:
P = pressure in atm
V = volume in liters
n = number of moles
R = gas constant (0.08206 L·atm/mol·K)
T = temperature in Kelvin
We can rearrange this equation to solve for V:
V = (nRT)/P
First, let's calculate the volume of 9.3 mol of an ideal gas at standard temperature and pressure (STP). STP is defined as 0°C (273.15 K) and 1 atm.
V = (9.3 mol * 0.08206 L·atm/mol·K * 273.15 K) / 1 atm
V = 20.2 L
Therefore, 9.3 mol of SO3 at STP occupies 20.2 liters.
Note that this assumes SO3 is an ideal gas, which may not be the case in reality.
Determine which statements are consistent with Daltons atomic theory
Answers
Answer:
Explanation:
Dalton's atomic theory had several postulates, including:
1. All matter is made up of atoms, which are indivisible and indestructible.
2. All atoms of a given element are identical in mass and properties.
3. Compounds are formed by a combination of two or more different kinds of atoms.
4. A chemical reaction is a rearrangement of atoms.
Based on these postulates, the following statements are consistent with Dalton's atomic theory:
1. Chemical reactions involve the rearrangement of atoms to form new compounds.
2. Elements are composed of small, indivisible particles called atoms.
3. Atoms of the same element have the same mass and chemical properties.
4. A compound is made up of two or more different elements.
5. Atoms can neither be created nor destroyed in a chemical reaction.
However, the following statements are not consistent with Dalton's atomic theory:
1. Atoms are divisible into smaller particles.
2. Isotopes of the same element have different chemical properties.
3. The mass of an atom is distributed uniformly throughout the atom.
It's worth noting that while Dalton's atomic theory has been largely superseded by modern atomic theory, it remains an important historical milestone in the development of atomic theory.
What is molecular weight of a substance given that 1.22g of the sample was vaporised in 100ml flask at 45°C and 687mmHg.
Answers
PV = nRT
where R is the gas constant. Rearranging this equation gives:
n = PV / RT
We can use this equation to calculate the number of moles of the substance in the vapor phase. Then, we can use the definition of molecular weight, which is the mass of one mole of the substance, to calculate the molecular weight.
First, let's convert the temperature and pressure to SI units:
Temperature: 45°C = 318.15 K
Pressure: 687 mmHg = 91.6 kPa
Next, let's calculate the number of moles of the substance in the vapor phase. We know that 1.22 g of the substance was vaporized in a 100 mL flask, which is equivalent to 0.1 L. Therefore, the concentration of the substance in the vapor phase is:
c = 1.22 g / 0.1 L = 12.2 g/L
We can convert this concentration to units of pressure using the ideal gas law:
PV = nRT
n/V = P/RT
n/V = (91.6 kPa) / (8.314 J/mol K * 318.15 K) = 0.0361 mol/L
Therefore, the number of moles of the substance in the vapor phase is:
n = (0.0361 mol/L) * 0.1 L = 0.00361 mol
Finally, we can calculate the molecular weight using the definition of molecular weight:
molecular weight = mass / moles
mass = 1.22 g
moles = 0.00361 mol
molecular weight = 1.22 g / 0.00361 mol = 337.4 g/mol
Therefore, the molecular weight of the substance is approximately 337.4 g/mol.
How many chloride ions are in 15.0 mL of a 2.5 molar solution of magnesium chloride? (MgCl2)
Answers
In 15.0 mL of a 2.5 molar magnesium chloride solution, there are approximately 4.51 × 10^{22} chloride ions.
How many chloride ions are there in MgCl_2?An inorganic compound composed of one magnesium ion and two chloride ions.
We must first determine the number of moles of MgCl_2 in the solution,
moles of solute = concentration × volume (in liters)
Converting the solution's volume from milliliters to liters,
15.0 mL = 15.0 × 10^{-3} L
moles of MgCl_2 = 2.5 mol/L × 15.0 × 10^{-3} L = 0.0375 moles
The solution contains the following number of chloride ions:
number of Cl^{-} ions = 2 × moles of MgCl_2
Substitute the value of moles of MgCl_2,
number of Cl^{-} ions = 2 × 0.0375 moles = 0.075 moles
by using Avogadro's number:
number of Cl^{-} ions = 0.075 moles × 6.02 × 10^{23} ions/mol ≈ 4.51 × 10^{22} ions
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Is my option correct?
Answers
Answer:
No, See explanation. Should be A.
Explanation:
No, the correct answer appears to be A. Particle B is past the required potential energy to react. This means that this particle has enough energy to react and does not require to be heated or energy imputed further.
Give the number of significant figures indicated.
0.066
Answers
please help this is dus today please help
Answers
A percentage expressing the variation between a calculated/measured value and the anticipated/real value is known as a "calculated percent error."
What does this mean?A positive numerical discrepancy implies that the assessed number exceeds what was expected, with negative values indicating the reverse.
When conducting an experiment, if an excessive percent error emerges it suggests potential errors or miscalculations in experimental design, data collection, or analysis.
This may mean unaccounted variables were present, and equipment/procedures require refinement. In opposition, lower percentages signal accuracy despite any limitations resulting from the human elements involved.
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Classify each of the following aqueous solutions as a nonelectrolyte, weak electrolyte, or strong electrolyte.
(a) PbCl2
(b) N(CH3)3
(c) CsOH
(d) H2S
(e) CrCl2
(f) Ni(CH3COO)2
Answers
Strong electrolyte: lead(II) chloride weak electrolyte trimethylamine Strong electrolyte: cesium hydroxide sulphide of hydrogen: a weak electrolyte, strong electrolyte chromium chloride Weak electrolyte: nickel(II) acetate
Is acetic acid an electrolyte that is weak?Response and justification The fact that acetic acid has a low dissociation constant suggests that it is a weak acid. Acetic acid's restricted ability to ionise in an aqueous solution is a result of its low dissociation constant. With this in mind, acetic acid can be regarded as a weak electrolyte.
Is acid phosphoric a type of electrolyte?Phosphoric acid is a useful electrolyte because of its low volatility, strong ionic conductivity, stability at high temperatures, carbon dioxide and carbon monoxide tolerance, and low flammability.
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A sample of helium gas, He(g), is placed in a rigid cylinder sealed with a movable piston.
The temperature of the helium is 25.0°C. The volume of the helium is 300. milliliters and
the pressure is 0.500 atmosphere.
16) State, in terms of the average distance between the helium atoms, why the density of the gas increases when the piston is pushed farther into the rigid cylinder. [1]
17) Determine the volume of the helium gas when the pressure is increased to 1.50 atm and
the temperature remains at 25.0°C. [1]
Answers
(16) When the piston is pushed farther into the rigid cylinder, the volume available for the helium gas decreases. This leads to an increase in density because the gas molecules are packed into a smaller volume, resulting in a shorter average distance between the helium atoms; (17) volume = 100mL.
What will be the effect on density on pushing the piston farther into the rigid cylinder?When the piston is pushed farther into the rigid cylinder, the volume available for the helium gas decreases. This causes the average distance between the helium atoms to decrease as well. Since the volume available for the gas molecules to move in is reduced, they will collide more frequently with each other and with the walls of the container, leading to an increase in density.
(17) To determine the new volume of the helium gas, we can use the combined gas law:
(P1V1)/T1 = (P2V2)/T2
where P1, V1, and T1 are the initial pressure, volume, and temperature of the helium gas, and P2 and T2 are the final pressure and temperature.
Substituting the given values, we get:
(0.500 atm)(300. mL)/(298.15 K) = (1.50 atm)(V2)/(298.15 K)
Solving for V2, we get:
V2 = (0.500 atm)(300. mL)/(1.50 atm) = 100. mL
Therefore, the volume of the helium gas when the pressure is increased to 1.50 atm and the temperature remains at 25.0°C is 100mL.
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What would the expected temperature change be (in °C) if a 0.5 gram sample of water released 0.0501 kJ of heat energy? The specific heat of liquid water is 4.184 J/g-°C.
Answers
Using the equation q = mcT, where q is the amount of heat energy released (0.0501 kJ), m is the sample's mass (0.5 g), c is the specific heat of liquid water (4.184 J/g-°C), and T is the temperature change, one can determine the anticipated temperature change for a 0.5 gramme sample of water that releases 0.0501 kJ of heat energy.
T = q / (mc) is the result of rearranging the equation. We calculate T = 0.0501 kJ / (0.5 g * 4.184 J/g-°C) = 0.6022 °C by plugging in the given variables.
As a result, a 0.5 gramme sample of water that releases 0.0501 kJ of heat energy should have an estimated temperature change of 0.6022 °C.
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Calculate the molality of a solution prepared by dissolving 19.9 g of KCl in 750.0 mL of water.
Answers
Answer:
Explanation:
First, we need to calculate the moles of KCl:
Calculate the molar mass of KCl:
KCl = 39.10 g/mol (atomic weight of K) + 35.45 g/mol (atomic weight of Cl)
= 74.55 g/mol
Calculate the moles of KCl:
moles of KCl = mass of KCl / molar mass of KCl
= 19.9 g / 74.55 g/mol
= 0.267 mol
Next, we need to convert the mass of the solvent (water) from milliliters to kilograms:
Convert mL to L:
750.0 mL = 750.0 mL * (1 L / 1000 mL)
= 0.7500 L
Calculate the mass of water:
mass of water = volume of water x density of water
= 0.7500 L x 1000 g/L
= 750.0 g
Convert the mass of water to kilograms:
mass of water = 750.0 g / 1000 g/kg
= 0.7500 kg
Now we can calculate the molality of the solution:
molality = moles of solute / mass of solvent (in kg)
molality = 0.267 mol / 0.7500 kg
= 0.356 mol/kg
Therefore, the molality of the solution is 0.356 mol/kg.
What is the energy change for the following equation?
Answers
Answer:
1) -572 kJ/mol
2) -2220.7 kJ/mol
Explanation:
Multiply the bond enthalpies to take into account the amount of moles of each compound in the reaction. Then, to get the total change in energy/enthalpy in the reaction, subtract the reactant energy from the product energy.
1) 2(-286) = -572 kJ/mol
2) 4(-286) + -393.5(3) - -103.8 = -2220.7 kJ/mol
The combustion of FeS forms Fe2O3 and SO2. The combustion of SO2 forms SO3. The SO3 can be treated with water to form sulfuric acid, H2SO+. How many grams of H2SO4 can be produced from 422 grams of iron ore containing 75.0% FeS.
Answers
Therefore, approximately 235.6 grams of Sulfuric Acid can be produced from 422 grams of iron ore containing 75.0% Iron(II) sulfide.
What function does Sulfuric Acid provide in iron estimation?
During such a titration, sulfuric acid is added to maintain the medium's acidity and meet the stoichiometric needs of the redox reaction. Additionally, extra amounts are injected to supply the protons (H+) needed for the redox reaction.
We can begin by figuring out how much Iron(II) sulfide there is in the 422 grammes of iron ore:
mass of FeS = 422 g x 0.75 = 316.5 g
We can see from the equation that everything balances out that 1 mol of FeS combines with 1.5 mol of sulfur dioxide to create 1 mol of Sulfuric Acid. Therefore, we must first determine the amount of Iron(II) sulfide in moles:
moles of Iron(II) sulfide = mass of Iron(II) sulfide / molar mass of Iron(II) sulfide
moles of Iron(II) sulfide = 316.5 g / 87.91 g/mol
moles of Iron(II) sulfide = 3.597 mol
Next, we can determine the quantity of moles of Sulfuric Acid generated using the stoichiometric ratios from the balanced equations:
1 mol sulfur dioxide : 1.5 mol sulfur dioxide : 1 mol Sulfuric Acid
3.597 mol : 5.3955 mol : x mol Sulfuric Acid
x mol Sulfuric Acid = (3.597 mol Iron(II) sulfide) x (1 mol Sulfuric Acid / 1 mol FeS) x (1.5 mol sulfur dioxide / 1 mol FeS) x (1 mol Sulfuric Acid / 1.5 mol SO2)
x mol H2SO4 = 2.398 mol
Finally, we can determine the mass of created Sulfuric Acid:
mass of Sulfuric Acid = moles of Sulfuric Acid x molar mass of Sulfuric Acid
mass of Sulfuric Acid = 2.398 mol x 98.079 g/mol
mass of Sulfuric Acid = 235.6 g.
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Explain to me please????
Answers
Answer:Non polar.
Explanation:because water is polar because of its shape
For which of the following reactions does ΔH
o
rxn
= ΔH
o
f
?
(a) H2(g) + S(rhombic) → H2S(g)
(b) C(diamond) + O2(g) → CO2(g)
(c) H2(g) + CuO(s) → H2O(l) + Cu(s)
(d) O(g) + O2(g) → O3(g)
Answers
ΔH° = ΔHf° is true for two reactions which are ,
H₂(g) + S(rhombic) → H₂S(g)
C(diamond) + O₂(g) → CO₂(g)
Hence, option a and b are correct.
∆Hf is the enthalpy of formation, is the enthalpy when product is made from its constituent elements. Generally the enthalpy of formation is described as the standard reaction enthalpy for the formation of the compound from its elements (which may include atoms or molecules) in their most stable reference states at the chosen temperature (298.15K) and at 1bar pressure.
In a part, H₂S is formed by H₂ and S means it's constituent elements.
So, a is correct.
In b also, CO₂ is formed by C and O₂, so b is also correct. Hence, option a and b are correct.
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Which is expected to have the largest dispersion forces?
Question 13 options:
N2
C2H6
CO2
C8H18
Answers
The molecule with the highest molecular weight and the largest number of electrons, C₈H₁₈, is expected to have the largest dispersion forces, option (d) is correct.
The dispersion forces, also known as London dispersion forces, are a type of intermolecular force that arises due to the temporary dipoles that occur in non-polar molecules. The molecule C₈H₁₈ with the largest number of electrons and the highest molecular weight is expected to have the largest dispersion forces.
This molecule has a larger number of electrons and a larger surface area for intermolecular interactions, which results in stronger dispersion forces compared to the other molecules, option (d) is correct.
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The complete question is:
Which is expected to have the largest dispersion forces?
a. N₂
b. C₂H₆
c. CO₂
d. C₈H₁₈
3. A solution of hydrochloric acid is made by dissolving hydrogen chloride gas in 100.0ml water. This solution neutralizes a 15ml sample of 0.10 mol/L sodium carbonate solution. a. What mass of hydrogen chloride gas was dissolved in 100.0ml of water? b. What volume of hydrogen chloride was this?
Answers
Okay, here are the steps to solve this problem:
a) To neutralize 15ml of 0.10 mol/L sodium carbonate solution, the hydrochloric acid solution must contain 0.015 moles of HCl.
Since the HCl solution is made by dissolving HCl gas in 100ml water, we can calculate the moles of HCl gas dissolved in 100ml:
0.015 moles HCl / 15ml sodium carbonate solution = X moles HCl gas / 100ml HCl solution
X = 0.015 * (100/15) = 0.01 moles HCl gas
b) Molar volume of HCl gas at STP is 24.45 L/mol.
So the volume of 0.01 moles HCl gas is: 0.01 moles * 24.45 L/mol = 0.2445 L
Since the solution is made with this gas in 100ml water, the volume of HCl gas dissolved in 100ml water is 0.2445 L.
So the final answers are:
a) 0.01 moles of HCl gas
b) 0.2445 L of HCl gas
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What quantity of heat (in J) would be required to convert 0.27 mol of a pure substance from a liquid at 50 °C to a gas at 113.0 °C?.
Cliquid = 1.45 J/mol C
Cgas = 0.65 J/mol *C
Tboiling = 88.5 °C
AHvaporization = 1.23 kJ/mol
Give your answer in Joules
Answers
Explanation:
First you have to heat the liquid from 50 to 88.5 C to get it boiling...
then you need to boil it all to a gas....the you have to heat it to 113 C
.27 mole *
( (88.5 -50 C)*1.45 J / (mole-C) <====heating the liquid
+ 1230 J / mole <====boiling to a gas
+ (113 - 88.5 C) * .65 J/(mole C) <=====heating the gas
= 351 .5 J <=====result
Sucrose (C12H22011) is combusted in air according to the following reaction:
C12H22011(s) + O2(g) = CO2(g) + H2O(l )
How many moles of carbon dioxide would be produced by the complete combustion of 38.5 grams of sucrose in the presence of excess oxygen?
Answers
Okay, let's break this down step-by-step:
1) The molecular formula for sucrose is C12H22011. This means each mole of sucrose contains 12 moles of carbon and 22 moles of hydrogen.
2) You are combusting 38.5 grams of sucrose. To convert grams to moles, we divide by the molar mass:
Molar mass of C12H22011 = 342.3 g/mol
So 38.5 g / 342.3 g/mol = 0.113 moles of sucrose
3) According to the combustion reaction, each mole of sucrose produces 12 moles of CO2.
So 0.113 moles of sucrose will produce 0.113 * 12 = 1.356 moles of CO2.
4) Round to the nearest whole number:
1 mole of CO2
Therefore, the complete combustion of 38.5 grams of sucrose in excess oxygen would produce 1 mole of carbon dioxide.
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Problem 1. Sea water contains dissolved salts at a total ionic concentration of about 1.13 mol×L–1. What pressure must be applied to prevent osmotic flow of pure water into sea water through a membrane permeable only to water molecules (at 25oC)?
Problem 2. What is the osmotic pressure of a solution prepared by adding 6.65 g of glucose to enough water to make 350 mL of solution at 35°C?
Problem 3. What is the osmotic pressure of a solution prepared by adding 9.0 g of glucose to enough water to make 450 mL of solution at 35°C?
Problem 4. What is the osmotic pressure of a solution prepared by adding 11.0 g of propanol to enough water to make 850 mL of solution at 25°C?
Problem 5. What is the osmotic pressure of a solution prepared by adding 65 g of glucose to enough water to make 35000 mL of solution at 15°C?
Answers
The osmotic pressure can be calculated using the equation:
π = CRT
Where π is the osmotic pressure, C is the total concentration of dissolved particles, R is the gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
In this case, the total concentration of dissolved particles is 1.13 mol/L, and the temperature is 25°C, or 298 K.
π = (1.13 mol/L) (0.0821 L·atm/(mol·K)) (298 K) = 24.7 atm
Therefore, a pressure of 24.7 atm must be applied to prevent osmotic flow of pure water into seawater.
Problem 2:
The osmotic pressure of a solution can be calculated using the equation:
π = MRT
Where M is the molar concentration of the solution, R is the gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
First, we need to calculate the molar concentration of the glucose solution:
molar mass of glucose = 180.16 g/mol
moles of glucose = 6.65 g / 180.16 g/mol = 0.0369 mol
volume of solution = 350 mL = 0.350 L
Molar concentration = moles of solute / volume of solution = 0.0369 mol / 0.350 L = 0.105 M
Now we can calculate the osmotic pressure:
π = (0.105 M) (0.0821 L·atm/(mol·K)) (308 K) = 2.87 atm
Therefore, the osmotic pressure of the glucose solution is 2.87 atm.
Problem 3:
Using the same equation as in Problem 2:
molar mass of glucose = 180.16 g/mol
moles of glucose = 9.0 g / 180.16 g/mol = 0.0499 mol
volume of solution = 450 mL = 0.450 L
Molar concentration = moles of solute / volume of solution = 0.0499 mol / 0.450 L = 0.111 M
π = (0.111 M) (0.0821 L·atm/(mol·K)) (308 K) = 3.05 atm
Therefore, the osmotic pressure of the glucose solution is 3.05 atm.
Problem 4:
In this case, propanol is the solute, so we need to first calculate the molar concentration:
molar mass of propanol = 60.10 g/mol
moles of propanol = 11.0 g / 60.10 g/mol = 0.183 mol
volume of solution = 850 mL = 0.850 L
Molar concentration = moles of solute / volume of solution = 0.183 mol / 0.850 L = 0.215 M
π = (0.215 M) (0.0821 L·atm/(mol·K)) (298 K) = 4.84 atm
Therefore, the osmotic pressure of the propanol solution is 4.84 atm.
Problem 5:
molar mass of glucose = 180.16 g/mol
moles of glucose = 65 g / 180.16 g/mol = 0.361 mol
volume of solution = 35000 mL = 35.00 L
Molar concentration = moles of solute / volume of solution = 0.
Problem 1:
The osmotic pressure (π) can be calculated using the van 't Hoff equation: π = iMRT, where i is the van 't Hoff factor (1 for water), M is the molar concentration, R is the gas constant, and T is the temperature in Kelvin.
In this case, the molar concentration is 1.13 mol/L, and the temperature is 25°C = 298 K. So,
π = iMRT = (1)(1.13 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(298 K)
π = 29.8 atm
Therefore, a pressure of 29.8 atm must be applied to prevent osmotic flow of pure water into sea water through a membrane permeable only to water molecules at 25°C.
Problem 2:
The osmotic pressure of a solution can be calculated using the van 't Hoff equation: π = iMRT, where i is the van 't Hoff factor, M is the molar concentration, R is the gas constant, and T is the temperature in Kelvin.
First, we need to find the molar concentration of glucose in the solution. The molecular weight of glucose is 180.16 g/mol. So,
Molar concentration = (mass/volume) / (molecular weight)
Molar concentration = (6.65 g/0.35 L) / 180.16 g/mol
Molar concentration = 0.104 mol/L
Now, we can calculate the osmotic pressure:
π = iMRT = (1)(0.104 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(308 K)
π = 2.44 atm
Therefore, the osmotic pressure of the solution prepared by adding 6.65 g of glucose to enough water to make 350 mL of solution at 35°C is 2.44 atm.
Problem 3:
Using the same process as in Problem 2, we can find the molar concentration of glucose in the solution:
Molar concentration = (mass/volume) / (molecular weight)
Molar concentration = (9.0 g/0.45 L) / 180.16 g/mol
Molar concentration = 0.44 mol/L
Now, we can calculate the osmotic pressure:
π = iMRT = (1)(0.44 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(308 K)
π = 10.2 atm
Therefore, the osmotic pressure of the solution prepared by adding 9.0 g of glucose to enough water to make 450 mL of solution at 35°C is 10.2 atm.
Problem 4:
Propanol (C₃H₇OH) is a non-electrolyte, so its van 't Hoff factor is 1.
First, we need to find the molar concentration of propanol in the solution. The molecular weight of propanol is 60.10 g/mol. So,
Molar concentration = (mass/volume) / (molecular weight)
Molar concentration = (11.0 g/0.85 L) / 60.10 g/mol
Molar concentration = 0.178 mol/L
Now, we can calculate the osmotic pressure:
π = iMRT = (1)(0.178 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹(298 K)
π = 3.67 atm
Therefore, the osmotic pressure of the solution prepared by adding 11.0 g of propanol to enough water to make 850 mL of solution at 25°C is 3.67 atm.
Problem 5:
Using the same process as in Problem 2 and Problem 3, we can find the molar concentration of glucose in the solution:
Molar concentration = (mass/volume) / (molecular weight)
Molar concentration = (65 g/35,000 mL) / 180.16 g/mol
Molar concentration = 0.00177 mol/L
Now, we can calculate the osmotic pressure:
π = iMRT = (1)(0.00177 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(288 K)
π = 0.0398 atm
Therefore, the osmotic pressure of the solution prepared by adding 65 g of glucose to enough water to make 35,000 mL of solution at 15°C is 0.0398 atm.
A 0.431-g sample of an unknown monoprotic acid was titrated with 0.108 M KOH and the resulting titration curve is shown here.
1. Determine the molar mass of the acid.
2. Determine the pKa of the acid.
Answers
1) The molar mass of the acid of the titration curve is 166 g/mol
2) The pKa of the acid of the titration curve is 4.97.
1) The molar mass of the acid can be determined from the equivalence point of the titration curve
Mass of acid sample = 0.431 g
Volume of KOH solution at equivalence point = 24.0 mL = 0.0240 L
Molarity of KOH solution = 0.108 M
Number of moles of KOH added at equivalence point = Molarity x Volume
= 0.108 M x 0.0240 L
= 0.00259 moles
Number of moles of acid = Number of moles of KOH added
= 0.00259 moles
Molar mass of acid = Mass of acid sample ÷ Number of moles of acid
= 0.431 g ÷ 0.00259 moles
= 166 g/mol
2) The pKa of the acid can be determined from the half-equivalence point of the titration curve.
Volume of KOH solution at half-equivalence point = 11.5 mL = 0.0115 L
Number of moles of KOH added at half-equivalence point = Molarity x Volume
= 0.108 M x 0.0115 L
= 0.00124 moles
Concentration of acid at half-equivalence point = Number of moles of acid remaining ÷ Volume of acid
= (0.431 g - (0.00124 moles x 56.1 g/mol)) ÷ 0.025 L
= 0.017 M
Concentration of conjugate base at half-equivalence point = Number of moles of KOH added ÷ Volume of KOH
= 0.00124 moles ÷ 0.0115 L
= 0.108 M
pKa = pH + log([conjugate base] ÷ [acid])
= 7.08 + log(0.108 ÷ 0.017)
= 4.97
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How many carbon atoms are in 11.5 g C2H5OH?
Answers
There are approximately 3.02 x 10²³ carbon atoms in 11.5 g of [tex]C_{2}H_{5} OH[/tex]
To calculate the number of carbon atoms present in the 11.5g of [tex]C_{2}H_{5}OH[/tex]which is ethanol, we need to find and calculate the number of moles of ethanol present and convert moles to the number of carbon atoms by using Avogadro's number.
The molecular weight of [tex]C_{2}H_{5}OH[/tex] is =
no. of atoms x its atomic weight. So the molecular weight is,
(2 atoms of carbon x 12 g/mol )+ (5 atoms of hydrogen x 1 g/mol) + (1 atom of oxygen x 16 g/mol) + (1 atom of hydrogen x 1 g/mol ) =
24+10+16+1 ≅ 47 g/mol
The number of moles of ethanol = given mass / molecular weight = 11.5g /47 g/mol = 0.2446 moles
Number of Carbon atoms = no. of moles of ethanol x Avogadro number x no. of carbon atoms in each molecule of ethanol = 0.2446 moles x 6.022 x 10²³ atoms/mol x 2 carbon atoms/molecule ≅ 3.02 x 10²³
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Question 22 of 25
Which of the following is a carboxylic acid?
O
НИИ
II
A. н-с-с-с-с
ИНИ
НИИ
в. н-с-с-с-с
III о-н
ИНН
ИНИ
I II
с. н-с-с-с-
Н
ННИ
ТТІ
O D. H-C-C-C-C
JIT
ИНН
OCH,
CH₂
Answers
The answer is B since the carboxylic acid group is COOH
Macmillan Learning
A 346.9 mL sample of carbon dioxide was heated to 373 K. If the volume of the carbon dioxide sample at 373 K is 596.2 mL,
what was its temperature at 346.9 mL?
T=. K
Answers
Answer:
Explanation:
Use the Equation: [tex]\frac{V1}{T1}=\frac{V2}{T1} \\[/tex]
V1 = 346.9 mL
T1 = ? K
V2 = 596.2 mL
T2 = 373 L
This assumption can only be made when pressure is held constant.
[tex]\frac{346.9}{V1}=\frac{596.2}{373}, solve for V1\\ V1 = 217.0 K[/tex]
Among the elements of the main group, the first ionization energy increases
from left to right across a period.
from right to left across a period.
when the atomic radius increases.
down a group
Answers
The first ionisation energy increases over time from left to right among the major group of elements. answer is option (a).
What is Ioniztion?When an element loses its valence electron, its oxidation number increases (a process known as oxidation), and this energy loss is known as ionisation (Ei).
Earth alkaline metals, which are located immediately next to alkaline metals, have higher ionisation energies than alkaline metals because they have two valence electrons, while alkaline metals, which are located far left in the main group, have the lowest ionisation energies and are easiest to remove.
Because they contain a large number of valence electrons, nonmetals are far to the right in the main group and have the highest ionisation energy.
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The complete question is,
Among the elements of the main group, the first ionization energy increases
a. from left to right across a period.
b. from right to left across a period.
c. when the atomic radius increases.
d. down a group.