Explanation:
To find the empirical formula of the compound, we need to determine the simplest whole-number ratio of the atoms in the compound.
First, we need to find the number of moles of Ag, N, and O relative to each other. The smallest number of moles is used as a reference point, and the other moles are divided by this value to get their ratios:
0.58 mol Ag : 0.58 mol N : 1.79 mol O
Dividing each mole value by 0.58 mol (the smallest number of moles), we get:
1 mol Ag : 1 mol N : 3.09 mol O
We can see that the ratio of Ag:N:O is 1:1:3.09. However, we need to express the ratio in whole numbers, so we divide each number by the smallest number in the set of ratios:
1 : 1 : 3.09/1.00 ≈ 3.
Rounding to the nearest whole number, we get the empirical formula of the compound:
AgN3O3
Therefore, the empirical formula of the compound is AgN3O3.
How many chloride ions are in 15.0 mL of a 2.5 molar solution of magnesium chloride? (MgCl2)
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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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
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]
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]
(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 is molecular weight of a substance given that 1.22g of the sample was vaporised in 100ml flask at 45°C and 687mmHg.
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)
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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Explain to me please????
Answer:Non polar.
Explanation:because water is polar because of its shape
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?
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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Watch the animation and select the interactions that can be explained by hydrogen bonding.
HF
is a weak acid neutralized by NaOH
.
HF
has a higher boiling point than HCl
.
CH4
molecules interact more closely in the liquid than in the gas phase.
Ice, H2O
, has a solid structure with alternating H−O
interactions.
H2Te
has a higher boiling point than H2S
The interactions that can be explained by hydrogen bonding are:
HF has a higher boiling point than HClIce, H₂O, has a solid structure with alternating H−O interactions.Hydrogen bonding is a type of intermolecular force that occurs between a hydrogen atom in a polar molecule and a highly electronegative atom, such as nitrogen, oxygen, or fluorine, in a neighboring molecule. It is a relatively strong force and can significantly affect the physical and chemical properties of substances.
In the case of HF and HCl, both molecules are polar, but HF has a higher boiling point due to the stronger hydrogen bonding between its molecules. In ice, the hydrogen bonding between water molecules creates a crystal structure with a characteristic lattice arrangement.
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Show that the following reaction obeys zero order kinetics, and find the rate constant and half-life. 10²x[A]/mol dm-3 13.80 11.05 8.43 5.69 3.00 Time/hr 0 10 20 30 40
The half life for the given reaction is 20s. The rate constant for the given zero order reaction can be given as 0.5mol L⁻¹s⁻¹.
According to the definition of a zero-order reaction, it is "a chemical reaction that occurs when the rate of response remains unchanged whether the percentage of the reactants increases or decreases." Since the rate increases according in proportion to the 0th magnitude of the reactant concentration, the rate at which these responses is always equivalent to the constant rate for the particular reactions.
k = 0.5mol L⁻¹s⁻¹
Half life is a/2K=3/2×40
= 20s
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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?
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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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 =
Answer:
216.9 K
Explanation:
We can use the combined gas law to solve this problem:
(P1V1/T1) = (P2V2/T2)
where P is pressure, V is volume, and T is temperature. We can assume that the pressure is constant, since the problem doesn't mention any changes in pressure.
Let's label the initial temperature T1 and the final temperature T2, and plug in the given values:
(P1V1/T1) = (P2V2/T2)
V1 = 346.9 mL
V2 = 596.2 mL
T2 = 373 K
(P1 * 346.9 mL / T1) = (P1 * 596.2 mL / 373 K)
Simplifying and solving for T1:
T1 = (P1 * 346.9 mL * 373 K) / (P1 * 596.2 mL)
T1 = (346.9 * 373) / 596.2
T1 = 216.9 K
Therefore, the temperature of the carbon dioxide sample at 346.9 mL was 216.9 K.
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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?
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.
Which of the following reactants lead to the fastest reaction?
A) Solid aluminum and oxygen
B) Sodium hydroxide and hydrogen chloride in water
C) Copper nitrate and ethanol in water
D) Solid barium chloride and sodium sulfate
Solid aluminum and oxygen are the reactant that lead to the fastest reaction. Therefore, the correct option is option A.
Chemical reaction, the transformation of one or more chemicals (the reactants) into one or more distinct compounds (the products). Chemical elements or chemical compounds make up substances.
Chemical reactions constitute a fundamental component of life itself, as well as technology and culture. Burning fuels, melting iron, creating glass or pottery, brewing beer, making wine. Solid aluminum and oxygen are the reactant that lead to the fastest reaction.
Therefore, the correct option is option A.
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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
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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Calculate the mass of water produced when 1.03 g
of butane reacts with excess oxygen.
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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What is the energy change for the following equation?
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
Calculation: 3. A new alloy of steel is 525 g at 100-C. It is dropped into 375 grams of water at 25 C. The final temperature changes to 55C, what is the specific heat of steel?
The specific heat capacity of the steel is 2 g/°C.
What is the specific heat capacity of the steel?To determine the specific heat capacity of a particular sample of steel, the sample can be heated to a known temperature and then placed in contact with a known amount of water at a lower temperature.
We know that the heat lost by the steel is equal to the heat gained by the water.
Thus we have that;
-(525 * c * (55- 100)) = (375 * 4.2 * (55- 25)
23625c = 47250
c = 47250/23625
c = 2 g/°C
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Under what conditions would one mole of methane gas, ch4 , occupy the smallest volume?
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.
Give the number of significant figures indicated.
0.066
Compare and contrast the four different types of solar activity above the photosphere. (Select all that apply)
A. Prominences are huge loops of the Sun’s ionized but cool material that are pushed by magnetic forces from the chromosphere into the corona.
B. There are regions of higher density and temperature than the surrounding material in the chromosphere called plages.
C. Flares are long lived phenomena above the photosphere beginning at the solar minimum and ending at solar maximum.
D. Coronal mass ejections happen when a flare is so violent that the material exceeds the escape speed of the sun. It is connected to sunspots.
There are up to four different kinds of features that can be seen when viewing the photosphere. Sunspots, faculae, granulation, and super-granulation are listed in decreasing order of observational ease. All the given statement are true.
Solar activityIn the chromosphere, plateaus are areas that are more dense and hotter than the surrounding material. Large, chilly, ionized loops of material known as prominences are slowly pushed into the corona by magnetic energy from the chromosphere. Flares are erratic, fiery, and intensely energetic bursts that last only a moment or two. When a flare is extremely powerful, its material is thrown into the solar system beyond the Sun's escape velocity and causes coronal mass ejections. These are all essentially connected to sunspots.For more information on solar activity above photosphere kindly visit to
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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?
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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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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Determine which statements are consistent with Daltons atomic theory
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.
How many carbon atoms are in 11.5 g C2H5OH?
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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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.
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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Select the correct answer. A certain reaction has this form:aA bB. At a particular temperature and [A]0 = 2.00 x 10-2 Molar, concentration versus time data were collected for this reaction and a plot of ln[A]t versus time resulted in a straight line with a slope value of -2.97 x 10-2 min-1. What is the reaction order and rate law for this reaction? A. first, rate = k[A] B. first, rate = k[A]2 C. second, rate = k[A] D. second, rate = k[A]2 E. third, rate = k[A]
The reaction order and rate law for this reaction is first, rate = k[A]. Option A is correct.
The reaction order and rate law for a reaction can be determined from the slope of a plot of ln[A] versus time.
Given that the plot of ln[A] versus time resulted in a straight line with a slope value of -2.97 x 10⁻² min⁻¹, we can determine the reaction order and rate law as follows;
If the slope is equal to 1, the reaction order is 1st order.
If the slope is equal to 2, the reaction order is 2nd order.
If the slope is equal to 3, the reaction order is 3rd order.
From the given slope of -2.97 x 10⁻² min⁻¹, we can conclude that the reaction order is 1st order, because the slope value is equal to -1 times the reaction order. So, the reaction order for this reaction is 1st order.
The rate law for a 1st order reaction will be given by;
rate = k[A]
where [A] is the concentration of the reactant, and k is the rate constant.
Therefore, first, rate = k[A].
Hence, A. is the correct option.
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Calculate the molality of a solution prepared by dissolving 19.9 g of KCl in 750.0 mL of water.
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 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
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
What is the temperature of 0.59 mol of gas at a pressure of 1.0 atm and a volume of 11.0 L? (Express the temperature in kelvins to two significant figures).
Answer:
T = 230 K
Explanation:
Simply use the equation: PV = nRT, where P is pressure, V is volume, n is moles, R is 0.0821, and T is temperature in kelvin
Place the values in
(1.0 atm)(11.0 L)=(0.59 moles)(0.0821 Lxatm/Kxmol)(T)
Solve for T = 230 K
Consider a sample containing 0.300 mol of a substance. How many atoms are in the sample if the substance is iron?
The number of atoms in a sample containing 0.300 mol of iron is 1.8066 x 10²³ atoms.
The number of atoms in a sample of a substance can be determined using Avogadro's number, which is equal to 6.022 x 10²³ particles per mole. The atomic mass of iron is 55.85 g/mol.
Therefore, 0.300 mol of iron will contain:
0.300 mol x 6.022 x 10³ atoms/mol = 1.8066 x 10²³ atoms
Avogadro's number is a fundamental constant in chemistry that represents the number of particles (atoms or molecules) in one mole of a substance.
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