Answer:
The particles decrease & they lose energy.
Explanation:
Answer:
When a substance is cooled, its internal energy decreases: the movement of its particles decreases. bonds between particles form when a substance condenses or freezes, or sublimes to form a solid from a gas
Explanation:
Acetylene gas (ethyne; HC≡CH) burns with oxygen in an oxyacetylene torch to produce carbon dioxide, water vapor, and the heat needed to weld metals. The heat of combustion for acetylene is −1259 kJ/mol. Calculate the C≡C bond energy.
The average bond energies (kJ/mol)
C―H 413
C―C 347
C═C 620
H―H 436
O―O 142
O═O 498
H―O 467
C―O 351
C═O 799
The C≡C bond energy in acetylene (HC≡CH) can be calculated by subtracting the sum of the bond energies in the products from the bond energies in the reactants. The bond energy of the C≡C bond is approximately 839 kJ/mol.
To calculate the C≡C bond energy in acetylene, we need to consider the bond energies of the various bonds involved in the reaction. The given bond energies are as follows:
C―H: 413 kJ/mol
C―C: 347 kJ/mol
C═C: ?
H―H: 436 kJ/mol
O―O: 142 kJ/mol
O═O: 498 kJ/mol
H―O: 467 kJ/mol
C―O: 351 kJ/mol
C═O: 799 kJ/mol
The balanced equation for the combustion of acetylene is:
2C2H2 + 5O2 → 4CO2 + 2H2O
Using the bond energies, we can calculate the energy change in the reaction:
ΔH = (4 × C═O) + (2 × O―H) - (2 × C≡C) - (5 × O═O) - (4 × C―O) - (2 × O―H)
Substituting the values:
ΔH = (4 × 799) + (2 × 467) - (2 × C≡C) - (5 × 498) - (4 × 351) - (2 × 467)
Simplifying:
ΔH = 3196 + 934 - 2C≡C - 2490 - 1404 - 934
ΔH = -1259 kJ/mol
To isolate the C≡C bond energy, we rearrange the equation:
2C≡C = 3196 + 934 - 2490 - 1404 - 934 + 1259
2C≡C = 351
C≡C = 351 / 2 ≈ 175.5 kJ/mol
Therefore, the approximate bond energy of the C≡C bond in acetylene is approximately 175.5 kJ/mol.
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Please answer fast! The density of aluminum is 2.70 g/cm? The volume of a solid piece of aluminum is 1.50 cm? Find its mass.
Answer:
19.3 g/cm3
Explanation:
is the best answer
A student measures a volume as 25 mL, whereas the correct volume is 23 mL. What is the percent error? * O 8.7% O 0.92% O 0.087% 08.0%
An error can be defined as a deviation from the correct value of a given measurement.
The percent error is 8.7%
The formula for percent error is given as:
Percent error = |Measured value – Exact Value|/Exact value * 100.
From the above:
Measured value = 25 ml
Exact Value = 23 ml
Hence:
Percent Error = | 25ml - 23 ml | / 23 ml * 100
Percent Error = 2/23 * 100
Percent Error = 8.7%
Therefore the percent error is 8.7%
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Which diagram illustrates that matter is always conserved during a chemical reaction?
A
B
C
D
Answer:
the photography is the C) is the correct
According to conservation of mass, the total mass in a chemical reaction is always conserved. Thus atoms lost or gained. In diagram D the number of atoms in the reactant side is equal to the number of atoms in the product side. Thus option D is correct.
What is mass conservation?Mass conservation is same as the law of conservation of energy. Thus total mass of a system is conserved. Therefore, mass can neither be created nor be destroyed.
According to mass conservation, for a chemical reaction, mass in the reactant side will be equal to the mass in the product side. Hence, no mass new atoms are created or existing atoms are lost.
The total number of atoms or molecules in the reactant side will be equal to total number of atoms in the product side in a balanced reaction, where all the elements are given in perfect stoichiometric proportions.
Only diagram D shows equal number of atoms in product side and reactant side and thus it is correct according to law of conservation of mass. Hence, option D is correct.
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ssign oxidation states to each atom in each of the following species.
1. N2
2. Fe3+
3. CuCl2
4. CH4
5. Cr2O72-
6. HCO3-
Oxidation states to each atom in each of the following species:
1. N₂: N has an oxidation state of 0.
2. Fe₃⁺: Fe has an oxidation state of +3.
3. CuCl₂: Cu has an oxidation state of +2, and Cl has an oxidation state of -1.
4. CH₄: C has an oxidation state of -4, and H has an oxidation state of +1.
5. Cr₂O₇⁻²: Cr has an oxidation state of +6, and O has an oxidation state of -2.
6. HCO₃⁻: H has an oxidation state of +1, C has an oxidation state of +5, and O has an oxidation state of -2.
1. N₂: Since it is an elemental form of nitrogen, the oxidation state of each nitrogen atom is 0.
2. Fe⁺³:
The oxidation state of Fe (iron) is +3, as indicated by the superscript of 3+.
3. CuCl₂:
The oxidation state of Cu (copper) is +2 because the overall charge of the compound is 0, and there are two Cl (chlorine) atoms with an oxidation state of -1 each.
4. CH₄:
The oxidation state of C (carbon) is -4 because hydrogen (H) typically has an oxidation state of +1, and there are four hydrogen atoms in CH4.
5. Cr₂O₇²⁻:
The oxidation state of Cr (chromium) is +6 because there are two oxygen atoms with an oxidation state of -2 each. The overall charge of the compound is -1, so the sum of oxidation states must be equal to -1. Therefore, each Cr atom must have an oxidation state of +6.
6. HCO₃⁻:
The oxidation state of H (hydrogen) is +1. The oxidation state of C (carbon) can be calculated by assuming the oxidation state of O (oxygen) is -2. Since there are three O atoms with an oxidation state of -2 each, the oxidation state of C can be determined as follows:
(+1) + (x) + (3 * -2) = -1
x - 6 = -1
x = +5
Thus, the oxidation state of C is +5 in HCO₃⁻.
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A block of concrete has a density of 9 kg/m^3 and a volume of 210 000 cm^3 What is its mass
Answer:
The answer is 1.89 kgExplanation:
The mass of a substance when given the density and volume can be found by using the formula
mass = Density × volumeFrom the question
density = 9 kg/m³
1 cm³ = 0.000001 m³
210,000 cm³ = 0.21 m³
So we have
mass = 9 × 0.21
We have the final answer as
1.89 kgHope this helps you
Which of the following information is primarily obtained from infrared spectroscopy? Select one: O a. arrangement of carbon and hydrogen atoms in a compound b. molecular weight of a compound O c. any conjugated n system present in a compound O d. functional groups present in a compound e. all of these Check
Answer: A
Explanation:
Methanol, ethanol, and n-propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is liberated as shown by the following data: (a) methanol (CH3OH), –22.6 kJ/g; (b) ethanol (C2H5OH), –29.7 kJ/g; (c) n-propanol (C3H7OH), –33.4 kJ/g. Calculate the heats of combustion of these alcohols in kJ/mol.
The heat of combustion of n-propanol is approximately -0.555 kJ/mol.To calculate the heats of combustion of these alcohols in kJ/mol, we need to convert the given data from kJ/g to kJ/mol.
The molar masses of methanol, ethanol, and n-propanol are 32.04 g/mol, 46.07 g/mol, and 60.10 g/mol, respectively. By dividing the given heat of combustion values by the molar mass of each alcohol, we can determine the heats of combustion in kJ/mol.
(a) For methanol:
Heat of combustion = -22.6 kJ/g
Molar mass of methanol = 32.04 g/mol
Heat of combustion per mole of methanol = (-22.6 kJ/g) * (1 g/mol) / (32.04 g/mol) = -0.706 kJ/mol
Therefore, the heat of combustion of methanol is approximately -0.706 kJ/mol.
(b) For ethanol:
Heat of combustion = -29.7 kJ/g
Molar mass of ethanol = 46.07 g/mol
Heat of combustion per mole of ethanol = (-29.7 kJ/g) * (1 g/mol) / (46.07 g/mol) = -0.645 kJ/mol
Therefore, the heat of combustion of ethanol is approximately -0.645 kJ/mol.
(c) For n-propanol:
Heat of combustion = -33.4 kJ/g
Molar mass of n-propanol = 60.10 g/mol
Heat of combustion per mole of n-propanol = (-33.4 kJ/g) * (1 g/mol) / (60.10 g/mol) = -0.555 kJ/mol
In summary, the heats of combustion for methanol, ethanol, and n-propanol are approximately -0.706 kJ/mol, -0.645 kJ/mol, and -0.555 kJ/mol, respectively. These values represent the amount of heat released per mole of alcohol when burned in air.
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11.16 (a| Which is generally stronger; intermolecular tions Or intramolecular interactions? (b) intetac Which kinds of interactions are broken when a liquid is hey 'gas? converted toa
Intermolecular interactions are generally weaker than intramolecular interactions. When a liquid is converted to a gas, intermolecular interactions are broken.
Intermolecular interactions refer to the forces of attraction between molecules, while intramolecular interactions are the forces holding atoms within a molecule together. Intramolecular interactions, such as covalent or ionic bonds, are generally stronger because they involve the sharing or transfer of electrons between atoms.
On the other hand, intermolecular interactions, such as van der Waals forces (London dispersion forces, dipole-dipole interactions, and hydrogen bonding), are weaker in comparison. These interactions occur between neighboring molecules and are responsible for determining many physical properties of substances, such as boiling points, melting points, and solubilities.
When a liquid is converted to a gas, the intermolecular interactions between the molecules need to be overcome. This requires an input of energy to break these intermolecular forces, allowing the molecules to move freely and form a gas. The energy provided, typically in the form of heat, increases the kinetic energy of the molecules, causing them to overcome the attractive forces and transition into the gaseous state.
In summary, intramolecular interactions are generally stronger than intermolecular interactions. When a liquid is converted to a gas, the intermolecular interactions are broken as the molecules gain enough energy to overcome these forces and transition to the gaseous state.
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need help 1-5 test please quick
The thing that occurs in a single, polar covalent bond c. Two atoms unequally share two electrons.
How to explain the informationThe thing that occurs in a triple, nonpolar covalent bond is b. Two atoms equally share six electrons.
The bond that requires the most energy to break the bond is c. Triple covalent bond.
A molecule of hydrogen monochloride is polar because d. The chlorine attracts the shared electrons more strongly than does the hydrogen atom.
Diatomic iodine (I) has a higher boiling point than diatomic bromine (Br₂) because diaton has a. Larger molecule.
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Which of the following statements is the best description of the nature of light?
The nature of light is best described as a dualistic phenomenon, encompassing both wave-like and particle-like properties, as demonstrated by its behavior in various experimental contexts.
The nature of light is best described as a dualistic phenomenon, exhibiting properties of both particles and waves. This concept, known as wave-particle duality, arises from the observation that light behaves as both discrete particles, called photons, and as a wave-like disturbance in the electromagnetic field.
From a wave perspective, light exhibits characteristics such as diffraction, interference, and polarization. These phenomena suggest that light propagates as a wave, with properties such as wavelength, frequency, and amplitude. The wave model successfully explains various optical phenomena, including the bending of light around obstacles and the formation of colorful interference patterns.
On the other hand, light also displays particle-like behavior. This is evident in the photoelectric effect, where light transfers discrete amounts of energy to electrons, behaving as discrete particles called photons. The energy of each photon is directly proportional to its frequency, as described by Max Planck's quantum theory.
Moreover, the behavior of light can be further understood through the framework of quantum mechanics, where photons are described by wavefunctions and exhibit probabilistic behavior. This probabilistic nature is captured by the wave-particle duality concept, which acknowledges that light can exhibit wave-like or particle-like characteristics depending on the experimental setup.
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Isotopes are best described as two atoms of the same element that
Question 9 options:
have the same mass but different chemical properties
give away electrons to become positively charged
differ in mass but are chemically alike
become electrically charged
Explanation:
they have different mass number but same atomic number
Fluctuations in the phosphorus cycle in aquatic ecosystems _______.
a.
can lead to changes in the growth rates of aquatic populations
b.
are never caused by human impact on the environment
c.
lead to a decrease in the amount of phosphorus found in sediment
d.
never occur naturally
Answer:
The answer is A
Explanation:
Fluctuations in the phosphorus cycle in aquatic ecosystems can lead to changes in the growth rates of aquatic populations.
What is an ecosystem?
Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.
Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.
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36) What is the mass of a 4.259 g/cm substance which takes up 250.00 cm of space?
Answer:
The answer is 1064.75 gExplanation:
The mass of a substance when given the density and volume can be found by using the formula
mass = Density × volumeFrom the question
volume of substance = 250 cm³
density = 4.259 g/cm³
We have
mass = 4.259 × 250
We have the final answer as
1064.75 gHope this helps you
A normal breath takes in about 1.0 L of air. Assuming that air has an average molar mass of 28.8 g/mol, and that its density is 0.97 g/L, how many molecules of air do you take in with each breath?
a. 2.0 x10^22 b. 2.2 10^22 c. 5.8 x 10^23 d. 1.7 x 10^25 e. 1.8 x10^25
With each breath, you take in approximately 2.0 x 10^22 molecules of air.
To calculate the number of molecules of air taken in with each breath, follow these steps:
Calculate the number of moles of air:
Moles of air = Volume of air / molar volume of air
Molar volume of air = molar mass of air / density of air
Calculate the number of molecules:
Number of molecules = Moles of air * Avogadro's number
Given that the volume of air taken in with each breath is 1.0 L, the molar mass of air is 28.8 g/mol, and the density of air is 0.97 g/L, we can calculate the number of molecules as follows:
Molar volume of air = (28.8 g/mol) / (0.97 g/L)
Moles of air = 1.0 L / Molar volume of air
Number of molecules = Moles of air * Avogadro's number
By performing these calculations, the number of molecules of air taken in with each breath is approximately 2.0 x 10^22.
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why is water an excellent solvent for most ionic compounds and polar covalent molecules but not for non-polar compounds?
Water is an excellent solvent for most ionic compounds and polar covalent molecules but not for non-polar compounds because of its content loaded with certain chemical properties that make it a highly effective solvent. Water molecules are polar, and due to their dipolar nature, the oxygen atom carries a negative charge while the hydrogen atoms carry a positive charge.
The polarity of the water molecule allows it to interact with and dissolve other polar and ionic substances.When an ionic compound is dissolved in water, the ions of the compound dissociate into individual charged species (cations and anions), and these charged species are solvated by water molecules. The polarity of the water molecule allows it to interact with the ions by attracting the positively charged ions to the negative end of the water molecule and vice versa.
The same is true for polar covalent molecules, which have a net dipole moment. Water molecules can interact with these molecules, forming a solvation layer around them. On the other hand, non-polar compounds lack a net dipole moment, so they don't interact with the water molecules. Instead, non-polar compounds interact with each other via van der Waals forces, making it more challenging for them to dissolve in water.
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The Ksp of beryllium hydroxide, Be(OH)₂, is 6.92 x 10-22. Calculate the solubility of this compound in grams per liter. solubility: 5.57 X10-8 Incorrect
The solubility of beryllium hydroxide, Be(OH)₂, can be calculated using the given Ksp value of 6.92 x 10⁻²².
Solubility is the maximum amount of a compound that can dissolve in a given solvent. To find the solubility in grams per liter (g/L), we use the Ksp equation and consider the concentration of Be(OH)₂ in moles per liter.
The Ksp expression becomes 4x³, where x represents the concentration of OH⁻ ions. Solving for x gives approximately 1.78 x 10⁻⁸ mol/L.
Multiplying this by the molar mass of Be(OH)₂ (43.03 g/mol) yields a solubility of around 7.66 x 10⁻⁷ g/L.
Therefore, the solubility of beryllium hydroxide is approximately 7.66 x 10⁻⁷ grams per liter.
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A 0. 750 M solution of ascorbic acid has a ph of
2. 11. Calculate the Ka of ascorbic acid.
To calculate the Ka (acid dissociation constant) of ascorbic acid, we can use the pH and the concentration of the acid solution.
Given:
pH of the solution = 2.11
Concentration of ascorbic acid = 0.750 M
Step 1: Convert the pH to the concentration of H⁺ ions.
pH = -log[H⁺]
2.11 = -log[H⁺]
[H⁺] = 10^(-pH)
[H⁺] = 10^(-2.11)
Step 2: Calculate the concentration of ascorbic acid that has dissociated.
Since ascorbic acid is a monoprotic acid, the concentration of H⁺ ions is equal to the concentration of the dissociated acid.
[H⁺] = [dissociated ascorbic acid]
Step 3: Calculate the concentration of undissociated ascorbic acid.
[undissociated ascorbic acid] = [total ascorbic acid] - [dissociated ascorbic acid]
[undissociated ascorbic acid] = 0.750 M - [H⁺]
Step 4: Write the expression for the acid dissociation reaction.
Ascorbic acid (HAsc) ⇌ H⁺ + Asc⁻
Step 5: Write the equilibrium expression for the acid dissociation.
Ka = ([H⁺] [Asc⁻]) / [HAsc]
Step 6: Substitute the concentrations into the equilibrium expression.
Ka = ([H⁺] [Asc⁻]) / [HAsc]
Ka = ([H⁺] [H⁺]) / [undissociated ascorbic acid]
Ka = ([H⁺]^2) / [undissociated ascorbic acid]
Ka = ([10^(-2.11)]^2) / [0.750 M - 10^(-2.11)]
Calculating the value of Ka requires substituting the values into the equation and performing the necessary calculations.
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the two methods of determining ph values (ph indicator versus ph meter) should show similar ph values for those solutions. what was different?
The two methods of determining pH values, pH indicator versus pH meter,, should show similar pH values for those solutions. However, there are some differences between the two methods. Here's what's different:The pH meter measures the electrical potential difference between the reference electrode and the sensing electrode of the device.
It offers a more accurate reading of the pH value. pH meters are used to measure a wide variety of samples, such as groundwater, drinking water, wastewater, and other solutions that have pH values outside of the range that pH paper can measure.
PH indicator paper, on the other hand, is an easy-to-use and inexpensive method for determining pH. They work on the principle of visual color changes, which are brought about by the presence of hydrogen ions (H+) or hydroxyl ions (OH-). pH paper can be used to measure pH values in various industries, including food production, scientific research, medical diagnosis, and wastewater treatment facilities. pH paper is ideal for low-volume and small-scale pH measurements.
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Calculate the ratio of moles of H2O to moles of anhydrous KAl(SO4)2 which has a mass of 3.5g.
The ratio of moles of H2O to moles of anhydrous KAl(SO4)2 can be calculated by considering the molar masses of water and anhydrous KAl(SO4)2. In this case, the molar mass of water is 18.015 g/mol and the molar mass of anhydrous KAl(SO4)2 is 258.21 g/mol.
By dividing the mass of anhydrous KAl(SO4)2 (3.5 g) by its molar mass, we can determine the number of moles of anhydrous KAl(SO4)2. Then, by comparing the stoichiometric ratio between water and anhydrous KAl(SO4)2 in the chemical formula, we can determine the ratio of moles of H2O to moles of anhydrous KAl(SO4)2.
To calculate the ratio of moles of H2O to moles of anhydrous KAl(SO4)2, we need to determine the number of moles of anhydrous KAl(SO4)2 first. We can do this by dividing the mass of anhydrous KAl(SO4)2 (3.5 g) by its molar mass (258.21 g/mol). This calculation gives us approximately 0.0136 moles of anhydrous KAl(SO4)2.
Next, we need to consider the stoichiometric ratio between water and anhydrous KAl(SO4)2. The chemical formula of anhydrous KAl(SO4)2 indicates that there are 2 moles of water for every 1 mole of anhydrous KAl(SO4)2. Therefore, for every mole of anhydrous KAl(SO4)2, there are 2 moles of water.
Based on this information, the ratio of moles of H2O to moles of anhydrous KAl(SO4)2 is 2:1. This means that for the given mass of 3.5 g of anhydrous KAl(SO4)2, there would be approximately 0.0136 moles of anhydrous KAl(SO4)2 and double that amount, approximately 0.0272 moles, of water.
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3. A rocket sled accelerates from 10 m/sec to 50 m/sec in 2 seconds. What O points
is the acceleration? *
A. 25 m/sec2
B. 20 m/sec2
Ci 10 m/sec2
D. 15 m/sec2
Answer:
b
Explanation:
by using formula
a=(v-u)/t
where a= acceleration
v = initial velocity = 10 m/s
u= final velocity =50 m/s
t = time taken = 2s
The acceleration of the rocket has been given by 20 m/[tex]\rm sec^2[/tex]. Thus option B is correct.
Acceleration has been the increase in the speed of the object with respect to the time. The acceleration can be given by:
a = [tex]\rm \dfrac{v\;-\;u}{t}[/tex]
Where,
a = acceleration
v = final velocity = 50 m/s
u = initial velocity = 10 m/s
t = time = 2 s
Substituting the values:
Acceleration = [tex]\rm \dfrac{50\;-\;10}{2}[/tex]
Acceleration = 20 m/[tex]\rm sec^2[/tex]
The acceleration of the rocket has been given by 20 m/[tex]\rm sec^2[/tex]. Thus option B is correct.
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Calculate the solubility of mg(oh)2 in 0. 50m nh4cl
The solubility of Mg(OH)2 in 0.50 M NH4Cl is calculated below: What is solubility? Solubility is the amount of solute that dissolves in a solvent at a specific temperature and pressure. Solubility is typically expressed as grams per litre (g/L) or moles per litre (mol/L).
Solubility is a chemical property that varies with temperature, pressure, and the presence of other substances (solutes) in a solution. What is Mg(OH)2? Magnesium hydroxide is a white solid that is odorless and non-volatile. It has a low solubility in water (12 mg/L at 25°C), making it a weak base that is primarily used as an antacid and laxative. Calculation: The reaction for Mg(OH)2 dissolving in water is: Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)Ksp = [Mg2+][OH-]2We'll need to calculate the concentrations of Mg2+ and OH- ions to calculate the solubility of Mg(OH)2 in 0.50 M NH4Cl. First, let's write the dissociation equation for NH4Cl:NH4Cl(s) ⇌ NH4+(aq) + Cl-(aq).
Since NH4Cl is a strong electrolyte, it completely dissociates into NH4+ and Cl- ions. When NH4+ ions interact with OH- ions from the Mg(OH)2 dissociation equation, they produce NH3 and water:NH4+(aq) + OH-(aq) ⇌ NH3(aq) + H2O(l). This reaction depletes the concentration of OH- ions in the solution, making it more difficult for Mg(OH)2 to dissolve. Because NH3 is a weak base, it does not significantly impact the concentration of OH- ions in the solution. To calculate the solubility of Mg(OH)2 in 0.50 M NH4Cl, we must first determine the concentration of OH- ions in the solution. To do so, we must first find the concentration of NH4+ ions in the solution.
Using the dissociation equation for NH4Cl, we know that [NH4+] = [Cl-] = 0.50 M (since NH4Cl is a strong electrolyte, it completely dissociates into NH4+ and Cl- ions).Using the equilibrium constant expression for the NH3 reaction, we can write the following expression for [OH-]:Kb = [NH3][H2O]/[NH4+] = 1.8 × 10-5pOH = 1/2(pKb - log([NH3]/[NH4+])) = 1/2(4.74 - log(0.50/0.50)) = 2.37[OH-] = 10-pOH = 4.47 × 10-3 M. Solving for [Mg2+] using the Ksp expression for Mg(OH)2 and the concentration of OH- ions, we obtain:[Mg2+] = Ksp/[OH-]2 = (5.6 × 10-12)/(4.47 × 10-3)2 = 2.47 × 10-7 M. Therefore, the solubility of Mg(OH)2 in 0.50 M NH4Cl is 2.47 × 10-7 M.
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True or False. A force is a push or pull exerted on an object.
Answer:
TRUE
Explanation:
remember newtons laws of motion. Every action has an equal and opposite reaction. LOL
Answer:
True.
Explanation:
the concentration of ammonia increases from 0.257 m to 0.815 m in 15.0 min. calculate the average rate of reaction over this time interval'
The average rate of reaction is the change in the concentration of reactants or products over a specified time interval. The average rate of reaction can be calculated from the initial and final concentrations of the reactants or products. To find the average rate of reaction between two concentrations, the formula is:
Average rate of reaction = Δ concentration / Δ timeIn the given problem, the concentration of ammonia increases from 0.257 m to 0.815 m in 15.0 min. To calculate the average rate of reaction over this time interval, we can use the above formula as follows:Δ concentration = (0.815 - 0.257) m = 0.558 mΔ time = 15.0 minAverage rate of reaction = Δ concentration / Δ time= 0.558 m / 15.0 min= 0.0372 m/minTherefore, the average rate of reaction over this time interval is 0.0372 m/min.
The rate of a chemical reaction is defined as the change in concentration of a reactant or product with respect to time. The rate of reaction is expressed in units of concentration per unit time, typically moles per liter per second (mol/L/s). The average rate of reaction is the change in concentration of reactants or products over a specified time interval. This is calculated by finding the slope of the concentration versus time graph during the specified time interval. If the reaction rate is constant over the specified time interval, the average rate of reaction will be equal to the instantaneous rate of reaction at any point in time during that interval.In the given problem, the concentration of ammonia increases from 0.257 m to 0.815 m in 15.0 min. The average rate of reaction over this time interval can be calculated using the formula:Average rate of reaction = Δ concentration / Δ timewhere Δ concentration is the change in concentration of ammonia, and Δ time is the time interval over which the concentration changes. In this case, Δ concentration = (0.815 - 0.257) m = 0.558 m and Δ time = 15.0 min. Substituting these values into the formula, we get:Average rate of reaction = Δ concentration / Δ time= 0.558 m / 15.0 min= 0.0372 m/minTherefore, the average rate of reaction over this time interval is 0.0372 m/min.
The average rate of reaction can be calculated from the change in the concentration of reactants or products over a specified time interval. In this problem, the average rate of reaction was calculated using the formula Δ concentration / Δ time, where Δ concentration was the change in concentration of ammonia and Δ time was the time interval over which the concentration changed. The average rate of reaction was found to be 0.0372 m/min.
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A sample of a solid organic compound is completely combusted in a calorimeter. The heat generated by combustion is transferred to the water, causing the temperature of the water to increase. Based on the data in the table above, which of the following is the best estimate of the heat of combustion, ΔHcomb, of the organic solid? (The specific heat of water is 4.2J/(g⋅°C) )
the best estimate of the heat of combustion (ΔHcomb) of the organic solid is 8400 J. To estimate the heat of combustion (ΔHcomb) of the organic solid.
we need to use the information provided in the table and the change in temperature of the water.
Specific heat of water (C) = 4.2 J/(g⋅°C)
Mass of water (m) = 100 g
Change in temperature of water (ΔT) = 20 °C
The heat transferred to the water can be calculated using the equation:
Q = m * C * ΔT
Plugging in the values:
Q = 100 g * 4.2 J/(g⋅°C) * 20 °C
Q = 8400 J
Since the heat of combustion is the heat released during the combustion process, we can estimate the heat of combustion (ΔHcomb) of the organic solid to be approximately equal to the heat transferred to the water, which is 8400 J.
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(ik this isn’t chemistry) if i had a neutral atom of oxegyn and removed an electron from it, what kind of ion would i get?
1)anion
2)cation
3)dogion
4)mouseion
Answer:
Cation
Explanation:
I'm pretty sure the last two options are a joke, but electrons are negative so losing one gives a positive charge.
what is the ph after 0.150 mol of hcl is added to the buffer from part a? assume no volume change on the addition of the acid.express the ph numerically to three decimal places.
A buffer is a solution that resists changes in pH. A buffer solution is composed of a weak acid (or a weak base) and its conjugate base (or acid), with the weak acid and the salt of the weak acid existing in roughly equal proportions. The addition of an acid or a base to a buffer has a minimal effect on its pH.
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log10([A-]/[HA]) where A- is the conjugate base of the weak acid HA, [A-] is the concentration of the conjugate base, [HA] is the concentration of the weak acid, and pKa is the acid dissociation constant. The given values for the buffer system from Part A are:[HA] = 0.0500 M[A-] = 0.0350 M (pKa for acetic acid is 4.74).
The first step is to calculate the initial pH of the buffer: pH = pKa + log10([A-]/[HA])pH = 4.74 + log10(0.0350/0.0500)pH = 4.74 - 0.093pH = 4.65When 0.150 mol of HCl is added, it will react with the acetate ion A- to form acetic acid HA: HCl + A- → HA + Cl-The reaction will consume some of the acetate ion A- and produce some hydronium ion H3O+. The new concentrations will be:[HA] = 0.0500 M[A-] = 0.0350 M - 0.150 M = 0.0344 M[H3O+] = 0.150 M. Now we can calculate the pH of the new solution: pH = pKa + log10([A-]/[HA])pH = 4.74 + log10(0.0344/0.0500)pH = 4.74 - 0.115pH = 4.63Therefore, the pH after 0.150 mol of HCl is added to the buffer from Part A is 4.63.
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How many grams does 0.500 moles of CuBr weigh
Answer:
72 grams
Explanation:
n=m/mr
so
m=n*mr
0.500*144
=72 g
There are 71.725 grams of CuBr is present in 0.500 moles of CuBr. It is required to know the substance's molar mass in order to calculate its weight given its number of moles.
The atomic masses of copper (Cu) and bromine (Br) in the compound may be added to determine the molar mass of CuBr.
The atomic mass of Cu = 63.55 grams/mol, and
The atomic mass of Br = 79.90 grams/mol.
Molar mass of CuBr = (atomic mass of Cu) + (atomic mass of Br)
= 63.55 g/mol + 79.90 g/mol
= 143.45 g/mol
Now, find the weight of 0.500 moles of CuBr:
Weight of CuBr = Number of moles × Molar mass
= 0.500 moles × 143.45 g/mol
= 71.725 grams
Thus, 0.500 moles of CuBr weighs 71.725 grams.
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Hurry please!!! Lithospheric plates move because of the movement of the ___________________ under them.
a. lower mantle
b. asthenosphere
c. outer core
d. inner core
Answer:
A. Lower mantle
Explanation:
Answer:
lower mantle
Explanation:
how many grams of nacl are needed to prepare 0.500 l of a 4.00 m nacl solution?select one:a.2.00 grams naclb.4.00 grams naclc.117 grams nacld.58.5 grams nacl
To determine the number of grams of NaCl required to prepare a 0.5 L of 4.00 M NaCl solution, we need to use the following formula;M = (mol / L)mol = M × Lmol = 4.00 mol/L × 0.5 Lmol = 2 mol. We have determined that we need 2 moles of NaCl for this solution.
To convert moles to grams, we need to use the following formula;moles = mass / molar massmass = moles × molar massThe molar mass of NaCl is 58.44 g/mol. Therefore;mass = 2 mol × 58.44 g/mol = 116.88 g
From the calculation above, the number of grams of NaCl required to prepare 0.5 L of a 4.00 M NaCl solution is 116.88 grams. Therefore, the correct option is option (c) 117 grams NaCl.The formula to calculate molarity (M) is M = mol/L, and the formula to calculate mass is mass = moles × molar mass. We can use these two formulas to calculate the grams of NaCl required for this solution.The first step is to determine the number of moles required. We were given a 0.5 L of a 4.00 M NaCl solution. Therefore, we will use M = mol/L to determine the number of moles.M = (mol / L)4.00 M = mol / 0.5 Lmol = 4.00 M × 0.5 Lmol = 2 molThe second step is to convert the number of moles to grams by using the formula;mass = moles × molar massThe molar mass of NaCl is 58.44 g/mol.mass = 2 mol × 58.44 g/molmass = 116.88 gTherefore, the answer is option (c) 117 grams NaCl.
Therefore, the number of grams of NaCl required to prepare a 0.500 L of a 4.00 M NaCl solution is 116.88 g or 117 g.
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