Answer : The final concentration of [tex]N_2O_5[/tex] is, 2.9 M
Explanation :
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]5.89\times 10^{-3}\text{ min}^{-1}[/tex]
t = time passed by the sample = 3.5 min
a = initial concentration of the reactant = 3.0 M
a - x = concentration left after decay process = ?
Now put all the given values in above equation, we get
[tex]3.5=\frac{2.303}{5.89\times 10^{-3}}\log\frac{3.0}{a-x}[/tex]
[tex]a-x=2.9M[/tex]
Thus, the final concentration of [tex]N_2O_5[/tex] is, 2.9 M
To determine the final concentration of N2O5 after 3.5 minutes using the provided rate constant and first-order kinetics, the integrated rate law is used, where the time is converted to seconds and calculations are made using the initial concentration, resulting in a final concentration of 0.17 M.
Explanation:The student is asking about the final concentration of N2O5 after a first-order decomposition reaction where the initial concentration and rate constant are provided. To find the final concentration of N2O5 after a period of 3.5 minutes, we use the first-order integrated rate law:
ln[A] = -k * t + ln[A0]
Where:
[A] is the final concentration of N2O5k is the rate constant (5.89 × 10⁻³ s⁻¹)t is the reaction time in seconds[A0] is the initial concentration of N2O5 (3.0 M)Since the time given is in minutes, we first convert it to seconds: t = 3.5 min × 60 s/min = 210 s. Now we can use the integrated rate law to find the final concentration:
ln[3.0] = -5.89 × 10⁻³ × 210 + ln[3.0]
Solving for [A], we find that the final concentration of N2O5 is 0.17 M (using appropriate exponential and logarithmic operations).
Which of the following statements about buffers is true? Group of answer choices A buffer composed of a weak acid of pKa = 5 is stronger at pH 4 than at pH 6. At pH values lower than the pKa, the salt concentration is higher than that of the acid. The strongest buffers are those composed of strong acids and strong bases. The pH of a buffered solution remains constant (exactly the same) no matter how much acid or base is added to the solution. When pH = pKa, the weak acid and salt concentrations in a buffer are equal.
The statement "When pH = pKa, the weak acid and salt concentrations in a buffer are equal." is true.
A solution or system that resists pH changes when small amounts of acid or base are introduced to it is called a buffer. A weak acid and its conjugate base, or a weak base and its conjugate acid, make up the substance. In order to maintain a steady pH environment for various chemical reactions or biological processes, buffers are frequently utilised in chemistry and biological sciences. They are essential for preserving homeostasis in biological systems like blood, where a steady pH is necessary for proper operation.
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Final answer:
The true statement about buffers is that they have their maximum buffering capacity when pH equals pKa, meaning the concentrations of the weak acid and its salt (conjugate base) are equal. Buffer solutions resist changes in pH effectively within a range of ± 1 pH unit from their pKa, and their buffering capacity depends on the buffer concentration.
Explanation:
The correct statement about buffers is: When pH = pKa, the weak acid and salt concentrations in a buffer are equal. This is because buffers consist of a weak acid and its conjugate base and exhibit maximum buffering capacity when the pH is numerically equal to the weak acid's pKa. At this pH, the buffer effectively resists changes in pH when small amounts of acid or base are added.
Buffer solutions are most effective within a pH range of ± 1 unit from their pKa. They are not invincible; their ability to maintain the pH level depends on the buffer concentration and the amount of strong acid or base added. The higher the concentration of the buffer components, the greater the buffer capacity, meaning more acid or base can be added before a significant change in pH occurs.
Buffers composed of weak acids work best for pH less than 7, while those composed of weak bases are more suitable for pH greater than 7. Moreover, buffers created from strong acids or bases are not effective, as they do not establish an equilibrium that is necessary for buffering action.
Using a spectrophotometer, and a cuvette with a path length of 1 cm you measure the absorbance (A275) of Guanosine to be 0.70. Calculate the concentration of guanosine in your sample
Answer : The concentration of guanosine in your sample is, [tex]8.33\times 10^{-5}M[/tex]
Explanation :
Using Beer-Lambert's law :
[tex]A=\epsilon \times C\times l[/tex]
where,
A = absorbance of solution = 0.70
C = concentration of solution = ?
l = path length = 1.00 cm
[tex]\epsilon[/tex] = molar absorptivity coefficient guanosine = [tex]8400M^{-1}cm^{-1}[/tex]
Now put all the given values in the above formula, we get:
[tex]0.70=8400M^{-1}cm^{-1}\times C\times 1.00cm[/tex]
[tex]C=8.33\times 10^{-5}M[/tex]
Thus, the concentration of guanosine in your sample is, [tex]8.33\times 10^{-5}M[/tex]
Predict the precipitate produced by mixing a(n) Al(NO3)3 solution with a(n) NaOH solution. Write the molecular equation for the reaction. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)
Answer:
Al(NO₃)₃ (aq) + NaOH(aq) → Al(OH)₃ (s)↓ + 3NaNO₃ (aq)
Explanation:
Al³⁺ cation can generate a precipitate when it bonds to OH⁻ from a strong base but it is important to the base, not to be in excess.
When the OH⁻is in excess, the produced aluminium hydroxide will be soluble.
Al³⁺(aq) + 3OH⁻(aq) ⇄ Al(OH)₃(s) ↓ Kps
How are combustion and cellular respiration different? How are combustion and cellular respiration different? Cellular respiration breaks down sugar, and combustion breaks down octane. Combustion produces heat, but cellular respiration does not. Cellular respiration produces carbon dioxide and water, but combustion does not. Cellular respiration requires oxygen, but combustion does not.
Answer:
Cellular respiration is the process by which the molecules of food are broken down into simple components due to the oxidation process resulting in the release of cellular energy in the form of ATP. Cellular respiration involves the oxidation of fats, carbohydrates (sugars), and proteins which are fuels for the cellular respiration process.
The combustion is also an exothermic reaction just like cellular respiration. Combustion is a process of burning, in this, the reactants are reacted with oxygen gas to produce carbon dioxide and water. For example, gasoline is octane and it burns to produce water and carbon dioxide as products.
Combustion and cellular respiration are energy-releasing processes with differences in mechanisms and outcomes. Cellular respiration generates ATP and happens in cells, while combustion releases energy as heat and light.
Combustion and cellular respiration are two processes that release energy through the breakdown of substances, but they differ significantly in their mechanisms and outcomes.
Key Differences
Substances Broken Down: Cellular respiration breaks down glucose, while combustion typically breaks down fuels like octane.Energy Release: Combustion produces heat and light, whereas cellular respiration primarily produces ATP, the energy currency of cells, with minimal release of heat.Byproducts: Both processes produce carbon dioxide and water, but cellular respiration focuses on energy conservation through ATP production, not intense heat.Oxygen Requirements: Cellular respiration requires oxygen just as much as combustion does.In summary, cellular respiration occurs in the mitochondria of cells and converts glucose into ATP using oxygen, whereas combustion breaks down various fuels with oxygen to release energy as heat and light.
Breads and other whole grain foods are composed of very large polysaccharide molecules which contain hydrogen, oxygen, and which other element?A) carbonB) ironC) nitrogenD) water
Final answer:
Bread and whole grain foods contain large polysaccharide molecules which are composed of carbon, hydrogen, and oxygen. Carbon is the element present in carbohydrates and is essential in forming biomolecules found in foods like bread. Glycogen is an example of a complex polysaccharide composed of these elements.
Explanation:
Breads and other whole-grain foods are composed of large polysaccharide molecules which include hydrogen, oxygen, and the other element is carbon. Carbohydrates, which are found in these foods, are made up of carbon (C), hydrogen (H), and oxygen (O) atoms. Polysaccharides like starch and glycogen are complex carbohydrates, which are made up of many monosaccharide units.
The nutrient that is part of carbohydrates, like bread, also present in proteins and nucleic acids that forms biomolecules, is carbon. The complex carbohydrate glycogen, for example, has a chemical formula of C24H42021, indicating it is composed of carbon, hydrogen, and oxygen elements. Glycogen is a polysaccharide, as its name suggests (poly- meaning many and saccharide meaning sugar), and not a monosaccharide because it consists of multiple sugar units.
What condition must occur when a chemical reaction is at equilibrium?
The condition is that the amounts of reactants or products do not change when a chemical reaction is in state of equilibrium.
Amounts of reactants or products do not change.
Chemical ReactionChemical equilibrium is a dynamic process in which the rate of product formation by the forward reaction equals the rate of product re-formation by the reverse reaction.
A chemical reaction is a process that results in the chemical change of one set of chemical substances into another set of chemical substances. Rust is an example of iron and oxygen mixing. Sodium acetate, carbon dioxide, and water are formed when vinegar and baking soda are combined.
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At chemical equilibrium, the forward and reverse reactions are happening at equal rates, resulting in no net change in the concentrations of reactants and products. This state is known as dynamic equilibrium. The system must also be closed for equilibrium to be maintained.
The correct answer is: The forward and reverse reactions are happening at equal rates.
Chemical equilibrium is the state in which both reactants and products are present in concentrations that remain constant over time. This occurs when the rate of the forward reaction equals the rate of the reverse reaction, meaning that there are no net changes in the concentrations of the reactants and products. This is known as dynamic equilibrium, indicating that reactions continue to occur in both directions at equal rates.
If you combine 270.0 mL of water at 25.00 ∘ C and 140.0 mL of water at 95.00 ∘ C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.
The final temperature of the mixture of cold water and hot water is 48.9 ⁰C.
The given parameters;
volume of the cold water = 270 mltemperature of the cold water = 25⁰Cvolume of the hot water, = 140 mltemperature of the hot water, = 95⁰Cdensity of water, = 1 g/mlThe mass of the cold water is calculated as follows;
[tex]m = \rho \times V\\\\m_c = 1 \ g/ml \ \ \times \ \ 270 \ ml \\\\m_c = 270 \ g[/tex]
The mass of the hot water is calculated as follows;
[tex]m_h = 1 \ g/ml \ \ \times \ \ 140 \ ml\\\\m_h = 140 \ g[/tex]
The final temperature of the mixture is determined by applying the principle of conservation of energy;
[tex]m_c C \Delta \theta_c = m_h C \Delta \theta_h\\\\ m_c \Delta \theta_c = m_h \Delta \theta_h\\\\m_c (t - 25) = m_h(95 - t)\\\\270(t- 25) = 140(95-t)\\\\270t - 6750 = 13,300 - 140t\\\\410t = 20,050\\\\t = \frac{20,050}{410} \\\\t = 48.9 \ ^0C[/tex]
Thus, the final temperature of the mixture of cold water and hot water is 48.9 ⁰C.
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The final temperature of the mixture is 55.00 °C.
To calculate the final temperature of the mixture, we can use the following equation:
T_final = (m_1T_1 + m_2T_2) / (m_1 + m_2)
where:
* T_final is the final temperature of the mixture (in °C)
* m_1 is the mass of the first water sample (in grams)
* T_1 is the temperature of the first water sample (in °C)
* m_2 is the mass of the second water sample (in grams)
* T_2 is the temperature of the second water sample (in °C)
We are given the following information:
* m_1 = 270.0 mL * 1.00 g/mL = 270.0 g
* T_1 = 25.00 °C
* m_2 = 140.0 mL * 1.00 g/mL = 140.0 g
* T_2 = 95.00 °C
Substituting these values into the equation above, we get the following:
T_final = (270.0 g * 25.00 °C + 140.0 g * 95.00 °C) / (270.0 g + 140.0 g)
T_final = 55.00 °C
Therefore, the final temperature of the mixture is 55.00 °C.
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When measuring Potassium with an ion-selective electrode by means of a liquid ion-exchange membrane, what antibiotic will be incorporated into the membrane?
Answer:
Valinomycin
Explanation:
This antibiotic is acquired from the Streptomyces species cells. Valinomycin is selective to potassium and inhibits sodium ions from entering the cell. This antibiotic allows the potassium ions to move down the electrochemical potential gradient of the lipid membranes.
Valinomycin is the antibiotic incorporated into the liquid ion-exchange membrane when measuring Potassium with an ion-selective electrode. It binds specifically to potassium ions, thus allowing accurate detection and measurement.
Explanation:When measuring Potassium with an ion-selective electrode by means of a liquid ion-exchange membrane, the antibiotic incorporated into the membrane is valinomycin.
Valinomycin is a relatively specific antibiotic used in these measurements because it binds selectively with potassium ions, allowing for accurate detection and measurement. The mechanism operates such that when the potassium is bound by the valinomycin, it causes a change in the membrane's potential. This change can be measured and is proportional to the concentration of potassium in the solution.
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What form does nitrogen take in the atmosphere?
N
NH4+
N2
NO3–
Answer:
N2
Explanation:
The nitrogen element exists as N₂ molecule in the atmosphere which is a major constituent of air.
What is an element?It is defined as a substance which cannot be broken down further into any other substance. Each element is made up of its own type of atom. Due to this reason all elements are different from one another.
Elements can be classified as metals and non-metals. Metals are shiny and conduct electricity and are all solids at room temperature except mercury. Non-metals do not conduct electricity and are mostly gases at room temperature except carbon and sulfur.
The number of protons in the nucleus is the defining property of an element and is related to the atomic number.All atoms with same atomic number are atoms of same element.Elements combine to give compounds.
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A 19.0 L helium tank is pressurized to 26.0 atm. When connected to this tank, a balloon will inflate because the pressure inside the tank is greater than the atmospheric pressure pushing on the outside of the balloon. Assuming the balloon could expand indefinitely and never burst, the pressure would eventually equalize causing the balloon to stop inflating. What would the volume of the balloon be when this happens
Answer:
The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.
The balloon expands by am additional 475 L.
Explanation:
Assuming Helium behaves like an ideal gas and temperature is constant.
According to Boyle's law for ideal gases, at constant temperature,
P₁V₁ = P₂V₂
P₁ = 26 atm
V₁ = 19.0 L
P₂ = 1 atm (the balloon is said to expand till the pressure matches the pressure of the atmpsphere; and the pressure of the atmosphere is 1 atm)
V₂ = ?
P₁V₁ = P₂V₂
(26 × 19) = 1 × V₂
V₂ = 494 L (it is assumed the balloon never bursts)
The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.
The balloon expands by am additional 475 L.
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Using the ideal gas law, the volume of the balloon when it stops inflating and the pressures equalize would be 494.0 L, as calculated from the initial pressure and volume of the tank and the atmospheric pressure.
Explanation:The question concerns the behavior of gases under different conditions and relates to the ideal gas law, which is a fundamental concept in chemistry. When a balloon is filled with helium from a tank with a pressure of 26.0 atm and a volume of 19.0 L, the balloon will inflate until the pressure inside the balloon equals the outside atmospheric pressure. At this point, the volume of the balloon could be derived using the ideal gas law, which states that for a fixed amount of gas at constant temperature, the product of the pressure and volume (P1V1) will be equal to the product of the final pressure and volume (P2V2). In this scenario, assuming the temperature remains constant and the atmospheric pressure is 1 atm, we apply the equation P1V1 = P2V2.
Given that P1 is 26.0 atm and V1 is 19.0 L, and P2 is 1.0 atm (atmospheric pressure), we can solve for V2 as follows: V2 = (P1V1/P2) = (26.0 atm * 19.0 L) / 1.0 atm = 494.0 L. The volume of the balloon when it stops inflating and the pressures equalize would be 494.0 L.
In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions), which affects lattice energy (a measure of the attractive force holding those ions together). Based on ion sizes, rank these compounds of their expected lattice energy..
Note: Many sources define lattice energies as negative values. Please rank by magnitude and ignore the sign. |Lattice energy| = absolute value of the lattice energy.
RbCl ,RbBr ,Rbl ,RbF
Answer:
RbF>RbCl>RbBr>RbI
Explanation:
The lattice energy is an indicator of the strength of an ionic bond. It is also a rough indicator of the probability that an ionic substance will dissolve in water. The higher the lattice energy, the more difficult it is for the substance to dissolve in water.
Lattice energy depends on the relative sizes of ions present in the substance. As already known, the order of increasing sizes of halogen atoms is F<Cl<Br<I. The lesser the size, the higher the lattice energy.
Since the cation size is constant, lattice energy is only affected by increasing anion size and follows the pattern highlighted in the paragraph above. Hence RbF has the highest lattice energy and RbI has the least lattice energy.
Paradichlorobenzene, C6H4Cl2, is a component of mothballs. A solution of 2.00 g in 22.5 g of cyclohexane boils at 82.39 ∘C. The boiling point of pure cyclohexane is 80.70 ∘C. Calculate Kb for cyclohexane.
Answer:
The Kb for cyclohexane is 2.79 °C/m
Explanation:
Step 1: Data given
Mass of Paradichlorobenzene = 2.00 grams
Mass of cyclohexane = 22.5 grams
Boiling point of the solution = 82.39 °C
Boiling point of pure cyclohexane = 80.70 °C
Molar mass of Paradichlorobenzene = 147 g/mol
Step 2: Calculate moles Paradichlorobenzene
Moles Paradichlorobenzene = mass / molar mass
Moles Paradichlorobenzene = 2.00 grams / 147 g/mol
Moles Paradichlorobenzene = 0.0136 moles
Step 3: Calculate molality
Molality = moles Paradichlorobenzene / mass cyclohexane
Molality = 0.0136 moles / 0.0225 kg
Molality = 0.605 molal
Step 4: Calculate Kb
Kb = change in boiling point / molality of solution
⇒ Change in boiling point = 82.39 - 80.70 = 1.69 °C
⇒ molality = 0.605 molal
Kb = 1.69 °C / 0.605 molal = 2.79 °C/m
The Kb for cyclohexane is 2.79 °C/m
Answer:
The Kb for cyclohexane is 2.80°C/m
Explanation:
delta T = 82.39 - 80.70 = 1.69 °C
Moles C6H4Cl2 = 2.00 g/ 147.0 g/mol= 0.0136
molality = 0.0136 mol / 0.0225 Kg = 0.604
1.69 = 0.604 x kf
kf = 2.80
The Kb for cyclohexane is 2.80°C/m
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Would it be more efficient to warm your bed on a cold night with a hot water bottle that contains 1 kg of water at 50 degrees C? Or with a 1-kg gold bar at 50 degrees C? Why?
Explanation:
Heat necessary to raise the temperature of a substance by unit mass of a given substance by a given amount is known as specific heat.
It is known that relation between heat energy, specific heat, and mass is as follows.
q = [tex]m \times C \times \Delta T[/tex]
As the specific heat of water is 4.18 [tex]J/g^{o}C[/tex] and the specific heat of gold is 0.129 [tex]J/g^{o}C[/tex]. Since, the specific heat of water is greater than the specific heat of gold.
Therefore, we can conclude that water is more efficient to warm your bed on a cold night with a hot water bottle that contains 1 kg of water at 50 degrees C.
A hot water bottle would be more efficient at warming your bed on a cold night. This conclusion is based on the specific heat properties of water and gold. Water, with a higher specific heat, can better retain and transfer heat than gold.
Explanation:The key to answering this question lies in understanding the concept of specific heat. This is the amount of heat per unit mass required to raise the temperature by one degree Celsius. Different substances have different specific heats, and this impacts how well they store and transfer heat. Water has a high specific heat, meaning it can absorb a lot of heat before its temperature rises. Gold, on the other hand, has a relatively low specific heat, meaning it heats up quickly but does not retain or transfer heat as well as water does.
So, to answer your question, a hot water bottle would be more efficient at warming your bed on a cold night. That's because the 1 kg of water at 50 degrees Celsius can store and transfer more heat than a 1-kg gold bar at the same temperature.
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The histogram shown below represents the weights (in kg) of 47 female and 97 male cats. Approximately % of these cats weigh less than 2.5kg. Approximately % of these cats weigh between 2.5 and 2.75kg. Approximately % of these cats weigh between 2.75 and 3.5kg.
Answer:
Approximately % of these cats weigh less than 2.5kg
the percentage = 61/144 × 100 = 6100/144 = 42.3611111111 ≈ 42.36%
Approximately % of these cats weigh between 2.5 and 2.75 kg
percentage = 20/144 ×100 = 2000/144 = 13.8888888889 ≈ 13.90%
Approximately % of these cats weigh between 2.75 and 3.5kg.
percentage = 54/144 × 100 = 5400/144 = 37.5 %
Explanation:
The picture below is the histogram used .
The horizontal is the weight of the cat . The vertical is the number of cat. The cats have 47 female and 97 male . The total cats is 47 + 97 = 144.
Approximately % of these cats weigh less than 2.5kg
Cat that weighs less than 2.5 kg is the sum of the first bar and the second bar. The sum is 29 + 32 = 61
61 cat weighs less than 2.5 kg
the percentage = 61/144 × 100 = 6100/144 = 42.3611111111 ≈ 42.36
Approximately % of these cats weigh between 2.5 and 2.75 kg
cat that weigh between 2.5 and 2.75 is 20
percentage = 20/144 ×100 = 2000/144 = 13.8888888889 ≈ 13.90
Approximately % of these cats weigh between 2.75 and 3.5kg.
The number of cat that fall under this category is 27 + 12 + 15 = 54
percentage = 54/144 × 100 = 5400/144 = 37.5
The approximately % of cats weigh less than 2.5 kg has been 42.36 %.
The approximately % of cats weighing between 2.5 and 2.75 kg has been 13.88 %.
The approximately % of cats weighing between 2.75 and 3.5 kg has been 37.5 %.
The histogram has been the representation of the data in a user defined condensed format. The total number of cats has been the sum of male and female cats.
The given male cats can be, [tex]C_M=97[/tex]
The given female cats, [tex]C_F=47[/tex]
The total cats (C) can be given as:
[tex]C=C_M\;+\;C_F\\C=97\;+\;47\\C=144[/tex]
The total number of cats has been 144.
The percentage of cats weigh less than 2.5 kg has been given as:From the histogram, the number of cats weighing less than 2.5 kg, [tex]C_>_2_._5=61[/tex]
The % of cats weighing less than 2.5 kg ([tex]C_>_2_._5\;\%[/tex]) has been:
[tex]C_>_2_._5\;\%=\dfrac{61}{144}\;\times\;100\\ C_>_2_._5\;\%=42.36\%[/tex]
The approximately % of cats weigh less than 2.5 kg has been 42.36 %.
The percentage of cats weigh between 2.5 and 2.75 kg has been given as:From the histogram, the number of cats weighing between 2.5 and 2.75 kg, [tex]C_2_._5_-_2_._7_5=20[/tex]
The % of cats weighing between 2.5 and 2.75 kg, [tex]C_2_._5_-_2_._7_5\%[/tex], has been:
[tex]C_2_._5_-_2_._7_5\%=\dfrac{20}{144}\;\times\;100\\ C_2_._5_-_2_._7_5\%=13.88\%[/tex]
The approximately % of cats weighing between 2.5 and 2.75 kg has been 13.88 %.
The percentage of cats weigh between 2.75 and 3.5 kg has been given as:From the histogram, the number of cats weighing between 2.75 and 3.5 kg, [tex]C_2_._7_5_-_3_._5=54[/tex]
The % of cats weighing between 2.75 and 3. 5 kg, [tex]C_2_._7_5_-_3_._5\%[/tex], has been:
[tex]C_2_._7_5_-_3_._5\%=\dfrac{54}{144}\;\times\;100\\ C_2_._7_5_-_3_._5\%=37.5\%[/tex]
The approximately % of cats weighing between 2.75 and 3.5 kg has been 37.5 %.
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How many milliliters of ammonium sulfate solution having a concentration of 0.218 M are needed to react completely with 62.6 ml of 2.31 M sodium hydroxide solution?
Answer:
330 mL of (NH₄)₂SO₄ are needed
Explanation:
First of all, we determine the reaction:
(NH₄)₂SO₄ + 2NaOH → 2NH₃ + 2H₂O + Na₂SO₄
We determine the moles of base:
(First, we convert the volume from mL to L) → 62.6 mL . 1L/1000 mL = 0.0626L
Molarity . volume (L) = 2.31 mol/L . 0.0626 L = 0.144 moles
Ratio is 2:1. Therefore we make a rule of three:
2 moles of hydroxide react with 1 mol of sulfate
Then, 0.144 moles of NaOH must react with (0.144 .1) /2 = 0.072 moles
If we want to determine the volume → Moles / Molarity
0.072 mol / 0.218 mol/L = 0.330 L
We convert from L to mL → 0.330L . 1000 mL/1L = 330 mL
The element that can act like a metal when it is under tremendous pressure and is probably responsible for Jupiter and Saturn's magnetism is:__________. 1. hydrogen 2. water 3. helium 4. unobtanium 5. gold
Answer:
1. Hydrogen
Explanation:
The interior of these planets contains liquid hydrogen (the Earth has liquid iron in it).
When this element is subjected to tremendous pressures (as in Jupiter and Saturn), the electrons of each hydrogen atom can "jump" to other atoms. This property allows liquid hydrogen to behave with a metal.
With the rotation of the planets and the large amount of constant energy emitted by the nucleus, currents are induced in liquid hydrogen, giving rise to magnetic fields that propagate for millions of kilometers in space (Jupiter's magnetic field is fourteen times stronger than Earth's, for example).
Enter your answer in the provided box. Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25°C, diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine ΔHrxn for C(diamond) → C(graphite) with equations from the following list: (1) C(diamond) + O2(g) → CO2(g) ΔH = −395.4 kJ (2) 2 CO2(g) → 2 CO(g) + O2(g) ΔH = 566.0 kJ (3) C(graphite) + O2(g) → CO2(g) ΔH = −393.5 kJ (4) 2 CO(g) → C(graphite) + CO2(g) ΔH = −172.5 kJ
Answer:
ΔHrxn = -1.9kJ
Explanation:
Considering the 3 elementary equations
Cancelling out O2 From equation 1 reactant and equation 2 product,
Also
Cancelling out CO2 From equation 1 and 3 product and equation 2 reactant..
Finally
Cancelling out 2CO From equation 2 product and 2CO from equation 3 reactant..
This will give us an overall equation that is due to arithmetic addition of equation 1,2&3
Hence
C(diamond) → C(graphite)
ΔHrxn= -395.4+566.0-172.5 = -1.9kJ
ΔHrxn = -1.9kJ
The enthalpy changes for the conversion of diamond to graphite at 1 atm and 25°C is [tex]\( \Delta H = -1.9 \text{ kJ} \)[/tex].
The reactions given are:
1. [tex]\( \text{C(diamond)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -395.4 \text{ kJ} \)[/tex]
2. [tex]\( 2 \text{CO}_2(g) \rightarrow 2 \text{CO}(g) + \text{O}_2(g) \quad \Delta H = 566.0 \text{ kJ} \)[/tex]
3. [tex]\( \text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393.5 \text{ kJ} \)[/tex]
4. [tex]\( 2 \text{CO}(g) \rightarrow \text{C(graphite)} + \text{CO}_2(g) \quad \Delta H = -172.5 \text{ kJ} \)[/tex]
To find [tex]\( \Delta H \)[/tex] for [tex]\( \text{C(diamond)} \rightarrow \text{C(graphite)} \)[/tex], we can reverse reaction 3 and add it to reaction 1. This will cancel out [tex]CO_2[/tex] (g) and [tex]O_2[/tex] (g), leaving us with the desired reaction:
[tex]\[ \text{C(diamond)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -395.4 \text{ kJ} \][/tex]
[tex]\[ -\text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = 393.5 \text{ kJ} \][/tex]
Adding these two equations gives us:
[tex]\[ \text{C(diamond)} + \cancel{\text{O}_2(g)} \rightarrow \cancel{\text{CO}_2(g)} \][/tex]
[tex]\[ -\text{C(graphite)} - \cancel{\text{O}_2(g)} \rightarrow \cancel{\text{CO}_2(g)} \][/tex]
[tex]\[ \text{C(diamond)} \rightarrow \text{C(graphite)} \][/tex]
And the enthalpy change for the desired reaction is the sum of the enthalpy changes for the two reactions:
[tex]\[ \Delta H = (-395.4 \text{ kJ}) + (393.5 \text{ kJ}) \][/tex]
[tex]\[ \Delta H = -1.9 \text{ kJ} \][/tex]
A piece of metal is heated by placing it in hot oil. It is removed from the hot oil and dropped into a beaker of cold water. The water heats up due to the transfer of heat from the metal. What happens to the temperature of the beaker? How is the heat that causes the temperature of the glass to rise accounted for in calorimetry
Answer:
a) The temperature of the beaker rises as this transfer of heat goes on.
b) Check Explanation.
Explanation:
a) The heat lost by the piece of metal is normally gained by the all the components that it comes in contact with after the heating procedure.
(Heat lost by piece of metal) = (Heat gained by the cold water) + (Heat gained by the beaker).
So, since heat is also gained by the Beaker, its temperature should rise under normal conditions.
That is essentially what the zeroth law of thermodynamics about thermal equilibrium talks about.
If two bodies are at thermal equilibrium with reach other and body 2 is in thermal equilibrium with a third body, then body 1 and body 3 are also in thermal equilibrium
Temperature of the piece of metal decreases, temperature of water rises and the temperature of the beaker rises as they all try to attain thermal equilibrium.
b) In calorimetry, the aim is usually for the water (in this case) to take up all of the heat supplied by the piece of metal. Hence, the calorimeter is usually heavily insulated (or properly called lagged). Thereby, reducing the amount of heat that the calorimeter would gain.
But in cases where the heat lost to the insulated calorimeter isn't negligible, the heat capacity of the calorimeter is usually obtained and included it is included in the heat transfer calculations.
Hope this Helps!!!
The temperature of the beaker of cold water will increase due to the transfer of heat from the hot metal. In calorimetry, the heat gained by the water and the beaker is equal to the heat lost by the metal.
When the hot metal is dropped into the beaker of cold water, a heat transfer process occurs. According to the principle of conservation of energy, energy cannot be created or destroyed, only transferred or transformed. In this case, heat energy is transferred from the higher temperature metal to the lower temperature water and beaker.
The specific heat capacity of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. The heat lost by the metal can be calculated using the formula:
[tex]\[ q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}} \][/tex]
where [tex]\( q_{\text{metal}} \)[/tex]is the heat lost by the metal, [tex]\( m_{\text{metal}} \)[/tex] is the mass of the metal, [tex]\( c_{\text{metal}} \)[/tex] is the specific heat capacity of the metal, and [tex]\( \Delta T_{\text{metal}} \)[/tex] is the change in temperature of the metal.
Similarly, the heat gained by the water can be calculated using the formula:
[tex]\[ q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \][/tex]
where [tex]\( q_{\text{water}} \)[/tex] is the heat gained by the water, [tex]\( m_{\text{water}} \)[/tex] is the mass of the water,[tex]\( c_{\text{water}} \)[/tex] is the specific heat capacity of water, and [tex]\( \Delta T_{\text{water}} \)[/tex] is the change in temperature of the water.
In calorimetry, assuming no heat is lost to the surroundings, the heat lost by the metal is equal to the heat gained by the water (and the beaker, if its heat capacity is considered):
[tex]\[ q_{\text{metal}} = q_{\text{water}} + q_{\text{beaker}} \][/tex][tex]\[ m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} + m_{\text{beaker}} \cdot c_{\text{beaker}} \cdot \Delta T_{\text{beaker}} \][/tex]
Here, [tex]\( m_{\text{beaker}} \)[/tex] is the mass of the beaker, [tex]\( c_{\text{beaker}} \)[/tex] is the specific heat capacity of the beaker material, and [tex]\( \Delta T_{\text{beaker}} \)[/tex] is the change in temperature of the beaker. The temperature change of the beaker and the water will be the same if they reach thermal equilibrium.
The temperature of the beaker will rise until thermal equilibrium is reached, at which point the temperature of the metal, water, and beaker will be the same. The heat that causes the temperature of the glass beaker to rise is accounted for in calorimetry by including the beaker's mass and specific heat capacity in the calculations.
Which of the following electron configurations for neutral atoms is correct? Li nitrogen: 1s 2, 2s 2, 2p 3 the third sub-level silicon: 1s 2, 2s 2, 3s 2, 2p 6 3p 2 helium: 1s 1, 1p 1
ans is helium.........
Answer:
The answer is: helium: 1s 1, 1p 1
Explanation:
Discuss what all the values of w and q should be if a system is exothermic. Then discuss what all the values of w and q should be if a system is endothermic.
Answer:
In an exothermic reaction, q is negative while w is positive.
In an endothermic reaction, q is positive while w is negative
Explanation:
An Exothermic reaction is one in which the heat content of the system is greater than the heat content of the surroundings. As a result of this, heat (measured by Q) is given off into the surroundings and the surroundings is hotter than than the system. The heat emitted does work against the surroundings, hence the value of work done W is positive.
An endothermic reaction is the opposite of an exothermic reaction.
During an endothermic reaction, the heat content of the reactants is lesser than the heat content of the products. The surroundings then does work on the system , resulting in heat (measured by Q) being absorbed from the surroundings into the system, making the system hotter than than the surroundings. The value of work done W in an endothermic reaction is negative because the system does work against the surroundings.
Electrolysis of molten MgCl2 is the final production step in the isolation of magnesium from seawater by the Dow process. Assuming that 38.0 g of Mg metal forms, answer the following questions. (a) How many moles of electrons are required? 2 mol e− (b) How many coulombs are required? 1.93 × 10 5 C Enter your answer in scientific notation. (c) How many amps will produce this amount in 3.50 h? A
For the electrolysis of molten MgCl2, 1.57 moles of electrons (or 3.03 x 10^5 Coulombs) are required to produce 38.0 g of Mg. This would require a current of approximately 24.0 Amperes over 3.5 hours.
Explanation:The electrolysis of molten MgCl2 for the isolation of magnesium from seawater by the Dow process involves a reaction where 1 mole of Mg metal is produced for every 2 moles of electrons involved. Therefore, if 38.0 g of Mg forms, which is approximately 1.57 moles (given magnesium's molar mass is 24.31 g/mol), the number of moles of electrons required is twice this amount, or approximately 3.14 moles of electrons.
Faraday's constant, which is equal to approximately 96485.332 Coulombs per mole of electrons, is used to find the amount of charge required. By multiplying the number of moles of electrons by Faraday's constant, you can find that approximately 3.03 x 10^5 Coulombs are required.
Current, measured in Amperes (A), is a measure of the amount of charge passing a point in a circuit per unit time. Therefore, to find the current necessary to produce this amount in 3.5 hours (or 12600 seconds), divide the total charge required by the time, giving approximately 24.0 A
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please help me with this chemistry question, image attached.
Based on Table F, which of these saturated solutions has the lowest concentration of dissolved ions? (Explain why shortly please)
1) NaCl (aq)
2) MgCl² (aq)
3) NiCl (aq)
4) AgCl (aq)
AgCl (aq)
Option: 4
Explanation:
AgCl has the lowest concentration of dissolved ions because it is insoluble. Hence, it does not allow any ions into the solution.
According to the Solubility rule, the salts that have Cl⁻ are usually soluble but Ag⁺ is an exception which shows that AgCl is insoluble. It is insoluble because the lattice structure of AgCl is very strong that it cannot be overcome by the forces that favor the formation of hydrated ions, Ag⁺(aq) and Cl⁻(aq). Solubility of AgCl in water is very low but however, it can precipitate in water.
The saturated solution that contains the concentration of dissolved ions is AgCl (aq)
Solubility rule:AgCl should contain a less concentration of dissolved ions due to insoluble. Due to this, it permits any ions into the solution. As per the above rule, the salts that have Cl⁻ are normally soluble however Ag⁺ should be an exception which represents that AgCl is insoluble. It is insoluble due to the lattice structure of AgCl should be very strong. The solubility of AgCl in water should be very low but it should be precipitated in water.
Hence, The saturated solution that contains the concentration of dissolved ions is AgCl (aq)
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What pressure does 3.54 moles of chlorine gas at 376 k exert on the walls of it 51.2 l container? I'm Lazy sooooo
Answer:
The answer to your question is P = 2.13 atm
Explanation:
Data
Pressure = ?
number of moles = 3.54
Temperature = 376 °K
Volume = 51.2 L
R = 0.08205 atm L/mol°K
Formula
PV = nRT
- Solve for P
P = nRT / V
- Substitution
P = (3.54)(0.08205)(376) / 51.2
- Simplification
P = 109.21 / 51.2
Result
P = 2.13 atm
Answer:
2.13
Explanation:
I just did the problem on acellus and got it right
A common laboratory preparation of oxygen gas is the thermal decomposition of potassium chlorate (KClO3). Assuming complete decomposition, calculate the number of grams of O2 gas that can be obtained from 24.82 g KClO3. (The products are KCl and O2).
Answer:
9.72 g of O₂ are obtained in the decomposition
Explanation:
Reaction of decomposition is:
2KClO₃ → 2KCl + 3O₂
Ratio is 2:3. First, we determine the moles of chlorate
24.82 g. 1mol/ 122.55g = 0.202 moles
2 moles of chlorate can decompose into 3 moles of oxgen
Therefore, 0.202 moles of chlorate will decompose into (0.202 .3)/2 = 0.303 moles of O₂
We determine the mass of formed oxygen:
0.303 mol . 32g / 1mol = 9.72 g
Final answer:
The number of grams of O2 gas that can be obtained from 24.82 g of KClO3 is 9.725 g. This is calculated by converting the mass of KClO3 to moles, using the stoichiometric relationship from the balanced equation, and then converting moles of O2 to grams.
Explanation:
To calculate the number of grams of O2 gas that can be obtained from 24.82 g of KClO3, we need to follow a series of stoichiometric conversions. Initially, we must ensure that we have a balanced chemical equation, which for the decomposition of potassium chlorate to potassium chloride and oxygen gas is:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
This tells us that from every 2 moles of KClO3 we get 3 moles of O2. We can use the molar masses to convert between grams and moles:
1 mol of KClO3 is 122.55 g/mol1 mol of O2 is 32.00 g/molDividing the total mass of KClO3 by its molar mass gives us:
24.82 g KClO3 × (1 mol KClO3 / 122.55 g) = 0.2026 mol KClO3
Next, we use the stoichiometry of the balanced equation to determine the moles of O2:
0.2026 mol KClO3 × (3 mol O2 / 2 mol KClO3) = 0.3039 mol O2
Finally, we convert the moles of oxygen to grams:
0.3039 mol O2 × (32.00 g/mol O2) = 9.725 g O2
Therefore, from 24.82 g of KClO3, we can theoretically obtain 9.725 g of oxygen gas, assuming complete decomposition.
If a student performs an exothermic reaction in a calorimeter, how does the calculated value of ΔH (Hcalc) differ from the actual value (Hactual) if the heat exchanged with the calorimeter is not taken into account?
Answer:
The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.
Explanation:
The amount of heat changed during this process at a fixed pressure is termed Enthalpy
enthalpy change ∆H = ∆E + P∆V
∆E = internal energy change
P = fixed pressure
∆V = change in volume
When energy is absorbed during reaction, it is called endothermic reaction.
Endothermic reaction carried out in the calorimeter and enthalpy change for the reaction. Since we have that
q(surrounding) = q(solution)+q(calorimeter)
Therefore, q(calorimeter) > 0(endothermic).
The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.
The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.
What is Enthalpy change?The amount of heat changed during this process at a fixed pressure is termed Enthalpy.
Enthalpy change ∆H = ∆E + P∆V
∆E = internal energy change
P = fixed pressure
∆V = change in volume
When energy is absorbed during reaction, it is called endothermic reaction.
Endothermic reaction carried out in the calorimeter and enthalpy change for the reaction. Since we have that
q(surrounding) = q(solution)+q(calorimeter)
Thus, q(calorimeter) > 0(endothermic).
The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.
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An x-ray beam with wavelength 0.170 nm is directed at a crystal. As the angle of incidence increases, you observe the first strong interference maximum at an angle 62.5 ∘. What is the spacing d between the planes of the crystal?
The distance between the scattering planes in the crystal is d = 0.95 A°
Explanation:
The Bragg's equation is given by
2d sinθ = nλ
where,
d is the distance between the scattering planes in the crystal.
θ is the angle of diffraction.
n is the order of diffraction.
λ is the wavelength of X rays.
Given λ = 0.17 nm = 1.7 A°, angle = 62.5
2 [tex]\times[/tex] d [tex]\times[/tex] sin(62.5) = 1 [tex]\times[/tex] 1.7 A°
d = 0.95 A°
The distance between the scattering planes in the crystal is d = 0.95 A°
The spacing (d) between the planes of the crystal is 0.096nm.
BRAGG'S LAW EQUATION:
The spacing or distance between the planes of the crystal can be calculated using Bragg's law equation as follows:2dsinθ = nλ
Where;
d = distance between the scattering planes in the crystalθ is the angle of diffractionn is the order of diffractionλ is the wavelength of X raysd = nλ ÷ 2sinθ
According to this question;
n = 1λ = 0.170nmθ = 62.5°d = (1 × 0.170) ÷ 2 × sin 62.5°d = 0.170 ÷ 1.77d = 0.096nmTherefore, the spacing (d) between the planes of the crystal is 0.096nm.https://brainly.com/question/13009361?referrer=searchResults
Plz help quick. Find the pH of a solution with an ion concentration [H+] = 3.8x 10^-3
Round to the nearest thousands.
Answer: The pH of the solution is 2.420
Explanation: pH = -Log [H+]
= - Log [ 3.8 x 10^-3]
= 3 - 0.579784
= 2.420
Given a diprotic acid, H 2 A , with two ionization constants of K a1 = 3.0 × 10 − 4 and K a2 = 4.0 × 10 − 11 , calculate the pH for a 0.117 M solution of NaHA.
Answer:
The pH for a 0.117 M solution of NaHA is 2.227
Explanation:
To solve the question we check the difference in the Ka values thus
Ka₁ / Ka₂ = 7500000 < 10⁸ so we are required to calculate each value as follows
We therefore have
H₂X→ H⁺¹+HX⁻¹ with Ka₁ = 3.0 × 10⁻⁴
Therefore
3.0 × 10⁻⁴ = (x²)/(0.117)
x² = 3.0 × 10⁻⁴ ×0.117 and x = 5.925 × 10⁻³ = [H⁺]
Similarly
Ka₂ = 4.0 × 10⁻¹¹
and
4.0 × 10⁻¹¹= (x²)/(0.117)
x²= 0.117× 4.0 × 10⁻¹¹
x= 2.16× 10⁻⁶
Total H⁺ = 5.925 × 10⁻³+2.16× 10⁻⁶ = 5.927 × 10⁻³
Since pH = -log of hydrogen ion concentration,
pH = - log 5.927 × 10⁻³ = 2.227
Which of the following is a legume?
wheat
clover
corn
oats
Answer:
The answer to your question is Clover
Explanation:
Legumes are plants or the seed of plants. Legumes are harvested for human consumption, for livestock forage and silage.
These plants are also important during the Nitrogen cycle due to they fix nitrogen.
Examples of legumes are:
Beans, alfalfa, clover, lentils, peas, mesquite, carob, tamarind, peanuts, soybeans, etc.
Clover
LegumeA legume is a plant or the fruit or seed of a plant of the Fabaceae family. The seed is also known as a pulse when utilized as a dry grain. Legumes are grown in agriculture for a variety of reasons, including human consumption, cattle fodder and silage, and soil-enhancing green manure.Clover is a legume.Find out more information about the Legume:
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