Explanation:
Endothermic animals are also known as warm-blooded, they have the capacity to regulate their body temperature independent of the environment. They have mechanisms to compensate if heat loss exceeds heat generation (shivers) Or if heat generation exceeds the heat loss (panting, sweating).
On the other hand, ectothermal animals are known as cold blooded organisms and depend on external sources, like sunlight, to regulate their body temperature, reptiles are ectothermals.
To determine if the animal of interest is endo or ectothermal you’ll have to consider that is a reptile, you’ll also observe that it consumes less food and finally it’ll have more difficulties to adapt to sudden temperature changes.
I hope you find this information useful and interesting! Good luck!
To determine if a tropical reptile is an endotherm or an ectotherm, observe its behavior in response to environmental temperature changes and test its metabolic rate and body temperature in various conditions.
To determine whether a large tropical reptile with a high and relatively stable body temperature is an endotherm or an ectotherm, one should observe the animal's behavior and response to changes in environmental temperature. If the reptile is an endotherm, it would generate its own heat through metabolic processes and maintain a stable body temperature regardless of the environment.
Endotherms demonstrate behaviors like shivering to generate heat when cold. Ectotherms, on the other hand, rely on the environment to regulate their body temperature, and this can be observed if the animal basks in the sun to warm up or seeks shade to cool down.
Most reptiles are known to be ectotherms and can exhibit behaviors such as brumation in response to cold temperatures. This state is similar to hibernation but doesn't rely on fat reserves. Therefore, if the reptile is less active in colder conditions and depends on external heat sources like sunlight, it is likely an ectotherm. However, if the reptile maintains an active metabolism and body temperature in colder environments without relying on external heat sources, it may be an endotherm.
Specific tests could include monitoring the reptile's body temperature in controlled environments with varying temperatures and measuring the metabolic rate under these conditions to see if there is internal heat production.
Which biogeochemical cycle is the only one that does not include the atmosphere?
Answer:
Phosphorus cycle
Explanation:
Biogeochemical cycle, any of the natural pathways by which essential elements of living matter are circlated.
There are four types of biogeochemical cycle, they are ; water cycle,carbon cycle,nitrogen cycle and phosphorous cycle
Carbon cycle is the cycle in which photosynthesis and cellular respiration take place.
Water cycle involves transpiration.
Nitrogen cycle Is the cycle that is dependent upon bacteria for nitrogen fixation and denitrification.
Phosphorus cycle is one of the slowest biogeochemical cycle. It does not stay in the atmosphere, because it is normally in a liquid state at room temperature. It does not include the atmosphere.
Final answer:
The phosphorus cycle is the biogeochemical cycle that does not include an atmospheric component, differentiating it from cycles like water, carbon, and nitrogen.
Explanation:
The biogeochemical cycle that does not include the atmosphere is the phosphorus cycle. Unlike the water, carbon, and nitrogen cycles, the phosphorus cycle does not involve the atmospheric pathway. Phosphorus mainly cycles through rocks, soil, water, and living organisms. It is essential for life as part of DNA, RNA, and ATP molecules, which are vital for genetic information and energy transfer in cells. However, phosphorus doesn't enter the atmosphere in a gaseous phase; hence, it is distinct amongst biogeochemical cycles.
The ΔHcomb value for anethole is -5539 kJ/mol. Assume 0.840 g of anethole is combusted in a calorimeter whose heat capacity (Calorimeter) is 6.60 kJ/°C at 20.6 °C. What is the final temperature of the calorimeter
Answer:
Final temperature of calorimeter is 25.36^{0}\textrm{C}
Explanation:
Molar mass of anethole = 148.2 g/mol
So, 0.840 g of anethole = [tex]\frac{0.840}{148.2}moles[/tex] of anethole = 0.00567 moles of anethole
1 mol of anethole releases 5539 kJ of heat upon combustion
So, 0.00567 moles of anethole release [tex](5539\times 0.00567)kJ[/tex] of heat or 31.41 kJ of heat
6.60 kJ of heat increases [tex]1^{0}\textrm{C}[/tex] temperature of calorimeter.
So, 31.41 kJ of heat increases [tex](\frac{1}{6.60}\times 31.41)^{0}\textrm{C}[/tex] or [tex]4.76^{0}\textrm{C}[/tex] temperature of calorimeter
So, the final temperature of calorimeter = [tex](20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}[/tex]
To determine the final temperature of the calorimeter, we need to calculate the heat released by the combustion of anethole using its enthalpy change. Then, we can use the equation q = mcΔT to find the change in temperature of the calorimeter. The final temperature is 16.65 °C.
Explanation:To determine the final temperature of the calorimeter, we can make use of the equation q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we need to consider the heat capacity of the calorimeter as well. We can start by calculating the heat released by the combustion of anethole using the given enthalpy change of -5539 kJ/mol. Then, we can use the equation to find the change in temperature of the calorimeter.
First, we need to calculate the moles of anethole. Using the molar mass of anethole, which is 178.26 g/mol, we can find:
moles of anethole = 0.840 g / 178.26 g/mol = 0.004716 molNext, we can calculate the heat released by combustion:
heat released = ΔHcomb x moles of anethole = -5539 kJ/mol x 0.004716 mol = -26.1 kJNow, we can consider the calorimeter's heat capacity:
heat absorbed by the calorimeter = heat released by combustioncalorimeter x ΔT = -26.1 kJΔT = -26.1 kJ / 6.60 kJ/°C = -3.95 °CSince the initial temperature of the calorimeter is 20.6 °C, the final temperature will be:
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A student carried out a titration using HC2H3O2(aq) and NaOH(aq). The net ionic equation for the neutralization reaction that occurs during the titration is represented above. The NaOH(aq) was added from a buret to the HC2H3O2(aq) in a flask. The equivalence point was reached when a total of 20.0mL of NaOH(aq) had been added to the flask. How does the amount of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compare to the amount of HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq), and what is the reason for this result?
With every addition of NaOH, the concentration of HC2H3O2 in the solution decreases due to the neutralization reaction. Thus, the quantity of HC2H3O2 after adding 5.0 mL of NaOH will be less than the amount after adding 1.0 mL of NaOH. This is due to the stoichiometry of the reaction, where one mole of added NaOH neutralizes one mole of acetic acid.
Explanation:In this titration analysis, the neutralization reaction involves acetic acid (HC2H3O2) and Sodium Hydroxide (NaOH), creating sodium acetate and water as products. When 1.0 mL of NaOH is added, it reacts with the acetic acid present in the flask, reducing its quantity. When 5.0 mL of NaOH is added, more acetic acid reacts, and therefore, the amount of HC2H3O2 in the flask decreases more significantly compared to when only 1 mL was added. The principle of this result is based upon the stoichiometry of the neutralization reaction, indicating that for each mole of NaOH added, one mole of acetic acid is neutralized. Therefore, as more NaOH is added, more HC2H3O2 is neutralized, hence contributing to a lower concentration of HC2H3O2 in the solution.
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The amount of HC2H3O2(aq) in the flask will be less after the addition of 5.0mL NaOH than after the addition of 1.0mL NaOH. This happens because the NaOH neutralizes the acid directly proportional to its quantity added. More the NaOH added, more the acidic substance HC2H3O2 gets neutralized.
Explanation:In a titration process involving HC2H3O2(aq) and NaOH(aq), following the neutralization equation, each acid molecule reacts with one base molecule. Hence, the quantity of HC2H3O2 will reduce in direct proportion to the amount of NaOH added. So if you add 5.0mL of NaOH, you will neutralize 5 times as much HC2H3O2 as when you add 1.0mL of NaOH.
Therefore, the quantity of HC2H3O2(aq) remaining in the flask after the addition of 5.0mL NaOH will be less than that after the addition of 1.0mL NaOH. The reason is that more NaOH means more HC2H3O2(aq) has been neutralized into water and the salt NaC2H3O2.
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Consider the titration of 30.0 ml of 0.050 m nh3 with 0.025 mhcl. Calculate the ph after the following volumes of titrant have been added.
Explanation of pH changes in a titration between NH₃ and HCl.
The pH in a titration is determined by the amount of acid and base before and after the equivalence point. In the given scenario, titrating NH₃ with HCl, the initial pH is determined by the NH₃ concentration before any HCl is added.
(a) 0.00 mL: Before any HCl is added, NH₃ is in excess, making the solution basic with a pH >7.
(b) 25.00 mL: At the equivalence point, the pH is determined by the salt formed during the reaction. For NH₃ titrated with HCl, the salt formed is NH₄Cl, making the pH acidic.
A 27.5 −g aluminum block is warmed to 65.6 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.0 ∘C. The aluminum and the water are allowed to come to thermal equilibrium.
Answer:
Final temperature T is 26.197 degrees celcius.
Explanation:
As the system is in thermal equilibrium, the heat lost by aluminium is the heat gained by water.
specific heat capacity of aluminium = 0.9 J/g per degree celcius
specific heat capacity of water = 4.186 J/g per degree celcius
Let the final common temperature attained be "T".
Heat lost by aluminium = [tex](m)(c)(65.6-T)[/tex]
Heat gained by water = [tex](m)(c)(T-22)[/tex]
by law of conservation of energy,
Heat lost by hot body = Heat gained by cold body
[tex](m)(c)(65.6-T)[/tex] = [tex](m)(c)(T-22)[/tex]
[tex](27.5)(0.9)(65.6 - T) = (55.5)(4.186)(T - 22)[/tex]
calculating the value of T comes out as 26.197 degrees celcius.
List the 4 major groups of inorganic chemicals common in cells:
Water, Carbon Dioxide, Acids, and Oxygen are the inorganic chemicals common in cells.
What are inorganic chemicals?A substance without both carbon and hydrogen is said to be an inorganic compound.Many inorganic substances, like water (H₂O) and the hydrochloric acid (HCl) secreted by your stomach, do contain hydrogen atoms. In contrast, carbon atoms are only present in a small number of inorganic molecules. An example of one of the few is carbon dioxide (CO₂).Water makes up as much as 70% of an adult's body weight. The cells that make up tissues and organs have water both inside of them and in the spaces between them.A substance that releases hydrogen ions (H⁺) in solution is referred to be an acid. Because it releases all of its H⁺ in the stomach's watery environment, hydrochloric acid (HCl), which is secreted by cells in the lining of the stomach, is a powerful acid. This strong acid helps with digestion and destroys ingested microorganisms.Carbon dioxide (CO₂) is expelled during respiration in exchange for oxygen (O₂) from the surrounding air. At the cellular level, O₂ is utilized to produce CO₂ while converting metabolic energy from foods into adenosine triphosphate (ATP).
Thus, inorganic chemicals like Water, Carbon Dioxide, Acids, and Oxygen
are essentail for cells.
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Glucose is an example of which carbon-based macromolecule?
Answer: Glucose is an example of carbon-based macromolecule known as carbohydrates
Explanation:
carbon based macromolecule are important cellular components and they perform a variety of functions necessary for growth and development of living organisms. There are 4 major types of carbon based molecules and these includes;
Carbohydrate
Lipids
Proteins and
Nucleic acids.
Carbon is the primary components of these macromolecules. Carbohydrate macromolecules are made up of monosaccharide, disaccharide and polysaccharides. Glucose is an example of a monosaccharide and it has two important types of functional groups: a carbonyl group and a hydroxyl group. I hope this helps. Thanks
Glucose is a simple sugar and is an example of a carbohydrate, which is a type of carbon-based macromolecule. Glucose is a monosaccharide that can join together with other sugars to form complex carbohydrates or polysaccharides. It serves as a vital energy source for cells.
Explanation:Glucose is a type of carbon-based macromolecule known as a carbohydrate. Carbohydrates are organic molecules composed of carbon, hydrogen, and oxygen atoms and they serve as a vital energy source for cells. The family of carbohydrates includes both simple and complex sugars. Glucose and fructose are examples of simple sugars, whereas starch, glycogen, and cellulose are examples of complex sugars, also called polysaccharides.
Monosaccharides like glucose are the simplest form of carbohydrates. These small molecules can combine to form larger, more complex structures. When many monosaccharides join together, they form polysaccharides such as starch and glycogen, which serve as energy storage in plants and animals respectively.
While glucose serves as the most common fuel for ATP production within cells, excess glucose is either stored for future energy needs in the liver and skeletal muscles as glycogen, or it's converted into fat in adipose cells.
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Distillation is a process of vaporization a substance and chilling the vapor to collect it back the liquid form. How much heat is removed from 74.2 grams of ethanol vapor at 83 °C (Tb = 78.37 °C) if the collected liquid ethanol has a temperature of 26ºC?
Answer:
72 kJ of heat is removed.
Explanation:
First, the ethanol vapor will reduce its temperature until the temperature of the boiling point, then it will occur a phase change from vapor to liquid, and then the temperature of the liquid will decrease. The total heat will be:
Q = Q1 + Q2 + Q3
Q1 = n*cv*ΔT1, Q2 = m*Hl, and Q3 = n*cl*ΔT2
Where n is the number of moles, cv is the specific heat of the vapor (65.44 J/K.mol, cl is the specific heat of the liquid (111.46 J/K.mol), Hl is the heat of liquefaction (-836.8 J/g), m is the mass, and ΔT is the temperature variation (final - initial).
Q = n*cv*ΔT1 + m*Hl + n*cl*ΔT2
The molar mass of ethanol is 46 g/mol, and the number of moles is the mass divided by the molar mass:
n = 74.2/46 = 1.613 moles
Q = 1.613*65.44*(78.37 - 83) + 74.2*(-836.8) + 1.613*111.46*(26 - 78.37)
Q = -72000 J
Q = -72 kJ (because it is negative, it is removed)
Select all that apply. Which two of the following equations represent an exothermic reaction? NH3(g) + 12.0 kcal → ½N2(g) + 3/2H2(g) C(graphite) → C(diamond), ΔH = -0.45 kcal/mole C C + 2S → CS2, ΔH = 27,550 cal CH4 + 2O2 → CO2 + 2H2O + 212,800 cal 2H2 O → 2H2 + O2, ΔH = +58 kcal/mole H2O
Answer:
C(graphite) → C(diamond), ΔH = -0.45
CH4 + 2O2 → CO2 + 2H2O + 212,800 cal
Explanation:
NH3(g) + 12.0 kcal → ½N2(g) + 3/2H2(g) in this reaction we see that energy is added to the reactants so this is an endorthermic process. It takes in energy
C(graphite) → C(diamond), ΔH = -0.45 kcal/mole The change in enthalpy for this reaction is listed and it is negative, telling us that this is an exothermic reaction.
C + 2S → CS2, ΔH = 27,550 cal The change of enthalpy for this reaction is listed and it is positive which tells us that the reaction is endothermic reaction
CH4 + 2O2 → CO2 + 2H2O + 212,800 cal In this reaction, the energy is part of the equation on the products side. It is given out. this tells us that the reaction is exothermic
2H2 O → 2H2 + O2, ΔH = +58 kcal/mole H2O In this reaction, the energy is part of the equation on the products side. It is given out. this tells us that the reaction is exothermic
the reaction of 50 mL of gas with 50 mL of gas via the equation: Cl2(g) + C2H4(g) ➔ C2H4Cl2 (g) will produce a total of __________ mL of products if pressure and temperature are kept constant.
Explanation:
The given data is as follows.
50 ml of [tex]Cl_{2}[/tex], 50 ml of [tex]C_{2}H_{4}[/tex]
And, it is known that at STP 1 mole of a gas occupies 22.4 L. Hence, moles present in 50 ml of gas are as follows.
[tex]\frac{50}{22.4 \times 1000}[/tex] (As 1 L = 1000 ml)
= [tex]2.23 \times 10^{-3}[/tex] moles
So, according to the given equation [tex]2.23 \times 10^{-3}[/tex] moles of [tex]Cl_{2}[/tex] reacts with [tex]2.23 \times 10^{-3}[/tex] moles of [tex]C_{2}H_{4}[/tex].
Hence, moles of [tex]C_{2}H_{4}Cl_{2}[/tex] is equal to the moles of [tex]C_{2}H_{4}[/tex] and [tex]Cl_{2}[/tex].
Therefore, moles of [tex]C_{2}H_{4}Cl_{2}[/tex] = [tex]2.23 \times 10^{-3}[/tex] moles
1 mole of [tex]C_{2}H_{4}Cl_{2}[/tex] = 22.4 L
[tex]2.23 \times 10^{-3}[/tex] moles = [tex]22.4 \times 2.23 \times 10^{-3} moles[/tex]
= 50 ml of product
Thus, we can conclude that 50 ml of products if pressure and temperature are kept constant.
Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 24. g of butane is mixed with 44.3 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 3 significant digits.
Answer:
Maximum amount of [tex]CO_{2}[/tex] can be produced is 37.5 g
Explanation:
Balanced equation: [tex]2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O[/tex]
Molar mass of butane ([tex]C_{4}H_{10}[/tex]) = 58.12 g/mol
Molar mass of [tex]O_{2}[/tex] = 32 g/mol
Molar mass of [tex]CO_{2}[/tex] = 44.01 g/mol
So, 24 g of butane = [tex]\frac{58.12}{24}mol[/tex] of butane = 2.422 mol of butane
Also, 44.3 g of [tex]O_{2}[/tex] = [tex]\frac{44.3}{32}mol[/tex] of [tex]O_{2}[/tex] = 1.384 mol of [tex]O_{2}[/tex]
According to balanced equation-
2 moles of butane produce 8 mol of [tex]CO_{2}[/tex]
So, 2.422 moles of butane produce [tex](\frac{8}{2}\times 2.422)moles[/tex] of [tex]CO_{2}[/tex] = 9.688 moles of [tex]CO_{2}[/tex]
13 moles of [tex]O_{2}[/tex] produce 8 mol of [tex]CO_{2}[/tex]
So, 1.384 moles of [tex]O_{2}[/tex] produce [tex](\frac{8}{13}\times 1.384)moles[/tex] of [tex]CO_{2}[/tex] = 0.8517 moles of [tex]CO_{2}[/tex]
As least number of moles of [tex]CO_{2}[/tex] are produced from [tex]O_{2}[/tex] therefore [tex]O_{2}[/tex] is the limiting reagent.
So, maximum amount of [tex]CO_{2}[/tex] can be produced = 0.8517 moles = [tex](44.01\times 0.8517)g=37.5 g[/tex]
To calculate the maximum mass of carbon dioxide produced, use the balanced equation and determine the limiting reactant. Then, calculate the moles of carbon dioxide produced using the mole ratio from the balanced equation and convert it to grams using the molar mass of carbon dioxide.
Explanation:To calculate the maximum mass of carbon dioxide that could be produced by the reaction between gaseous butane (C4H10) and gaseous oxygen (O2), we need to use the balanced equation for the reaction:
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
First, we need to determine the limiting reactant by comparing the moles of butane and oxygen. The molar mass of butane is 58.1 g/mol, so 24 g of butane is equal to 24/58.1 moles. The molar mass of oxygen is 32 g/mol, so 44.3 g of oxygen is equal to 44.3/32 moles.
Next, we calculate the moles of carbon dioxide produced using the mole ratio from the balanced equation. Since the mole ratio between butane and carbon dioxide is 1:4, the moles of carbon dioxide produced is 4 times the moles of butane. Finally, we convert the moles of carbon dioxide to grams by multiplying by the molar mass of carbon dioxide (44 g/mol).
Using this information, we can calculate the maximum mass of carbon dioxide that could be produced. The answer should be rounded to 3 significant digits to match the rounding specified in the question.
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Choose the molecule or compound that exhibits dipole-dipole forces as its strongest intermolecular force. a) BCl3 b) N2 c) H2 d) SO2 e) CBr4.
Answer:
SO₂
Explanation:
The dipole-dipole force is not only determined by the electron density around each atoms in the molecule (dependent of electronegativity difference), but also how the atoms in the molecules are arranged. In general, the more symmetric a molecule is, the less dipole force it exerts as each dipole moments cancels each other out.
Now let's examine each answer
b and c, N₂ and H₂ are composed of same type of atoms, therefore, no dipole moment occurs, and no dipole-dipole forces are exerted a and e, BCl₃ and CBr₄ are composed of atoms with different electronegativity, but are symmetric. With BCl₃ having trigonal planar structure and CBr₄ has tetragonal structure, each B-Cl and C- Br bond cancels out each other dipole moment, and thus, no dipole moments were generated.d, SO₂ structure is not linear, but is a little bent, this allows net dipole moment to occurs in this molecule as dipole moment from each S=O bond do not cancels each other outLight with a wavelength of 190 nm is incident on a metal surface. The most energetic electrons emitted from the surface are measured to have 4.0 ev of kinetic energy. Which of the metals in the table is the surface most likely to be made of?
The question deals with the photoelectric effect, where light ejects electrons from a metal's surface. To identify the metal, one must calculate the work function using the kinetic energy of the emitted electrons and the frequency of the incident light.
Explanation:The question concerns the photoelectric effect, a phenomenon in physics where electrons (called photoelectrons) are ejected from a metal surface when it is exposed to light of a certain frequency or wavelength that is above the metal's threshold frequency. You've given a light with a wavelength of 190 nm incident on an unknown metal, and the most energetic electrons emitted have 4.0 eV of kinetic energy.
To identify which metal this could be, one needs to calculate the work function of the metal. This involves using the equation for the photoelectric effect: KE = hf - Φ, where KE is the kinetic energy of the ejected electrons, h is Planck's constant, f is the frequency of the incident light, and Φ represents the work function (also known as the binding energy) of the metal.
First, we need to convert the wavelength of 190 nm into frequency using the equation c = λf, where c is the speed of light, λ is the wavelength, and f is the frequency. Once we have the frequency, we can plug it into Planck's equation to find the energy of the photons. Subsequently, we'll use the initial energy of the photons to determine the work function. By comparing this calculated work function with the known work functions of various metals, we can identify the most likely metal of the surface.
Problem Page A chemist adds of a calcium bromide solution to a reaction flask. Calculate the mass in grams of calcium bromide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answer:
see explanation below
Explanation:
First, you are not providing any data of the bromide solution to calculate the mass. So, in order to help you, I will take some random values from a similar exercise, so you can solve this later with your data.
Let's suppose you add 360 mL of a 1.45 mol/L of a calcium bromide solution into the flask. To calculate the mass it was added, you need to calculate first the moles added. This can be done with the following expression:
M = n/V
Where:
M: molarity of solution
n: moles of solution
V: volume (in liters) of solution
here, you have to solve for n, so:
n = M*V
replacing the above data you have:
n = 1.45 * (0.360) = 0.522 moles
Now that we have the moles, you can calculate the mass by the following expression:
m = n * MM
Where MM it's the molar mass of calcium bromide. The reported MM of calcium bromide is 199.89 g/mol, so replacing:
m = 199.89 * 0.522
m = 104.34 g
And this is the mass that was added of the solution. As I stated before, use your data in this procedure, and you should get an accurate result.
A bath tub is half full of warm water. How could the thermal energy of the water be increased? A) Add hot water to the tub. B) Heat the water in the tub without adding more water. C) Add more water at the same temperature to the tub. D) Transfer the water to a smaller container.
Answer:
A, B, & C
Explanation:
got it right on edu
The thermal energy of the bathtub can be increased by adding hot water, heating the water, and adding more water to the tub at the same temperature. Thus, options A, B, and C are correct.
What is thermal energy?Thermal energy is the increase in the collision and the kinetic energy of particles of matter due to a rise in the temperature of the substance. This movement produces energy called thermal energy.
The thermal energy of the substance can be increased by providing them with the same energy molecules. The addition of the hot water will increase the energy due to increased collision of the particles of the hot water.
Heating the water will increase the kinetic energy and the rate of a collision resulting in increased thermal energy. Also, water with the same temperature can be added to increase energy.
Therefore, in options A, B, and C adding hot water and heating water increases the thermal energy.
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The energy change, ∆H, associated with the following reaction is +81 kJ. NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g) What is the expected energy change for the reverse reaction of nine moles of HOBr and two moles of NH3?
Answer:
162 kJ
Explanation:
The reaction given by the problem is:
NBr₃(g) + 3 H₂O(g) → 3 HOBr(g) + NH₃(g) ∆H = +81 kJIf we turn it around, we have:
3 HOBr(g) + NH₃(g) → NBr₃(g) + 3 H₂O(g) ∆H = -81 kJIf we think now of HOBr and NH₃ as our reactants, then now we need to find out which one will be the limiting reactant when we have 9 moles of HOBr and 2 moles of NH₃:
When we have 1 mol NH₃, we need 3 mol HOBr. So when we have 2 moles NH₃, we need 6 moles HOBr. We have more than 6 moles HOBr so that's our reactant in excess, thus NH₃ is our limiting reactant.-81 kJ is our energy change when there's one mol of NH₃ reacting, so we multiply that value by two when there's two moles of NH₃ reacting. The answer is 81*2 = 162 kJ.
Final answer:
The energy change for the reverse reaction of NBr3 and H2O to form HOBr and NH3 is -81 kJ. When considering nine moles of HOBr and two moles of NH3, thrice the stoichiometric amounts, the expected energy change would be three times as much, resulting in -243 kJ.
Explanation:
The energy change, ΔH, for a reaction indicates how much energy is absorbed or released during the reaction. In the given reaction, NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g), the energy change is +81 kJ, meaning 81 kJ of energy is absorbed in the forward reaction. Since energy is a state function and the magnitude of energy change is the same for a reaction whether it goes in the forward or reverse direction, the energy change for the reverse reaction would be -81 kJ. The reaction in reverse would release 81 kJ of energy.
However, the question asks for the expected energy change for reversing the original reaction with nine moles of HOBr and two moles of NH3. The energy change provided (+81 kJ) is for the stoichiometric amounts as per the original equation. For nine moles of HOBr and two moles of NH3, which are thrice the stoichiometric amounts indicated by the original reaction, the energy released would be three times as much. Therefore, the expected energy change for the reverse reaction under these conditions would be 3 × (-81 kJ) = -243 kJ.
A possible mechanism for the overall reaction represented above is the following.
NO(g) + NO(g) → N2O2(g) slow
N2O2(g) + O2(g) → 2NO2(g) fast
Which of the following rate expressions agrees best with this possible mechanism?
A) Rate = k[NO]2
(D) Rate = k[NO]2[O2]
(B) Rate = k[NO]
(E) Rate = k[N2O2][O2][O2]
(C) Rate = k[NO]2[O2]
Answer:
A
Explanation:
The slow step of the reaction is the rate determining step. This means to get the rate law for a chemical activity with slow and fast steps, we have to consider the slow step if we are to write the rate law successfully.
Since the compound NO is reacting with itself, we have to raise the value of the concentration in the square bracket by 2. That is why we have the concentration squared.
Answer: it’s c
Explanation:
Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to the following reaction. Molar Mass Fe2O3 = 159.7 g/mol
Answer:
The percent yield of the reaction is 82%
Explanation:
First step: make the chemist equation.
2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)
As the statement says that aluminun is in excess, the limiting reactant is the Fe2O3
Second step: Find out the moles in the reactant.
Molar weight Fe2O3: 159.7 g/m
Mass / Molar weight = moles
50 g /159.7 g/m = 0.313 moles
Third step: Analyse the reaction. 1 mol of Fe2O3 makes 2 moles of Fe.
1 mol Fe2O3 ____ 2Fe
0.313 mol Fe2O3 ____ 0.626 moles
Molar weight Fe = 55.85 g/m
Moles . molar weight = mass
55.85g/m . 0.626m = 34.9 grams
This will be the 100% yield of the reaction but we only made 28.65 g
34.9 g ____ 100%
28.65 g ____ 82.09 %
Answer:
83%
Explanation:
We first get the chemical reaction :
2 Al(s) + Fe2O3(s) ------> 2Fe(s) + Al2O3(s)
From the reaction we can see that one mole of the oxide yielded 2 moles of iron.
Firstly, we need to calculate the theoretical yield of the iron. This is done as follows. The number of moles of the oxide equals the mass of the oxide divided by the molar mass of the oxide = 50g ÷ 159.7 = 0.313moles
From the first relation, one mole oxide yielded 2 moles iron, hence 0.313 mole oxide will yield 2 × 0.313 mole iron = 0.616 moles
The mass of iron thus generated = 0.616 × 56 = 34.496g
% yield = Actual yield/theoretical yield × 100%
%yield = 28.65/34.396 × 100% = 83%
Draw the intermediate (in the first box) that forms in the reaction of 1-methylcyclopentene with br2 in ccl4, then draw the expected product (in the second box). do not provide stereochemistry. interactive 3d display mode
Answer:
The product of the reaction is 1,2 di-bromo-1-methylcyclopentane.
Explanation:
In the reaction of 1-methylcyclopentene with Br2 in CCl4, a bromonium ion intermediate forms, followed by nucleophilic attack by Cl-, resulting in the product 1-chloromethylcyclopentane.
In the reaction of 1-methylcyclopentene with Br2 in CCl4, a bromonium ion intermediate is formed. The bromonium ion is a three-membered ring with a bromine atom attached to two adjacent carbon atoms. Due to the lack of stereochemistry information, the structure of the intermediate is a general representation without specifying the stereochemical arrangement.
Following the formation of the bromonium ion, it undergoes a nucleophilic attack by the chloride ion (Cl-) from the solvent CCl4. The chloride ion opens the bromonium ring, leading to the substitution of bromine by chloride on one of the adjacent carbons. The product of this reaction is 1-chloromethylcyclopentane. The chlorine atom is now attached to the carbon that originally carried the methyl group.
This reaction mechanism is a classic example of halogenation of alkenes, where the bromine adds to the alkene in a concerted manner, forming a cyclic intermediate. The subsequent nucleophilic attack by the chloride ion results in the replacement of bromine with chlorine, leading to the final halogenated product.
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Two different ionic compounds each contain only copper and chlorine. Both compounds are powders, one white and one brown. An elemental analysis is performed on each powder. Which of the following questions about the compounds is most likely to be answered by the results of the analysis?
A. What is the density of each pure compound?
B. What is the formula unit of each compound?
C. What is the chemical reactivity of each compound?
D. Which of the two compounds is more soluble in water?
Answer:
B.
Explanation:
Elemental analysis involves the actual percentage compositions of each element in the sample of the compounds. We know that there are two different compounds here that both contain copper and oxygen. What we don't know is the actual number of copper or oxygen composed in the compounds.
An elemental composition of these compounds would thus tell us this and will give an idea of the chemical formula of each since we now know quite well the percentage composition of each of these elements
The question about the compounds which is most likely to be answered by the results of the analysis is "What is the formula unit of each compound?"
According to the law of multiple proportions, when two elements combine to form different compounds, the proportion of the elements in the different compounds are in simple ratio.
In accordance to this law, the proportion of copper and chlorine in the white and brown compounds vary.
One question that can be answered by an elemental analysis is the question; "What is the formula unit of each compound?"
The elemental analysis shows the amount of each element contained in each of the compounds from which its formula unit can be obtained.
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0.450 mol of aluminum hydroxide is allowed to react with 0.550 mol of sulfuric acid; the reaction which ensues is: 2Al(OH)3(s) + 3H2SO4(aq) -------->Al2(SO4)3(aq) + 6H2O(l) How many moles of H2O can form under these conditions?
Answer:
The answer to your question is 1.1 moles of water
Explanation:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
0.45 mol 0.55 mol ?
Process
1.- Calculate the limiting reactant
Theoretical proportion
Al(OH)₃ / H₂SO₄ = 2/3 = 0.667
Experimental proportion
Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81
From the proportions, we conclude that the limiting reactant is H₂SO₄
2.- Calculate the moles of H₂O
3 moles of H₂SO₄ ---------------- 6 moles of water
0.55 moles of H₂SO₄ ----------- x
x = (0.55 x 6) / 3
x = 3.3 / 3
x = 1.1 moles of water
Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in 1783, Jacques Charles used a balloon that contained approximately 31100 L of H2. He generated the H2 using the reaction between iron and hydrochloric acid: Fe(s)+2HCl(aq)→FeCl2(aq)+H2(g)
How many kilograms of iron were needed to produce this volume of H2 if the temperature was 21 ∘C? (For simplicity, assume that the pressure experienced by the balloon will be 1.00 atm.)
Answer:
[tex]m_{Fe}=71.982 kg[/tex]
Explanation:
First of all, we need to calculate the moles of H2. Assuming the H2 is an ideal gas:
[tex]n=\frac{P*V}{T*R}[/tex]
[tex]n=\frac{1atm*31100L}{294K*0.082 L*atm*mol^{-1}*K^{-1}}[/tex]
[tex]n=1290 mol[/tex]
Now, to produce 1 mol of H2 is required 1 mol of Fe:
[tex]m_{Fe}=n*M=1290mol*\{55.8 g}{mol}[/tex]
[tex]m_{Fe}=71982 g=71.982 kg[/tex]
The study of chemicals and bonds is called chemistry. There are two types of elements are there and these are metals and nonmetals.
The correct answer is 71.982kg.
What is ideal gas law?The ideal gas law, also called the general gas equation, is the equation of the state of a hypothetical ideal gas. It is a good approximation of the behavior of many gases under many conditions, although it has several limitations.The formula used as:-
[tex]n=\frac{P*V}{T*R} \\\\n=\frac{1*31100}{294*0.082} \\\\\\n=1290moles[/tex]
The mass will be:-
[tex]M= n*M\\1290*55.8\\\\=71.982[/tex]
Hence, the correct answer is 71.982.
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Name the structure. CHC H.CH2CH2i CHC CH3
A) 7-chloro-3-ethyl-4-methyl-3-heptene
B) 1-chloro-5-ethyl-4-methyl-3-heptene x
C) 1-chloro-3-pentenyl-2-pentene x
D) cis-7-chloro-3-ethyl-4-methyl-3-heptene
E) trans-7-chloro-3-ethyl-4-methyl-3-heptene
The question is incomplete. The complete question is attached below.
Answer : The correct option is, (A) 7-chloro-3-ethyl-4-methyl-3-heptene
Explanation :
The rules for naming of alkene are :
First select the longest possible carbon chain.
The longest possible carbon chain should include the carbons of double bonds.
The naming of alkene by adding the suffix -ene.
The numbering is done in such a way that first carbon of double bond gets the lowest number.
The carbon atoms of the double bond get the preference over the other substituents present in the parent chain.
If two or more similar alkyl groups are present in a compound, the words di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.
The given compound name will be, 7-chloro-3-ethyl-4-methyl-3-heptene.
The structure of given compound is shown below.
The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene.
Explanation:The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene. The name is determined by identifying the longest carbon chain, the substituents attached to it, and their positions. In this case, the longest carbon chain has 7 carbons, with a chlorine atom attached at position 7. There is an ethyl group at position 3 and a methyl group at position 4. The presence of double bonds is indicated by the -ene ending.
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Ammonia, NH3, acts as an Arrhenius base because it Ammonia, N H 3, acts as an Arrhenius base because it blank the concentration of hydroxide ion, \rm OH^-, in aqueous solution. the concentration of hydroxide ion, OH−, in aqueous solution.
Answer:
Ammonia acts as an Arrhenius base because it increases the concentration of OH⁻ in aqueous solution.
Explanation:
The acid-base theory of Arrhenius explains that in aqueous solutions both acid and base dissociate, releasing ions in the solution. The acid release the ion H⁺ and some anion, and the base release the ion OH⁻ and some cation.
In water, the reaction of ammonia is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Because of that, ammonia is an Arrhenius base.
Ammonia acts as an Arrhenius base because it increases the concentration of hydroxide ions in solution through a reaction with water molecules where it accepts a proton, forming NH4+ and OH-. The Arrhenius definition is focused on hydroxide ion production; however, broader definitions like Brønsted-Lowry and Lewis provide a more general understanding of acid-base chemistry.
Ammonia, NH3, acts as an Arrhenius base because it increases the concentration of hydroxide ion, OH-, in aqueous solution by accepting a proton from water. This reaction can be represented by:
NH3(aq) + H2O(l) ⇒ NH4+(aq) + OH-(aq)
In this reaction, ammonia takes a hydrogen ion (H+) from a water molecule to produce an ammonium ion (NH4+) and a hydroxide ion (OH-). Unlike what might be initially assumed, ammonia does not release hydroxide ions into a solution directly; rather, the concentration of hydroxide ions increases due to this proton transfer.
Additionally, more fundamental definitions of acids and bases, beyond the Arrhenius model, include the Brønsted-Lowry theory, where bases are substances that accept hydrogen ions (H+), and the Lewis model, where a base is an electron pair donor.
Overall, while the Arrhenius model was originally seen as limited, it helps to explain how substances like ammonia function as bases without containing hydroxide ions in their formula.
Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed.
If no reaction occurs, write "NO REACTION."
a. potassium carbonate and lead(II) nitrate
b. lithium sulfate and lead(II) acetate
c. copper(II) nitrate and magnesium sulfide
d. strontium nitrate and potassium iodi
Answer:
a. K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)
b. Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)
c. Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)
d. NO REACTION
Explanation:
For the reactions, the cation and the anion of the compounds will be replaced. The reaction will occur if at least one of the products is insoluble and will form a precipitated.
a. Potassium carbonate = K₂CO₃
Lead(II) nitrate = Pb(NO₃)₂
Products = KNO₃ and PbCO₃.
According to the solubility rules, all K⁺ ions are soluble, with no exceptions, so KNO₃ is soluble. All CO₃⁻² ions are insoluble, and Pb⁺² is not an exception, so PbCO₃ will be insoluble and will form a precipitated, so the reaction happen:
K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)
b. Lithium sulfate = Li₂SO₄
Lead(II) acetate = Pb(C₂H₃O₂)₂
Products = Li(C₂H₃O₂) and PbSO₄
All Li⁺ are solubles, without exceptions, so Li(C₂H₃O₂) is soluble, and all SO₄⁻² are soluble, but Pb⁺² is an exception, so PbSO₄ is insoluble and will form a precipitated, then the reaction happens:
Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)
c. Copper(II) nitrate = Cu(NO₃)₂
Magnesium sulfide = MgS
Products = CuS and Mg(NO₃)₂
All NO₃⁻ are soluble, with no exceptions, so Mg(NO₃)₂ is soluble, and all S⁺² are insoluble, and Cu⁺² is not an exception, so CuS is insoluble, and will form a precipitated, then the reaction happens:
Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)
d. Strontium nitrate = Sr(NO₃)₂
Potassium iodi = KI
Products = K(NO₃)₂ and SrI₂
All K⁺ are soluble, with no exceptions, so K(NO₃)₂ is soluble, and all I⁻ are soluble, and Sr⁺² are not an exception, then SrI₂ is soluble. Therefore, no precipitated is formed and the reaction doesn't happen.
Precipitation reactions occur for some pairs of aqueous solutions, while others do not.
Explanation:a. The combination of potassium carbonate and lead(II) nitrate will result in the formation of a precipitate. The balanced molecular equation for this reaction is:
Pb(NO3)2 + K2CO3 → PbCO3 + 2 KNO3
b. The combination of lithium sulfate and lead(II) acetate will not result in a precipitation reaction.
c. The combination of copper(II) nitrate and magnesium sulfide will result in the formation of a precipitate. The balanced molecular equation for this reaction is:
Cu(NO3)2 + MgS → CuS + Mg(NO3)2
d. The combination of strontium nitrate and potassium iodide will not result in a precipitation reaction.
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One difference between a chemical reaction and a nuclear reaction is that in a nuclear reaction:
a) atoms retain their identity.
b) only the valence electrons are involved.
c) atoms often change from one element to another.
d) only small amounts of energy are absorbed or emitted.
Answer:
C
Explanation:
A nuclear reaction is one in which 2 atoms combine or an atom divides to give rise to entirely different atoms. This is basically the difference it has with a chemical reaction. In a chemical reaction, only new compounds are formed but not new elements. In the case of a nuclear reaction however, even if new elements are not formed, a different atom would be formed.
It takes 208.4 kJ of energy to remove 1 mole of electrons from 1 mole of atoms on the surface of rubidium metal. How much energy does it take to remove a single electron from an atom on the surface of solid rubidium?
Answer:
3.46*10⁻²² kJ
Explanation:
By the Avogadro's number, 1 mole of electrons at 1 mole of atoms correspond to 6.02x10²³ electrons. So it's necessary 208.4 kJ to remove 6.02x10²³ electrons. To remove a single electron:
6.02*10²³ electrons ---------------- 208.4 kJ
1 electron ---------------- x
By a simple direct three rule:
6.02*10²³ x = 208.4
x = 3.46*10⁻²² kJ
Most chemical digestion and nutrient absorption is performed in the __________.
You sealed an Erlenmeyer flask that was determined to have a volume of 272.2 mL with a stopper. The density of air at 25°C is 0.001185 g/mL. Air has an effective molar mass of 28.96 g/mol. What amount of air was sealed in the flask?
Answer:
The amount of air is 11.1 mmol
Explanation:
Density shows the relation between mass and volume, so you can know the mass which has been sealed in the Erlenmayer.
Density = mass/ volume
Density . volume = mass
0.001185 g/ml . 272.2 ml = 0.322 g
We have the mass now, so the molar weight determinates the mols
Mass/molar weight = mol
0.322g/ 28.96 g/m = 0.0111 mol
To get a better number we can inform on milimols
Mols . 1000 = milimols
The amount of air sealed in the 272.2 mL Erlenmeyer flask is approximately 0.0111 mol, calculated using the given density and molar mass of air at 25°C.
Explanation:To determine the amount of air sealed in the flask, we need to use the density of the air (given as 0.001185 g/mL at 25°C), the volume of the flask and the molar mass of air.
First, using the density and the volume of the flask, we can calculate the mass of the air in the flask with the formula density = mass/volume. We multiply the given density of 0.001185 g/mL by the volume of 272.2 mL to find the mass of the air to be around 0.3224 g.
Next, to convert this mass to moles, we use the given molar mass of air (28.96 g/mol). Moles = mass / molar mass, hence the air in the flask is approximately 0.0111 mol. Thus, there is approximately 0.0111 mol of air sealed in the Erlenmeyer flask.
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Assuming each solution to be 0.10 M , rank the following aqueous solutions in order of decreasing pH.
Rank the solutions from the highest to lowest pH. To rank items as equivalent, overlap them
a.N2H2
b.Ba(OH)2
c.HOCLd.NAOH
e.HCL
Final answer:
The aqueous solutions ranked from highest to lowest pH are: NaOH (most basic), Ba(OH)2, N2H2, HOCl, and HCl (most acidic).
Explanation:
To rank the following aqueous solutions in order of decreasing pH, from highest pH (most basic) to lowest pH (most acidic), we need to understand the nature of each compound in water. The ions or molecules that each solution releases in water determine their pH.
NaOH - Sodium hydroxide is a strong base, and will completely dissociate in water, releasing OH- ions, which leads to a high pH.
Ba(OH)2 - Barium hydroxide is another strong base, and it also fully dissociates in water, but being a dibasic base, it releases twice as many OH- ions per formula unit as NaOH, potentially leading to an even higher pH.
N2H2 - Hydrazine is a weak base, it does not completely dissociate in water, but will still increase the pH to some extent.
HOCl - Hypochlorous acid is a weak acid, so it only partially dissociates in water.
HCl - Hydrochloric acid is a strong acid and fully dissociates in water, releasing H+ ions and resulting in a low pH.
The solutions in order of decreasing pH (highest pH to lowest pH) are: NaOH, Ba(OH)2, N2H2, HOCl, HCl.