When the balloon is submerged in water at a depth where the pressure is 4.7 atm and the temperature is 15 °C, its volume will be approximately 0.995 L.
From the ideal gas equation, we can use the combined gas law formula, which is:
(P1 × V1) / T1 = (P2 × V2) / T2
Here, P1 = 1.1 atm (initial pressure), V1 = 3.7 L (initial volume), T1 = 30 °C (initial temperature), P2 = 4.7 atm (final pressure), and T2 = 15 °C (final temperature). We need to find V2 (final volume).
First, convert the temperatures to Kelvin by adding 273.15:
T1 = 30 + 273.15 = 303.15 K
T2 = 15 + 273.15 = 288.15 K
Now, plug in the values into the combined gas law formula and solve for V2:
(P1 × V1) / T1 = (P2 × V2) / T2
(1.1 × 3.7) / 303.15 = (4.7 × V2) / 288.15
(4.07) / 303.15 = (4.7 × V2) / 288.15
Now, solve for V2:
V2 = (4.07 × 288.15) / (303.15 × 4.7)
V2 ≈ 0.995 L
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2. What factors determine the rate at which a substance dissolves?
Answer:
Temperature
Agitation
Particle size
Explanation:
A solution of lithium carbonate reacts with a solution of potassium iodide to produce solid potassium carbonate and aqueous lithium iodide. You measure 3. 9 moles of potassium carbonate produced by the reaction. How much lithium iodide was produced?
The quantity of lithium iodide produced in the reaction was determined to be 7.8 moles.
The balanced chemical equation for the reaction between lithium carbonate (Li₂CO₃) and potassium iodide (KI) is:
2 Li₂CO₃ + 2 KI → 2 K₂CO₃ + 4 LiI
From the balanced equation, we can see that for every 2 moles of Li₂CO₃ reacted, 4 moles of LiI are produced.
Therefore, if we have 3.9 moles of K₂CO₃, we can calculate the moles of LiI produced as:
3.9 moles K₂CO₃ × (4 moles LiI / 2 moles Li₂CO₃) = 7.8 moles LiI
Therefore, 7.8 moles of lithium iodide were produced in the reaction.
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2. write out the balanced chemical equation for the reaction of 2 moles of naoh with 1 mole of h3po4
2NaOH + H₃PO₄ → Na₂HPO₄ + 2H2O is the balanced chemical equation for the reaction between 2 moles of NaOH and 1 mole of H3PO4.
It is clear from the balanced chemical equation that the reaction between 2 moles of NaOH and 1 mole of H₃PO₄ is an acid-base reaction, commonly referred to as a neutralization reaction.
In this reaction, phosphoric acid (H₃PO₄) acts as the acid and sodium hydroxide (NaOH) as the base. Na₂HPO₄ and H2O are created when the base (NaOH) and acid (H₃PO₄) react. Since all the reactants are completely consumed in the reaction and no excess of either reactant is left over, the stoichiometric balance of the number of moles of the acid and base is demonstrated by the balanced chemical equation.
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what do you think determines these traits in the lobsters? How could these traits change?
The traits in lobsters are determined by their genetic makeup and environmental factors.
Natural selection can play a role in changing traits over time.
Which genetic factors are at play?Genetic factors include inherited traits from their parents such as color, size, and shell density. Environmental factors such as water temperature, salinity, and availability of food can also impact these traits.
For example, lobsters in warmer water tend to grow faster and larger than those in cooler water. Changes in habitat or pollution can also impact the availability of food and water quality, leading to changes in growth rates and physical traits.
Lobsters with advantageous traits, such as stronger shells or better camouflage, are more likely to survive and pass on their genes to the next generation. Over time, these beneficial traits may become more common in the population.
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How many ions would form from the dissociation of ammonium dichromate (nh4)2cr2o7?
One formula unit of ammonium dichromate, three ions would form upon dissociation in water.
When ammonium dichromate (NH₄)₂Cr₂O₇ dissociates in water, it breaks down into two ammonium ions (NH₄⁺) and one dichromate ion (Cr₂O₇²⁻). The dissociation is represented by the following chemical equation:
(NH₄)2Cr₂O₇ → 2NH₄⁺ + Cr₂O₇²⁻
Therefore, a total of three ions would be formed from the dissociation of ammonium dichromate in water. The two ammonium ions would have a positive charge, while the dichromate ion would have a negative charge.
These ions can interact with other ions in the solution and participate in various chemical reactions. The dissociation of ammonium dichromate is important in various industrial processes, as well as in chemical education for demonstrating chemical reactions and properties of ions.
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I NEED HELP ON THIS ASAP!!!
Answer: I believe it's A
Source: Trust me bro
Silo measure d 640 grams of sulphur which occupies 540ml of container at 47 degree celsius.find the pressure of the gas.
The pressure of the gas is 76.8 atm.
The pressure of the gas can be calculated using the ideal gas law formula:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to calculate the number of moles of sulfur:
molar mass of sulfur = 32 g/mol
number of moles = mass/molar mass
= 640 g/32 g/mol
= 20 mol
Next, we need to convert the volume from milliliters to liters and the temperature from Celsius to Kelvin:
V = 540 ml = 0.54 L
T = 47°C + 273.15 = 320.15 K
Finally, we can plug in the values and solve for pressure
P = nRT/V
= 20 mol x 0.08206 L atm mol⁻¹ K⁻¹ x 320.15 K / 0.54 L
= 76.8 atm
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13. Lab Analysis: You forgot to label your chemicals and do not know whether your unknown solution is strontium nitrate or magnesium nitrate. You use the solutions potassium carbonate and potassium sulfate in order to determine your mistake. unknown + potassium carbonate & unknown + potassium sulfate . What do you observe when the unknown solution is mixed with potassium sulfate? (Can you see the shape underneath?)
if the unknown solution is mixed with potassium carbonate, the reaction will proceed differently depending on whether the unknown solution is strontium nitrate or magnesium nitrate.
Mixing the unknown solution with potassium sulfate will not provide any useful information to identify whether the unknown solution is strontium nitrate or magnesium nitrate. This is because neither strontium nor magnesium sulfate has distinctive properties that allow them to be easily distinguished from one another.
However, When mixed with strontium nitrate, potassium carbonate will form a white precipitate of strontium carbonate, while no reaction will occur when mixed with magnesium nitrate. Therefore, the presence of a white precipitate after mixing with potassium carbonate indicates that the unknown solution is strontium nitrate.
In summary, to identify whether the unknown solution is strontium nitrate or magnesium nitrate, the solution should be mixed with potassium carbonate. If a white precipitate forms, the solution is strontium nitrate. If no reaction occurs, the solution is magnesium nitrate. Mixing the unknown solution with potassium sulfate will not provide any useful information.
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Naturally occurring potassium consists of potassium-39 and potassium-41. calculate the percentage of each isotope present if theaverage is 39.1.
When the average is 39.1, naturally occurring potassium consists of 50% potassium-39 and 50% potassium-41.
An isotope is a variant of an element that has the same number of protons but a different number of neutrons in its nucleus. Potassium has two naturally occurring isotopes: potassium-39 and potassium-41. To calculate the percentage of each isotope present when the average is 39.1, we can use the following formula:
% of potassium-39 = (39.1 - 41) / (39 - 41) x 100%
% of potassium-41 = 100% - % of potassium-39
Using this formula, we can first calculate the percentage of potassium-39:
% of potassium-39 = (39.1 - 41) / (39 - 41) x 100%
% of potassium-39 = -1 / (-2) x 100%
% of potassium-39 = 50%
This means that potassium-39 makes up 50% of the naturally occurring potassium. To calculate the percentage of potassium-41, we simply subtract the percentage of potassium-39 from 100%:
% of potassium-41 = 100% - 50%
% of potassium-41 = 50%
Therefore, potassium-41 also makes up 50% of the naturally occurring potassium. In summary, when the average is 39.1, naturally occurring potassium consists of 50% potassium-39 and 50% potassium-41.
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Which words would be under the subheading "Ingredients"?
(Heading) Old Hunting Recipe for Rhinoceros Stew
(Subheading) Ingredients:
hair
broth
pepper
rhinoceros
hare
salt
water
onions
The words listed under the subheading "Ingredients" for the recipe "Old Hunting Recipe for Rhinoceros Stew" would be: Rhinoceros, Hare, Onions, Water, Broth, Salt, Pepper, and Hair.
What word would be listed?Under the subheading "Ingredients" for the recipe "Old Hunting Recipe for Rhinoceros Stew," the following words would be listed:
RhinocerosHareOnionsWaterBrothSaltPepperHair (Note: this is an unusual ingredient and may be questioned as to its necessity in the recipe)
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What is the volume of 0.78 mol of CO₂ gas at STP?
Answer: 17.49L
Explanation:
STP is 1atm and 273.15K
V=nRT/
V=(0.78)(0.0821)(273.15)/1
V= 17.49L
NEED HELP FAST!!!! Please answer both questions
There is a 0.37 M molarity.
1.71 m molality is present.
Molarity: What is it?The quantity of a solute in a solution is measured in terms of molarity, a unit of concentration. It is described as the quantity of solutes that dissolve in one liter of solution, or mol/L. Molarity, in other words, reveals how many moles of solute there are in a liter of solution.
To determine molarity, use the following formula:
Molarity (M) is calculated as moles of solute divided by the liters of solution.
100g/180 g/mol * 1/1.5 L is the molarity.
= 0.37 M,
Molality = 200g/58.5g/mol * 1/2 Kg
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Review this reaction:
H2SO4+NaOH->?.
What are the products?
the process in which an atom or ion experiences a decrease in its oxidation state is _____________.
Answer:
Reduction
Explanation:
when an atom or ion decreases in oxidation state
The process in which an atom or ion experiences a decrease in its oxidation state is called reduction.
Reduction is the opposite of oxidation, which is the process in which an atom or ion experiences an increase in its oxidation state. In a redox (reduction-oxidation) reaction, one species undergoes reduction while the other undergoes oxidation.
In the process of reduction, the species gains electrons, resulting in a decrease in its oxidation state. The reducing agent is the species that donates electrons, while the oxidizing agent is the species that accepts electrons.
Reduction reactions are important in many chemical and biological processes, including metabolism, photosynthesis, and corrosion. The study of redox reactions is important in understanding the behavior of chemicals in natural and industrial processes.
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You're given an unknown acid and told that it will donate one proton per molecule. When 1. 0 g of this acid is dissolved in water, the resulting solution requires 50. 0 ml of a 0. 25 M solution of NaOH for neutralization. What's the molecular mass of the unknown acid? Explain. (Hint: Find the moles of acid present)
The molecular mass of the unknown acid is 100 g/mol.
To find the molecular mass, first determine the moles of acid present. Since 50.0 mL of 0.25 M NaOH is required for neutralization, calculate the moles of NaOH using the formula: moles = Molarity × Volume (in L).
Moles of NaOH = 0.25 mol/L × (50.0 mL × 0.001 L/mL) = 0.0125 mol
Since the acid donates one proton per molecule, the moles of acid present equal the moles of NaOH: 0.0125 mol.
Next, find the mass of one mole of the unknown acid. You have 1.0 g of the acid, so divide the mass by the moles to get the molecular mass:
Molecular mass = Mass / Moles = 1.0 g / 0.0125 mol = 100 g/mol
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Choosy moms choose JIF! Your mom is making PB & J sandwiches for you and her. When she looks in the cupboard, she realizes she has 3 slices of bread, 1 jar of peanut butter, and 1/2 jar of jelly. What is the limiting reactant?
In this scenario, the limiting reactant is the ingredient that will run out first and limit the number of sandwiches that can be made.
Assuming that each sandwich requires two slices of bread, one serving of peanut butter, and one serving of jelly, we can see that we have enough bread and jelly to make a maximum of 1.5 sandwiches. However, since we only have one serving of peanut butter, we can only make one sandwich.
Therefore, the peanut butter is the limiting reactant. It is important to identify the limiting reactant in chemical reactions to determine the maximum amount of product that can be formed and to avoid wasting resources.
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How many joules are required to raise the temperature of 100.0 grams of water from -269 degrees celsius to 1500 degrees celsius
About 739,982.4 Joules energy is required to raise the temperature of 100.0 grams of water from -269 degrees Celsius to 1500 degrees Celsius.
The formula for the change in heat is,
Q = mcΔT, the amount of energy required is Q, m is the mass of water, specific heat capacity of water is c, the change in temperature is ΔT,
ΔT = 1500°C - (-269°C)
ΔT = 1769°C
Next, we can look up the specific heat capacity of water, which is 4.184 J/g°C. Then, we can substitute the values into the formula,
Q = 100.0 g * 4.184 J/g°C * 1769°C
Q = 739,982.4 J
Therefore, it would require 739,982.4 Joules of energy to raise the temperature of 100.0 grams of water from -269 degrees Celsius to 1500 degrees Celsius.
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A sample of 0. 0400 mol potassium hydroxide, KOH was dissolved in water to yield 20. 0 mL of solution. What is the molarity of the solution?
0. 4M
250M
2. 0M
2. 00x 10-3M
The molarity of the potassium hydroxide solution is 2.0 M.
We know that, Molarity (M) = moles of solute (mol) / volume of solution (L)
We have 0.0400 mol of KOH dissolved in 20.0 mL of water.
Volume of the solution= 20.0 mL = 20.0 / 1000 = 0.0200 L
Therefore, molarity = (0.0400 mol) / (0.0200 L) = 2.0 M
So, the molarity of the potassium hydroxide solution is 2.0 M.
Thus, option 3 is the correct answer.
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26. Over time an open soft drink will lose carbonation (dissolved CO2
Which of these allows the CO2 to remain in solution the longest?
O Reduced air pressure
Exposure to direct sunlight
Increased air currents
O Cooler temperatures
Out of the options provided, cooler temperatures allow CO2 to remain in solution the longest.
When a soft drink is opened, CO2 molecules start to escape from the solution and accumulate in the headspace of the container. Higher temperatures increase the kinetic energy of the CO2 molecules, causing them to move faster and escape the solution more quickly. In contrast, cooler temperatures reduce the kinetic energy of the CO2 molecules, slowing down their movement and allowing them to remain in solution for a longer period of time.
Reduced air pressure, exposure to direct sunlight, and increased air currents can all further accelerate the loss of carbonation in an open soft drink. Reduced air pressure lowers the atmospheric pressure, which decreases the solubility of CO2 in the liquid. Exposure to direct sunlight increases the temperature of the drink and causes the CO2 molecules to escape more quickly. Increased air currents can also cause CO2 to escape more quickly by creating turbulence in the liquid.
Therefore, to keep a soft drink fizzy for longer, it is best to store it in a cool place and avoid exposing it to direct sunlight, reducing air pressure or increased air currents.
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use a sheet of paper to answer the following question. take a picture of your answers and attach to this assignment. treatment of pentanedioic (glutaric) anhydride with ammonia at elevated temperature leads to a compound of molecular formula c5h7no2. what is the structure of this product? [hint: you need to think about the reactivity not only of acid anhydrides but also of amides and carboxylic acids]
The structure of product is shown.
When pentanedioic (glutaric) anhydride reacts with ammonia at high temperature, it undergoes an amide formation reaction to produce a compound with the molecular formula C₅H₇NO₂. The amide formation reaction involves the nucleophilic attack of the ammonia molecule on one of the carbonyl carbon atoms of the anhydride, leading to the formation of an intermediate product called an amide.
This amide then undergoes further reactions to form the final product with the given molecular formula. The presence of both carboxylic acid and amide functional groups in the molecule indicates that it contains both the original anhydride and the product of its reaction with ammonia.
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Three ions that contain the element phosphorus are phosphate (PO43–), hydrogen phosphate (HPO42–), and dihydrogen phosphate (H2PO4–). Compare the formulas of these three ions. Also notice any other instances in which hydrogen is added to a polyatomic ion from the table. Then complete the description of the pattern you see. Select the correct answer from each drop-down menu
The three ions containing phosphorus are phosphate (PO₄³⁻), hydrogen phosphate (HPO₄²⁻), and dihydrogen phosphate (H₂PO₄⁻).
The pattern observed is that adding hydrogen atoms successively reduces the negative charge of the ion by one unit.
1. Observe the formulas of the three ions: PO₄³⁻, HPO₄²⁻, and H₂PO₄⁻.
2. Notice that hydrogen atoms are added successively: 0, 1, and 2.
3. Observe the charges of the ions: -3, -2, and -1.
4. Recognize the pattern: adding hydrogen atoms reduces the negative charge by one unit.
In other instances where hydrogen is added to polyatomic ions, a similar pattern occurs. The negative charge decreases as more hydrogen atoms are added. This pattern is consistent across various polyatomic ions containing hydrogen.
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For #12 - #14, write the balanced chemical equation and identify each by type of reaction.
12. copper + chlorine → copper(II) chloride
13. calcium chlorate → calcium chloride + oxygen
14. lithium + water → lithium hydroxide + hydrogen
Answer:
12. Balanced chemical equation: Cu + Cl2 → CuCl2
Type of reaction: Combination or synthesis reaction
13. Balanced chemical equation: 2Ca(ClO3)2 → 2CaCl2 + 3O2
Type of reaction: Decomposition reaction
14. Balanced chemical equation: 2Li + 2H2O → 2LiOH + H2
Type of reaction: Single displacement or substitution reaction
Explanation:
Hope it helps^^
Draw the major organic product for each reaction. Assume a one to one ratio of starting material to reagent. H3Cl Cl --> AlCl3 ;
The major organic product for the reaction between hydrogen chloride (HCl) and aluminum chloride [tex](AlCl_3)[/tex] is aluminum chloride [tex](AlCl_3)[/tex].
When hydrogen chloride (HCl) reacts with aluminum chloride [tex](AlCl_3)[/tex], the reaction is exothermic and produces aluminum chloride [tex](AlCl_3)[/tex] as the major product. Hydrogen chloride (HCl) is a strong acid that dissociates fully in water, releasing chloride ions (Cl-). Aluminum chloride [tex](AlCl_3)[/tex] is a strong base that reacts with hydrogen chloride (HCl) to form aluminum hydroxide [tex](AlCl_3)[/tex] and hydrochloric acid (HCl). The ratio of starting material to reagent is one to one.
The balanced equation for the reaction between hydrogen chloride (HCl) and aluminum chloride [tex](AlCl_3)[/tex] is:
HCl + [tex](AlCl_3)[/tex] → [tex]Al(OH)_3[/tex] + HCl
Therefore, the major organic product for this reaction is aluminum hydroxide [tex]Al(OH)_3[/tex], which is an inorganic compound.
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what is the major organic product from the addition reaction of hbr to 2-methyl-2-butene? group of answer choices 2-bromopentane 2-bromo-2-methylbutane 1-bromo-2-methylbutane 1-bromo-3-methylbutane 2-bromo-3-methylbutane
The addition of HBr to 2-methyl-2-butene is an example of an electrophilic addition reaction. The correct answer is (2)
The double bond in 2-methyl-2-butene is attacked by the electrophilic H+ ion from HBr, leading to the formation of a carbocation intermediate. The bromide ion (Br-) then attacks the carbocation, leading to the formation of a new carbon-bromine bond. The major organic product obtained from the addition reaction of HBr to 2-methyl-2-butene is 2-bromo-2-methylbutane, which is also known as t-butyl bromide. This is because the addition of HBr occurs at the tertiary carbon, leading to the formation of a tertiary carbocation intermediate, which is relatively stable. Therefore, the correct answer is (2) 2-bromo-2-methylbutane.
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--The complete Question is, what is the major organic product from the addition reaction of hbr to 2-methyl-2-butene? group of answer choices
1. 2-bromopentane 2-bromo-2-methylbutane
2. 1-bromo-2-methylbutane
3. 1-bromo-3-methylbutane
4. 2-bromo-3-methylbutane=--
A 54.2 g sample of Magnesium has an initial temperature of 55°C and a final temperature of 78°C, and the specific heat of Magnesium is 1.023 J/g°C. If the sample absorbs 1300 J of heat energy, what is the change in temperature?
Change in temperature of the Magnesium sample is calculated as 23.7°C.
What is meant by heat energy?Heat energy is a form of energy that is transferred between objects or systems due to temperature difference. It flows from hotter to cooler objects, and its amount is measured in joules.
As we know; Q = m c ΔT
Q is the amount of heat absorbed, m is mass of the object, c is specific heat, and ΔT is change in temperature.
ΔT = Q / (m * c)
ΔT = 1300 J / (54.2 * 1.023 )
ΔT = 23.7°C
Therefore, the change in temperature of the Magnesium sample is 23.7°C.
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Blackworms were collected from an environment with an acidic pH, and the pulse rates were measured. Predict the outcome of the measurements. [2 pt] The pH of the nevironment would have no effect on pulse rate. The pulse rate would be increased to minimize the effects of acidosis. The pulse rate would be increased to minimize the effects of alkalosis. The pulse rate would be decreased to minimize the effects of acidosis
The pulse rate of blackworms collected from an environment with an acidic pH would be increased to minimize the effects of acidosis.
Acidosis occurs when there is an excess of acid in the body, which can lead to a decrease in blood pH. To compensate for this, the body increases pulse rate to improve blood circulation and oxygen delivery.
Blackworms are no exception to this mechanism and would experience an increase in pulse rate to counteract the acidic environment. It is important to note that the increase in pulse rate would not be enough to completely eliminate the effects of acidosis, but rather to minimize them.
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An equilibrium mixture at 425°C is found
to consist of 1.83 × 10-3 mol/L of H2,
3.13 × 10-3 mol/L of I2, and 1.77 × 10-2 mol/L
of HI. Calculate the equilibrium constant, K, for
the reaction H2(g) + I2(g) ⇄ 2HI(g).
The equilibrium constant, K, for the reaction H2(g) + I2(g) ⇄ 2HI(g) can be calculated using the expression K= [HI]2/([H2][I2]). Since the concentrations of H2, I2, and HI are given in the question, we can calculate the equilibrium constant, K, for the reaction.
K = [HI]2/([H2][I2]) = (1.77 × 10-2)2/((1.83 × 10-3)(3.13 × 10-3)) = 4.43 × 104. Therefore, the equilibrium constant, K, for the reaction H2(g) + I2(g) ⇄ 2HI(g) at 425°C is 4.43 × 104.
Using the specified concentrations of H2, I2, and HI, it appears that you have correctly calculated the equilibrium constant, K, for the reaction H2(g) + I2(g) 2HI(g) at 425°C. The ratio of the concentrations of the reactants and products at equilibrium, K, is represented by each concentration being raised to the power of its stoichiometric coefficient.
The concentration of the product, HI, is preferred above the concentrations of the reactants, H2 and I2, at equilibrium, as shown by the value of K = 4.43 104 in this instance. This suggests that at equilibrium, the forward reaction—the creation of HI—is preferred.
It is significant to remember that the equilibrium constant, K, is temperature-dependent, and that temperature changes affect K's value.
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Use the drop-down menus to rank the boiling points of the following hydrocarbons. Use a "1" to indicate the compound with the lowest boiling point.
The boiling points of the hydrocarbons can be ranked as follows;
1. 4
2. 2
3. 3
4. 1
What controls the boiling points of the hydrocarbons?The size of the molecules and the nature of the intermolecular interactions between the molecules essentially determine the boiling points of hydrocarbons.
Because they have more electrons and a larger surface area available for intermolecular interactions like Van der Waals forces, larger hydrocarbon molecules typically have higher boiling points.
Additionally, polar hydrocarbons and those that can form hydrogen bonds have higher boiling points than non-polar hydrocarbons because of stronger intermolecular forces.
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which nutrients are most responsible for eutrophication?
a. carbon and oxygen
b. carbon and phosphorus
c. nitrogen and oxygen
d. nitrogen and phosphorus
Answer:
C. Nitrogen and Phosphorous.
Explanation:
Classify each type bifunctional molecule as being a material used in the synthesis of polyesters, nylons, both, or neither.
dialcohol
diester
dinitro
diacid
diamine
diether
Dialcohol and diacid are bifunctional molecules that can be used in the synthesis of polyesters. In a polyester, the dialcohol and diacid react to form an ester bond, resulting in a long chain polymer.
Diesters can also be used in the synthesis of polyesters, as they can be hydrolyzed to form two carboxylic acid groups and two alcohol groups, which can then react to form an ester bond. Therefore, diesters are also a material used in the synthesis of polyesters.
Diamines and diacids can react to form nylon, so they are bifunctional molecules used in the synthesis of nylons. The reaction between a diamine and diacid forms an amide bond, which leads to a long-chain polymer.
Dinitro and diether are bifunctional molecules that are neither used in the synthesis of polyesters nor nylons. Dinitro compounds typically have nitro groups attached to each of the functional groups, making them more reactive and often used as explosives.
Diethers, on the other hand, can be used in organic synthesis as protecting groups for alcohols or carbonyl groups, but they do not have a direct role in the synthesis of polyesters or nylons.
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