Answer:
The equilibrium will change in the direction of the reactants
Explanation:
If the balloon described in Question 10.24 is released into the air and rises to an altitude of 10,000 ft where the atmospheric pressure is 523 torr and the temperature is 7.50ºC, then what is its new volume, in L?
This question relates to Boyle's Law from Physics, which deals with the inverse relationship of gas pressure and volume. However, without the initial volume and pressure of the balloon, we can't calculate the new volume.
Explanation:The subject matter pertains to the concept of Boyle's Law from Physics, specifically gas laws. It appears that this question is about a balloon that is released into the air and reaches an altitude where the atmospheric pressure and temperature are known. The student is required to calculate the new volume of the balloon. Boyle's Law states that the pressure and volume of a gas have an inverse relationship when temperature is held constant. In this case, as the balloon rises and atmospheric pressure drops, its volume should increase. However, we cannot proceed without necessary data such as the initial volume and pressure of the balloon.
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False: The highest oxidation state for hydrogen in this reaction is 0 False: The oxidation state of chromium in dichromate is 3. False: The iron half-reaction takes place in an acidic solution. True: The oxidation state of oxygen does not change. False: The oxidizing agent is Cr3 (aq) False: The reducing agent is Fe2 (aq)
The question addresses the oxidation states and redox reaction balancing in chemistry, involving chromium's oxidation state in dichromate and identification of the oxidizing agent.
Explanation:The question involves oxidation states and balancing redox reactions in chemistry. Regarding the oxidation state of hydrogen, it is generally +1 in compounds, not 0. For chromium in dichromate, the oxidation state is +6 (chromium(VI)), not +3. The iron half-reaction might indeed take place in acidic solution, depending on the specific reaction being discussed. The chromate half-reaction needs to have oxygen atoms balanced with water, and hydrogen atoms balanced with hydrogen ions, as indicated in the steps provided. The dichromate ion is a strong oxidizing agent, especially in acidic solution; Cr3+ does not act as the oxidizing agent in typical reactions with dichromate ions.
A mixture of krypton and hydrogen gases, at a total pressure of 692 mm Hg, contains 20.9 grams of krypton and 0.133 grams of hydrogen. What is the partial pressure of each gas in the mixture?
Answer:
Partial pressure Kr → 546.1 mmHg
Partial pressure H₂ → 145.9 mmHg
Explanation:
To determine the partial pressure of each gas in the mixture we apply the mole fraction concept.
Mole fraction → Moles of gas / Total gas = Partial pressure / Total pressure
In this mixture: Partial pressure Kr + partial pressure H₂ = 692 mmHg
We determine the moles of each:
20.9 g / 83.80 g/mol = 0.249 moles Kr
0.133 g / 2 g/mol = 0.0665 moles H₂
Total moles: 0.249 moles Kr + 0.0665 moles H₂ = 0.3155 moles
Mole fraction Kr → 0.249 mol / 0.3155 mol = 0.789
Partial pressure Kr → 0.789 . 692mmHg = 546.1 mmHg
Partial pressure H₂ → 692mmHg - 546.1 mmHg = 145.9 mmHg
NEED HELP ASAP!!! I WILL GIVE 100 POINTS!!!
Describe the investigations from the readings that are being conducted to learn more about how much water flows in major rivers around the world. Explain why the researchers think that data from the rivers is related to climate change. Cite facts from the texts in your response.
Need to make at least a 3 paragraph essay. 4 sentences in each paragraph pls.
Heres the article:
BOULDER, Colorado. A new study has revealed that the flow of water from Earth's largest rivers has declined over the past 50 years, due primarily to climate change.
A team of researchers at the National Center for Atmospheric Research (NCAR) in Boulder, Colorado, discovered the reduction in water flow after analyzing the discharge from 925 major rivers around the world between 1948 and 2004. Among Earth's dwindling rivers are Asia's Changjiang (or Yangtze), Mekong, Ganges, Irrawaddy, Amur, and Xijiang. Also included are Africa's Congo and Niger rivers and South America's Amazon River. North America's Mackenzie and Columbia rivers are also affected.
Researchers assessed the overall effect of the shrinking rivers on the various oceans into which they drain. They used cubic kilometers as their unit of measurement. The researchers found that the overall trend over the years was a decrease in the rates of discharge. Over the past half-century, they found, the yearly amount of fresh water flowing into the Indian Ocean fell by about 3 percent, or 140 cubic kilometers. The annual river discharge into the Pacific Ocean dropped about 6 percent, or 526 cubic kilometers. Researchers noted that in terms of quantity, the decline in streamflow into the Pacific is equal to shutting off the Mississippi River.
Meanwhile, researchers discovered only a negligible variation in the annual river flow into the Atlantic Ocean. There, decreases in the Amazon River were offset by increases in South America's Parana River and our own Mississippi River. Due to increased rainfall, the Mississippi saw an increase in its flow of more than 22 percent over the period studied.
Aiguo Dai was the study's lead researcher. According to Dai, the only ocean showing a significant increase in the freshwater flow it received was the Arctic Ocean. There, annual river streamflow rose about 10 percent, or 460 cubic kilometers.
According to Dai, the change in river flows is primarily a result of climate change. Earth's warming temperatures have caused flooding in some regions. Other areas have experienced reduced precipitation and droughts. Droughts place even greater demands on the already shrinking rivers. They increase the need for dams and other diversions of river water for agriculture and industry. For example, decreased precipitation and increased water diversion caused a 14 percent reduction in the flow in the northwestern U.S.'s Columbia River.
"Freshwater resources will likely decline in the coming decades...due to climate changes," Dai said.
In the Arctic, climate change is responsible for accelerating the melting of snow and ice. This accounts for the increased discharge into the Arctic Ocean.
"[Rapidly] disappearing mountain glaciers in [higher latitudes] will make matters worse," said Dai.
In some regions, the effects of climate change on Earth's rivers are compounded by regularly occurring weather phenomena. Take the El Niño phenomenon, for example. An El Niño is a periodic cooling of sea surface waters in the tropical Pacific. Researchers noted that the El Niño phenomenon led to lower flows in the Amazon River and higher ones in the Mississippi River when it was in effect.
Researchers are concerned about the reduction in rivers, particularly the Yellow River in northern China, the Ganges in India, the Niger in West Africa, and the Colorado in the southwestern U.S. These rivers flow through densely populated areas, where water is in greater demand than in areas with smaller populations.
"These changes will have impacts on humans and ecosystems," said Margaret A. Palmer. Palmer is director of the Chesapeake Biological Laboratory of the University of Maryland Center for Environmental Science. "Many of these regions have large populations and drought-stressed ecosystems."
There are other reasons to be concerned about the diminished streamflow. One is that rivers supply oceans with essential nutrients. Another is that changes in streamflows can induce changes in ocean temperatures and salt levels. These changes are known to affect global ocean circulation patterns.
NCAR scientist Kevin Trenberth is concerned about climate change and its effects on rivers—and ultimately humans. "As climate change inevitably continues in coming decades," Trenberth said, "we are likely to see greater impacts on many rivers and water resources that society has come to rely on."
Answer:
In the first paragraph state how climate change is the earth warming. Tell them how the warmer atmosphere has lead to the melting of snow and ice in the arctic and sever droughts in other areas of the world.
In the second paragraph talk about how large rivers are becoming smaller in regards to water volume and that the amount of water discharged into the ocean has also decreased greatly.
In the final paragraph write about how weather phenomena impact global warming. State that El Niño has led to lower floods in some place whereas other areas have experienced an increase in floods. Finally tell the reader how the lower water volumes and discharge leads to a change in the global ocean circulation.
Explanation:
N2 +3H2 → 2NH3 What volume of ammonia at STP is produced if 30.0 g of nitrogen gas is reacted with an excess of hydrogen gas?
Answer:
V= 48L
Explanation:
Moles of N2 = m/M = 30/28= 1.07moles
From the equation of reaction
N2 +3H2 → 2NH3
1mole of N2 produces 2mole of NH3
Hence
1.07moles will produce 1.07×2= 2.14 moles
At STP, 1mole occupy 22.4L
Hence volume of N2 produced = 2.14×22.4= 48 L
A chemical processing plant has a network of pipes that are used to transfer liquid chemicalproducts from one part of the plant to another. The numbers next to the arcs in the network indicatethe maximum amount of chemicals that can flow through the various pipelines (measured in barrelsper minute). Formulate a LP model that will maximize flow to location 7 from location
Answer:
Explanationis the long-distance transportation of a liquid or gas through a system of pipes—a pipeline—typically to a market area for consumption. The latest data from 2014 gives a total of slightly less than 2,175,000 miles (3,500,000 km) of pipeline in 120 countries of the world.[1] The United States had 65%, Russia had 8%, and Canada had 3%, thus 75% of all pipeline were in these three countries.[1]
Pipeline and Gas Journal's worldwide survey figures indicate that 118,623 miles (190,905 km) of pipelines are planned and under construction. Of these, 88,976 miles (143,193 km) represent projects in the planning and design phase; 29,647 miles (47,712 km) reflect pipelines in various stages of construction. Liquids and gases are transported in pipelines and any chemically stable substance can be sent through a pipeline.[2] Pipelines exist for the transport of crude and refined petroleum, fuels – such as oil, natural gas and biofuels – and other fluids including sewage, slurry, water, beer, hot water or steam for shorter distances. Pipelines are useful for transporting water for drinking or irrigation over long distances when it needs to move over hills, or where canals or channels are poor choices due to considerations of evaporation, pollution, or environmental impact.:
Select the following processes that represent a decrease in entropy: Group of answer choices Water is heated in the microwave from 23 oC to 90 oC. 2 NO2(g) → N2O4(g) The sublimation of I2 CO2 gas is dissolved in water to make a carbonated beverage
Answer:
The answers that represent a decrease in entropy are:
-CO2 gas is dissolved in water to make a carbonated beverage
-NO2(g) → N2O4(g)
Explanation:
Since in carbonated drinks, carbon dioxide is in its liquid form, since a pressure is applied to said gas, in this way, when carbon dioxide dissolves in a liquid, its entropy decreases, thus producing a negative change in entropy (the entropy of a gas is higher than the entropy of a liquid).
The number of gaseous reagents is equal to 1-2 = -1. Since it is negative, therefore, entropy is also negative.
A variation of the acetamidomalonate synthesis can be used to synthesize serine. The process involves the following steps: Ethoxide ion deprotonates diethyl acetamidomalonate, forming enolate anion 1; Enolate anion 1 makes a nucleophilic attack on formaldehyde, forming tetrahedral intermediate 2; Protonation of the oxyanion forms alcohol 3; Acid hydrolysis yields dicarboxyamino alcohol 4; Decarboxylation leads to the final amino acid. Write out the mechanism on a separate sheet of paper, and then draw the structure of dicarboxyamino alcohol 4.
Answer:
See detailed mechanism in the image attached
Explanation:
The mechanism shown in detail below is the synthesis of serine in steps.
The first step is the attack of the ethoxide ion base on the diethyl acetamidomalonate substrate giving the enolate and formaldehyde.
The second step is the protonation of the oxyanion from (1) above to form an alcohol as shown.
Acid hydrolysis of the alcohol formed in (3) above yields a tetrahedral intermediate, a dicarboxyamino alcohol.
Decarboxylation of this dicarboxyamino alcohol yields serine, the final product as shown in the image attached.
1) Draw the structure of only the Zaitsev product that would be obtained from the dehydration of 1-methylcyclohexanol. 2) Write out a detailed mechanism of the acid catalyzed dehydration of cyclohexanol. Use only hydronium as your acid source. 3) With respect to GC, what effect
Question:
Question 3 is not complete. here is the complete question;
3. With respect to GC, what effect would raising the column temperature have on the retention time?
Answer:
See the attached file for the structures
Explanation:
Find attached of question 1 and 2.
3. Raising the column temperature enhances the internal energy of the analyse. Therefore, it will escape from the liquid phase faster and predominantly be present in the vapour phase. On increasing temperature, retention time will be DECREASED, because analyte would not be able to interact with stationary phase efficiently.
The Zaitsev product from the dehydration of 1-methylcyclohexanol would have the double bond between the carbon connected to the OH group and the adjacent more substituted carbon. The mechanism of acid-catalyzed dehydration of cyclohexanol involves protonation, departure of a water molecule forming a carbocation, and subsequent loss of a proton which forms a pi bond. Changes in temperature and pressure in Gas Chromatography (GC) can impact analysis outcomes.
Explanation:The basis of your question involves the Zaitsev product from the dehydration of 1-methylcyclohexanol, the mechanism of the acid-catalyzed dehydration of cyclohexanol using hydronium, and effects in relation to Gas Chromatography (GC).
1) When dehydrating 1-methylcyclohexanol, the Zaitsev rule tells us that the most substituted alkene will be the major product. Because you are dehydrating 1-methylcyclohexanol, the Zaitsev product will have the double bond between the carbon connected to the OH group and the adjacent carbon which is more substituted.
2) The second part of the question relates to the acid-catalyzed dehydration of cyclohexanol.
Firstly, in the presence of acid, the hydroxyl group on cyclohexanol is protonated forming a good leaving group. The water molecule leaves creating a carbocation. A proton is then lost resulting in the pi bond of the cyclohexene. Water acts as a base to receive this proton.
3) With respect to Gas Chromatography (GC), changes in temperature and pressure can affect the results. Higher temperatures can increase resolution but at the risk of over-separation.
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When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+CO2(g) How many grams of calcium chloride will be produced when 25.0 g of calcium carbonate is combined with 14.0 g of hydrochloric acid? mass of CaCl2 :
Answer:
Mass of CaCl₂ produced from the reaction = 21.31 g
Explanation:
The balanced equation for the reaction
CaCO₃(s) + 2HCl(aq) ⟶ CaCl₂(aq) + H₂O(l) + CO₂(g)
How many grams of calcium chloride will be produced when 25.0 g of calcium carbonate is combined with 14.0 g of hydrochloric acid
First of, we need to recognize the limiting reagent in this reaction.
The limiting reagent is the reagent that is totally used up in the chemical reaction; it determines the amount of other reactants that react and the amount of products that will be formed.
To know the limiting reagent, we convert the masses of the reactants given to number of moles.
Number of moles = (mass)/(molar mass)
25.0 g of calcium carbonate
Molar mass of CaCO₃ = 100.0869 g/mol
Number of moles of CaCO₃ present at the start of the reaction = (25/100.0869) = 0.250 moles
14.0 g of hydrochloric acid
Molar mass of HCl = 36.46 g/mol
Number of moles of HCl present at the start of the reaction = (14/36.45) = 0.384 moles
But from the stoichiometric balance of the reaction,
1 mole of CaCO₃ reacts with 2 moles of HCl
If CaCO₃ was the limiting reagent,
0.25 moles of CaCO₃ at the start of the reaction would require 0.50 moles of HCl; which is more than the available number of moles of HCl available at the start of the reaction (0.384 moles)
So, CaCO₃ isn't the limiting reagent.
If HCl is the limiting reagent,
0.384 moles of HCl would require 0.192 moles of CaCO₃ to react with. This is within the limit of CaCO₃ present at the start of the reaction. (0.250 moles)
Hence, HCl is the limiting reagent, it is the reactant that is used up in the reaction and determines the amount of products formed.
Again, from the stoichiometric balance of the reaction,
2 moles of HCl gives 1 mole of CaCl₂
0.384 moles of HCl will give (0.384/2) moles of CaCl₂; that is, 0.192 moles of CaCl₂
We then convert this number of moles to mass.
Mass = (number of moles) × (molar mass)
Molar mass of CaCl₂ = 110.98 g/mol
Mass of CaCl₂ produced by the reaction = 0.192 × 110.98 = 21.30816 g = 21.31 g
Hope this Helps!!!
Answer: 21.09g of Calcium chloride is produced
Explanation: Please see the attachments below
Given the equation representing a system at equilibrium:
PCl5(g) + energy ⇌ PCl3(g) + Cl2(g)
Which change will cause the equilibrium to shift to the right?
(1) adding a catalyst (2) adding more PCl3(g)
(3) increasing the pressure (4) increasing the temperature
Answer:
increasing the temperature
Explanation:
Now look carefully at the reaction equation, notice the inclusion on an energy term on the left hand side;
PCl5(g) + energy ⇌ PCl3(g) + Cl2(g)
The inclusion of an energy term means that the reaction is endothermic. Energy is absorbed as the reaction goes from left to right.
Since energy is absorbed, increasing the temperature (supplying energy in the form of heat) will favour the forward reaction over the reverse reaction in accordance with Le Chatelier's principle. Hence the answer.
Answer:
Option (4) increasing the temperature
Explanation:
PCl5(g) + energy ⇌ PCl3(g) + Cl2(g)
The reaction above clearly indicates endothermic reaction since heat is required for the reaction to proceed to product.
As the heat is supplied, the temperature of reaction increases thereby making the reaction to proceed forward at a much faster rate and hence the equilibrium position will shift to the right. This is in accordance with Le Chatelier's principle will explained that for an endothermic reaction, an increase in temperature will cause the equilibrium position to shift to the right.
Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory as a primary standard. It has the unwieldy formula of KHC8H4O4. This is often written in shorthand notation as KHP. If 34.10 mL of a barium hydroxide solution are needed to neutralize 2.050 grams of KHP, what is the concentration (mol/L) of the barium hydroxide solution
Answer:
0.1472 mol/L is the concentration of the barium hydroxide solution.
Explanation:
[tex]2KHC_8H_4O_4+Ba(OH)_2\rightarrow Ba(KC_8H_4O_4)_2+2H_2O[/tex]
Mass of potassium hydrogen phthalate = 2.050 g
Molar mass of potassium hydrogen phthalate = [tex]\frac{2.050 g}{204.2 g/mol}=0.01004 mol[/tex]
According to reaction , 2 moles of potassium hydrogen phthalate reacts with 1 mole of barium hydroxide, then 0.01004 moles of potassium hydrogen phthalate will :
[tex]\frac{1}{2}\times 0.01004 mol=0.005020 mol[/tex] of barium hydroxide
Moles of barium hydroxide = 0.005020 mol
Volume of the barium hydroxide solution = 34.10 mL = 0.03410 L
1 mL = 0.001 L
[tex]Molarity=\frac{Moles}{Volume(L)}[/tex]
Molarity of the barium hydroxiude silution :
[tex]=\frac{0.005020 mol}{0.03410 L}=0.1472 mol/L[/tex]
0.1472 mol/L is the concentration of the barium hydroxide solution.
A 27g pancake (61 Calories) is burned in a calorimeter that contains 2.500 kg of water. How much does the temperature of the water increase if all of the energy from the pancake is transferred into the water?
If all of the energy from the pancake is transferred into the water, the temperature of the water will increase by 0.02 °C
Data obtained from the questionFrom the question given above, the following data were untainted:
Mass of pancake = 27 gEnergy of package released (Q) = 61 caloriesMass of water (M) = 2.5 Kg = 2.5 × 1000 = 250 g Specific heat capacity of water (C) = 1 cal/gºC Change in temperature of water (ΔT) =? How to determine the increase in temperatureThe increase in temperature can be obtained as follow:
Q = MCΔT
Divide both side by MC
ΔT = Q / MC
ΔT = 61 / (2500 × 1)
ΔT = 61 / 2500
ΔT = 0.02 °C
Thus, the temperature of the water will increase by 0.02 °C
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For each reaction, calculate how many moles of the barium product you will produce using stoichiometry and the balanced reactions.since there are two reactants, calculate the moles of product using each reactant, and then use the number of moles which is less) this comes from the limiting reactant).
Moles of Ba(NH2SO3)2.
Answer:
Explanation:
given that
mass of Ba(NO3)2 = 1.40g
mass of NH2SO3H = 2.50 g
1)to determine the mole of Ba(NO3)2
2) to determine the mass of all three product formed in the reaction
reaction
Ba(NO3)2 + 2NH2SO3H → Ba(NH2SO3)2 + 2HNO3
Solution
we calculate the molar mass of each species by using their atomic masses
BA = 137.33g/mol
N = 14g/mol
O= 16g/mol
H = 1g/mol
S = 32g/mol
calculation
Ba(NO3)2 = Ba + 2N + 6O
= 137.33 + 2X 14 + 6 X 16
= 261.33g/mol
NH2SO3H = N + 3H + S+ 3O
=14 + 3X1 + 32 + 3X 16
= 97g/mol
Ba(NH2SO3)2 = Ba + 2N + 4H +2S +6O
= 137.33 + 2 X 14 + 4 X1 + 2X32 + 6 X 16
= 329.33g/mol
HNO3 = H + n + 3O
= 1 + 14 + 3 X 16
= 63g/mol
Calculate the number of moles of sugar in 800 ml of a 2.6 M solution
Answer:
2.08moles
Explanation:
Given parameters:
Volume of solution = 800mL = 0.8L
Molarity = 2.6M
Unknown:
Number of moles of sugar in the solution = ?
Solution:
We need to understand the relationship between number of moles, volume and molarity of a solution.
The molarity of a solution is the number of moles of solute of a substance dissolved in a given solvent.
Molarity = [tex]\frac{number of moles }{volume}[/tex]
Number of moles of solute = molarity x volume of solution
Now input the parameters and solve;
Number of mole of solute = 2.6 x 0.8 = 2.08moles
The number of moles of sugar is 2.08moles
Consider the following half-reactions: Half-reaction E° (V) I2(s) + 2e- 2I-(aq) 0.535V 2H+(aq) + 2e- H2(g) 0.000V Zn2+(aq) + 2e- Zn(s) -0.763V (1) The weakest oxidizing agent is: Zn2+(aq) enter formula (2) The strongest reducing agent is: Zn(s) (3) The strongest oxidizing agent is: I2(s) (4) The weakest reducing agent is: I-(aq) (5) Will I-(aq) reduce Zn2+(aq) to Zn(s)? (6) Which species can be reduced by H2(g)? If none, leave box blank.
Answer:
The weakest oxidizing agent is Zn^2+(aq)
The strongest reducing agent is Zn(s)
The strongest oxidizing agent is I2(s)
The weakest reducing agent is I^-(aq)
I^- cannot reduce Zn^2+ to Zn(s)
I2(s) can be reduced by hydrogen gas
Explanation:
In looking at oxidizing and reducing agents, our primary guide is the reduction potentials of each specie. The more negative the reduction potential of a specie, the better its function as a reducing agent. Zn has a very negative reduction potential hence it a very good reducing agent. Similarly, iodine has a very positive reduction potential hence it is a good oxidizing agent.
Only a specie having a more negative reduction potential than zinc can reduce it in aqueous solution. Similarly, the reaction potential of hydrogen is less than that of iodine hence hydrogen gas can reduce iodine.
Final answer:
In electrochemistry, the strongest reducing agent is Zn(s) and the weakest oxidizing agent is Zn2+(aq). I2(s) is the strongest oxidizing agent, while I-(aq) is the weakest reducing agent. I-(aq) cannot reduce Zn2+(aq) to Zn(s), and none of the species provided can be reduced by H2(g).
Explanation:
To determine the oxidizing and reducing agents, we refer to their standard electrode potentials (E°). A strong oxidizing agent has a higher positive E° value, indicating a greater tendency to gain electrons and be reduced. Conversely, a strong reducing agent has a lower (more negative) E° value, reflecting a higher tendency to lose electrons and be oxidized.
The weakest oxidizing agent is Zn²+(aq) because it has the most negative E° value (-0.763V), meaning it is the least likely to gain electrons. The strongest reducing agent is Zn(s) because zinc in its solid state is more willing to be oxidized (lose electrons) as indicated by its half-reaction (Zn(s) → Zn²+(aq) + 2e⁻).
Similarly, the strongest oxidizing agent is I₂(s) due to its higher E° value (0.535V), which demonstrates its greater ability to take up electrons. The weakest reducing agent in this set is I⁻(aq) because it is derived from I₂(s), which is a strong oxidizer, thus its conjugate, I⁻, would be the weakest reducer.
Regarding whether I⁻(aq) can reduce Zn²+(aq) to Zn(s), the answer is no, because I⁻(aq) is the weakest reducing agent and Zn²+(aq) is not a strong oxidizer. Lastly, H₂(g) has an E° value of 0.000V, making it neutral in this context, and therefore it will not reduce any of the species provided in the question.
You are given mixture made of 290 grams of water and 14.2 grams of salt. Determine the % by mass of salt in the salt solution.
Answer:
Solution is 4.67% by mass of salt
Explanation:
% by mass is the concentration that defines the mass of solute in 100g of solution.
In this case we have to find out the mass of solution with the data given:
Mass of solution = Mass of solute + Mass of solvent
Solute: Salt → 14.2 g
Solvent: Water → 290 g
Solution's mass = 14.2 g + 290g = 304.2 g
% by mass = (mass of solute / mass of solution) . 100
(14.2 g / 304.2g) . 100 = 4.67 %
(h) The student made observations related to the contents of the Erlenmeyer flask during the titration. Identify an observation that could have led the student to conclude that a chemical change took place during the titration.
Answer:
Explanation:
During titration indicators are often used to identify chemical changes between reacting species.
For colorless solutions in which no noticeable changes can easily be seen, indicators are the best bet. Most titration processes involves a combination of acids and bases to an end point.
Indicators are substances whose color changes to signal the end of an acid-base reaction. Examples are methyl orange, methyl red, phenolphthalein, litmus, cresol red, cresol green, alizarin R3, bromothymol blue and congo red.
Most of these indicators have various colors when chemical changes occur.
Also, there are heat changes that accompanies most of these reactions. These are also indicators of chemical changes.
Final answer:
An observation indicative of a chemical change during titration could include temperature change, light emission, unexpected color change, or the formation of bubbles signaling the production of gas, any of which suggest that a new substance has been formed.
Explanation:
An observation that could have led the student to conclude that a chemical change took place during the titration includes the following:A temperature change such as the solution in the Erlenmeyer flask becoming warmer or cooler, indicating an exothermic or endothermic reaction.The emission of light which could suggest an energy release during a chemical reaction.An unexpected color change indicating the formation of a new substance.The formation of bubbles not caused by boiling, suggesting the production of a gas during the reaction.For instance, if the student added an indicator to the acid and after titrating with a base, noticed a color change, this would be evidence of a chemical change. Similarly, if the student observed the solution fizzing but it was not reaching boiling temperature, it could indicate the formation of a gaseous product.
A 50.0 mL sample containing Cd^2+ and Mn^2+ was treated with 42.8 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA required 17.8 mL of 0.0250 M Ca^2+. The Cd^2+ was displaced from EDTA by the addition of an excess of CN Titration of the newly freed EDTA required 27.1 mL of 0.0250 M Ca^2+. What are the concentrations of Cd^2+ and Mn^2+ in the original solution?
Answer:
See explaination
Explanation:
EDTA is a chemical that binds and holds on to (chelates) minerals and metals such as chromium, iron, lead, mercury, copper, aluminum, nickel, zinc, calcium, cobalt, manganese, and magnesium. When they are bound, they can't have any effects on the body and they are removed from the body.
Check attachment for the step by step solution of the given problem.
when the iron(III) oxide reacts with hydrochloric acid, iron (III) chloride and water are formed. How many gras of iron (III) chloride are formed from 10.0g of iron (III) oxide
Answer:
10.1 g of FeCl₃ are formed by the reaction
Explanation:
First step is to determine the reaction where the reactants are Fe₂O₃ and HCl in order to produce FeCl₃ and H₂O.
Equation is: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O
We assume the acid is in excess, so the limiting reagent will be the oxide.
Let's work with mass:
1 mol of Fe₂O₃ is 159.7 g
2 mol of FeCl₃ is 162.2 g
So now we propose a rule of three:
159.7 g of oxide can produce 162.2 grams of chloride
Then, 10 g of oxide will produce (10 . 162.2) / 159.7 = 10.1 g of FeCl₃
Answer:
20.3 grams of FeCl3 will be formed
Explanation:
Step 1: Data given
iron(III) oxide = Fe2O3
hydrochloric acid = HCl
iron (III) chloride = FeCl3
water = H2O
Mass of Fe2O3 = 10.0 grams
Molar mass Fe2O3 = 159.69 g/mol
Step 2: The balanced equation
Fe2O3 +6HCl → 2FeCl3 + 3H2O
Step 3: Calculate moles Fe2O3
Moles Fe2O3 = mass Fe2O3 / molar mass Fe2O3
Moles Fe2O3 = 10.0 grams/ 159.69 g/mol
Moles Fe2O3 = 0.0626 moles
Step 4: Calculate moles of FeCl3
For 1 mol Fe2O3 we need 6 moles HCl to produce 2 moles FeCl3 and 3 moles H2O
For 0.0626 moles Fe2O3 we'll have 2*0.0626 = 0.1252 moles FeCl3
Step 5: Calculate mass FeCl3
Mass FeCl3 = moles FeCl3 * molar mass FeCl3
Mass FeCl3 = 0.1252 moles * 162.2 g/mol
Mass FeCl3 = 20.3 grams
20.3 grams of FeCl3 will be formed
When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of potassium bromide are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for potassium bromide in . Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.
Answer:
The factor is 2
Explanation:
Van't Hoff factor is defined as the ratio between the species of a solute before the addition to the solvent and particles produced when the substance is dissolved. It is used, principally, in colligative properties.
Before solution, potassium bromide, KBr, has just one specie, that is, KBr. When KBr is dissolved (As a salt):
KBr(aq) → K⁺(aq) + Br⁻(aq)
There are produced two species, K⁺ and Br⁻. By definition of Van't Hoff factor, for this salt, the factor is 2.
Write a balanced half-reaction for the reduction of dichromate ion Cr2O−27 to chromium ion Cr+3 in acidic aqueous solution. Be sure to add physical state symbols where appropriate.
Answer:
14 H⁺(aq) + Cr₂O₇²⁻(aq) + 6 e⁻ → 2 Cr⁺³ + 7 H₂O(l)
Explanation:
Let's consider the half-reaction for the reduction of dichromate ion Cr₂O₇²⁻ to chromium ion Cr⁺³ in acidic aqueous solution.
Cr₂O₇²⁻(aq) → Cr⁺³
Step 1: Perform the mass balance adding H₂O(l) and H⁺(aq) where appropriate.
14 H⁺(aq) + Cr₂O₇²⁻(aq) → 2 Cr⁺³ + 7 H₂O(l)
Step 2: Perform the charge balance adding electrons where appropriate.
14 H⁺(aq) + Cr₂O₇²⁻(aq) + 6 e⁻ → 2 Cr⁺³ + 7 H₂O(l)
A mixture of He
, N2
, and Ar
has a pressure of 13.6
atm at 28.0
°C. If the partial pressure of He
is 1831
torr and that of Ar
is 997
mm Hg, what is the partial pressure of N2
?
A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spherical air tank that measures 74.0 wide. The biologist estimates she will need of air for the dive. Calculate the pressure to which this volume of air must be compressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to significant digits.
Complete Question
A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spherical air tank that measures 74.0 wide. The biologist estimates she will need 2600 L of air for the dive. Calculate the pressure to which this volume of air must be compressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to significant digits.
Answer:
The pressure required is [tex]P_2= 12.2 \ atm[/tex]
Explanation:
Generally the volume of a sphere is mathematically denoted as
[tex]V_s = \frac{4}{3} * \pi r^3[/tex]
Substituting [tex]r = \frac{d}{2} = \frac{74}{2} = 37cm[/tex]
[tex]V_s = \frac{4}{3} * 3.42 * (37)^2[/tex]
[tex]V_s = 2.121746 *10^5 cm^3[/tex]
Converting to Liters
[tex]V_s = \frac{2.121746 *10^5}{1000}[/tex]
[tex]V_s= 212.1746L[/tex]
Assume that the pressure at which the air is given to the diver is 1 atm when the air was occupying a volume of 2600L
So
From Charles law
[tex]P_1V_1 = P_2 V_s[/tex]
Substituting [tex]V_1 =2600 L[/tex] , [tex]P_1 = 1 atm[/tex] , [tex]V_s =212.1746L[/tex] , and making [tex]P_2[/tex] the subject we have
[tex]P_2 = \frac{P_1 * V_1}{V_s}[/tex]
[tex]= \frac{1 * 2600}{212.1746}[/tex]
[tex]P_2= 12.2 atm[/tex]
Draw one product structure for the following Diels–Alder reaction. For ONLY the chirality centers with D and C(=O)H groups, specify the stereochemistry via wedge-and-dash bonds. (D is deuterium, an isotope of hydrogen. Include it in your drawing by either double clicking on an atom and typing \"d\" or by choosing D in the bottom row of the atoms menu.)
Answer:
See explaination
Explanation:
We can define the Diels-Alder reaction as a conjugate addition reaction of a conjugated diene to an alkene (the dienophile) to produce a cyclohexene.
Please kindly check attachment for the step by step solution of the given problem.
The Diels–Alder reaction is an organic chemistry reaction involving a cyclic rearrangement of atoms to form a six-membered ring. The product structure from such a reaction should specify stereochemistry with wedge-and-dash notation for the chirality centers involving carbon atoms and Deuterium (D). A wedge implies the bond is positioned towards you, while a dash implies the bond is positioned away from you.
This is a Chemistry problem dealing with the Diels–Alder reaction, a fundamental reaction in organic chemistry, which is a very key method in the synthesis of complex molecules. In essence, the Diels–Alder reaction involves a cyclic rearrangement of atoms rearranging to form a six-membered ring. Let's take a Diels–Alder reaction template where:
two carbon atoms involved in the double bond are labeled as X and Y,
the carbonyl carbon is labeled as Z,
the specific hydrogen atom replaced by Deuterium (D) is labeled as W.
Thus, when a Diels–Alder reaction is performed, the product structure will maintain relative positions. Carbon atoms X and Z will have wedge orientation while Carbon atoms Y and the Deuterium (D) on W will have dash orientation.
Remember, in wedge-and-dash notation, wedge implies the bond is coming up out of the plane (positioned towards you), while dash implies the bond is going back into the plane (positioned away from you).
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What volume of ammonia gas, measured at 547.9 mmHg and 27.6oC, is required to produce 8.98 g of ammonium sulfate according to the following balanced chemical equation? 2NH3(g) + H2SO4(aq) → (NH4)2SO4(s)
Answer:
4.65 L of NH₃ is required for the reaction
Explanation:
2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(s)
We determine the ammonium sulfate's moles that have been formed.
8.98 g . 1mol / 132.06 g = 0.068 moles
Now, we propose this rule of three:
1 mol of ammonium sulfate can be produced by 2 moles of ammonia
Therefore, 0.068 moles of salt were produced by (0.068 . 29) / 1 = 0.136 moles of NH₃. We apply the Ideal Gases Law, to determine the volume.
Firstly we do unit's conversions:
27.6°C +273 = 300.6 K
547.9 mmHg . 1 atm / 760 mmHg = 0.721 atm
V = ( n . R . T ) / P → (0.136 mol . 0.082 L.atm/mol.K . 300.6K) / 0.721 atm
V = 4.65 L
Answer:
4.66 L of ammonia gas will be produced
Explanation:
Step 1: Data given
The pressure of ammonia gas = 547.9 mmHg = 0.72092116 atm
Temperature = 27.6 °C = 300.75 K
Mass of ammonium sulfate produced = 8.98 gramms
Molar mass of ammonium sulfate = 132.14 g/mol
Step 2: The balanced equation
2NH3(g) + H2SO4(aq) → (NH4)2SO4(s)
Step 3: Calculate moles (NH4)2SO4
Moles (NH4)2SO4 = mass (NH4)2SO4 / molar mass
Moles (NH4)2SO4 = 8.98 grams / 132.14 g/mol
Moles (NH4)2SO4 = 0.0680 moles
Step 4: Calculate moles NH3
For 1 mol (NH4)2SO4 we need 2 moles NH3
For 0.0680 moles (NH4)2SO4 we need 2*0.0680 = 0.136 moles NH3
Step 5: Calculate volume NH4
p*V=n*R*T
V = (n*R*T)/p
⇒with V = the volume of NH3 = TO BE DETERMINED
⇒with n = the number of moles NH3 = 0.136 moles NH3
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 300.75 K
⇒with p = the pressure of the gas = 0.72092116 atm
V = (0.136 * 0.08206 * 300.75) / 0.72092116
V = 4.66 L
4.66 L of ammonia gas will be produced
Balance the following unbalanced redox reaction (assume acidic solution if necessary): Cr2O72- + Cl- → Cl2 + Cr3+ Indicate the coefficient that will be used for Cl2 (g) in this reaction
Answer:
14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O
The coefficient that will be used for Cl₂ in this reaction is 3
Explanation:
We use the method of electron-ion to the balance.
We assume that the redox reaction is happening at acidic medium.
Cr₂O₇²⁻ + Cl⁻ → Cl₂ + Cr³⁺
Chloride is raising the oxidation state from -1 in the chloride, to 0 in the chloride dyatomic. This is the half reaction of oxidation
2Cl⁻ → Cl₂ + 2e⁻ Oxidation
In the dichromate anion, chromium acts with +6 in oxidation state, and we have 2 Cr, so the global charge of the element is +12. To change to Cr³⁺ it has release 3 electrons, but we have 2 Cr, so it finally released 6 e-. The oxidation state was decreased, so this is the reduction half reaction.
14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O Reduction
As we have 7 O in the product side, we add 7 water, to the opposite place. In order to balance the H (protons) we, add the amount of them, in the opposite side, again.
(2Cl⁻ → Cl₂ + 2e⁻) ₓ3
(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) ₓ1
We multiply the half reactions, in order to remove the electrons and we sum, the equations:
14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 6Cl⁻ → 3Cl₂ + 6e⁻ + 2Cr³⁺ + 7H₂O
Now, that we have the same amount of electrons, they can cancelled, so the balanced redox reaction is:
14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O
To balance the given redox reaction, follow the steps: assign oxidation states, balance the elements undergoing oxidation or reduction, balance the elements not involved in redox, balance the charge, balance oxygen atoms, combine H2O molecules, and finally, balance the equation. The coefficient for Cl2 (g) is 3.
Explanation:To balance the redox reaction Cr2O72- + Cl- → Cl2 + Cr3+ in an acidic solution, we need to follow these steps:
Assign oxidation states to each element in the reaction. Chromium (Cr) in Cr2O72- is assigned an oxidation state of +6 on the left side and +3 on the right side. Chlorine (Cl) in Cl- has an oxidation state of -1 on the left side and 0 in Cl2 on the right side.Balance the elements that are undergoing oxidation or reduction first. In this case, Cr is being reduced from +6 to +3, so we need to add 3 electrons to the left side of the equation.Next, balance the elements that are not involved in redox reactions. In this case, Cl- stays the same on both sides.Finally, balance the charge by adding H+ ions to the side that needs it. In this case, we need to add 14 H+ ions to the left side.Balancing the oxygen atoms by adding H2O molecules. There are 7 oxygen atoms on the left side, so we add 7 H2O molecules to the right side.Combine the H2O molecules on the right side to form OH- ions, then cancel out the same ions present on both sides.Now, the equation is balanced in terms of atoms and charges. The coefficient for Cl2 (g) is 3.Learn more about Balancing redox reaction here:https://brainly.com/question/32431557
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Suppose the surface-catalyzed hydrogenation reaction of an unsaturated hydrocarbon has a rate constant of 0.725 M/min. The reaction is observed to follow zero-order kinetics. If the initial concentration of the hydrocarbon is 5.90 M, what is the half-life of the reaction in seconds? *Please report 3 significant figures. Numbers only, no unit. No scientific notation.
Answer : The half-life of the reaction in seconds is, 244
Explanation :
The expression used for zero order reaction is:
[tex]t_{1/2}=\frac{[A_o]}{2k}[/tex]
where,
[tex]t_{1/2}[/tex] = half-life of the reaction = ?
[tex][A_o][/tex] = initial concentration = 5.90 M
k = rate constant = 0.725 M/min
Now put all the given values in the above formula, we get:
[tex]t_{1/2}=\frac{5.90}{2\times 0.725}[/tex]
[tex]t_{1/2}=4.069min=244.14s\approx 244s[/tex]
conversion used : (1 min = 60 s)
Thus, the half-life of the reaction in seconds is, 244
Which of the following aqueous solutions are good buffer systems? . 0.37 M hydrocyanic acid + 0.24 M sodium cyanide 0.14 M sodium nitrite + 0.27 M nitrous acid 0.21 M hypochlorous acid + 0.13 M potassium hypochlorite 0.20 M nitric acid + 0.22 M potassium nitrate 0.32 M ammonium bromide + 0.31 M ammonia
Answer : The good buffer systems are, (a), (b), (c) and (e)
Explanation :
Buffer : It is a solution that prevent any changes in the pH of the solution on the addition of an acidic and basic components.
Or, it is a solution that maintain the pH of the solution by adding the small amount of acid or a base.
There are two types of buffer which are acidic buffer and basic buffer.
Acidic buffer : It is the solution that have the pH less than 7 and it contains weak acid and its salt. For example : Acetic acid (weak acid) and sodium acetate (salt).
Basic buffer : It is the solution that have the pH more than 7 and it contains weak base and its salt. For example : Ammonia (weak base) and ammonium chloride (salt).
The conditions for a good buffer system is:
(1) a weak acid and its conjugate base.
(2) a weak base and its conjugate acid.
(a) 0.37 M hydrocyanic acid + 0.24 M sodium cyanide
It is a combination of weak acid and its conjugate base. So, it is a good buffer.
(b) 0.14 M sodium nitrite + 0.27 M nitrous acid
It is a combination of weak acid and its conjugate base. So, it is a good buffer.
(c) 0.21 M hypochlorous acid + 0.13 M potassium hypochlorite
It is a combination of weak acid and its conjugate base. So, it is a good buffer.
(d) 0.20 M nitric acid + 0.22 M potassium nitrate
It is a combination of strong acid. So, it will not form buffer solution.
(e) 0.32 M ammonium bromide + 0.31 M ammonia
It is a combination of weak base and its conjugate acid. So, it is a good buffer.
Hence, the good buffer systems are, (a), (b), (c) and (e)
Good buffer systems are those that consist of a weak acid and its conjugate base or a weak base and its conjugate acid. The given combinations that are good buffer systems include hydrocyanic acid and sodium cyanide, sodium nitrite and nitrous acid, hypochlorous acid and potassium hypochlorite, and ammonium bromide and ammonia.
Explanation:The aqueous solutions that are good buffer systems from the provided options are:
0.37 M hydrocyanic acid + 0.24 M sodium cyanide0.14 M sodium nitrite + 0.27 M nitrous acid0.21 M hypochlorous acid + 0.13 M potassium hypochlorite0.32 M ammonium bromide + 0.31 M ammoniaA good buffer system consists of a weak acid and its conjugate base or a weak base and its conjugate acid. These pairs are capable of neutralizing small amounts of added acid or base, thus maintaining a relatively stable pH in the solution. In contrast, solutions like 0.20 M nitric acid + 0.22 M potassium nitrate do not form a buffer because nitric acid is a strong acid and does not create a conjugate weak acid-base pair needed for buffering.
Recall from Chapter 12.6 that chlorofluorocarbons (CFCs) catalyze the decomposition of ozone in the stratosphere. Infrared light is absorbed by CFCs while visible light is transmitted. Ultraviolet light can break the bonds in these molecules. Answer the following questions about CFCs. A) Draw the three-dimensional structure of trichlorofluoromethane, CC13F. B) How will CFCs interact with solar radiation in the troposphere? C) How will CFCs interact with solar radiation in the stratosphere?
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