in your experiment, sodium bisulfite (nahso3) in water is used to destroy any unreacted bromine (br2) or to trap the br2 and not allow it to escape from the reaction setup. the reaction is shown below but can't be described using conventional organic curved-arrow pushing. after adding sodium bisulfite in your procedure, why is the resulting mixture put into acid waste?

Answers

Answer 1
When sodium bisulfite is added to an experiment, why is the resulting mixture put into acid waste?

In the experiment, sodium bisulfite (NaHSO3) in water is used to destroy any unreacted bromine (Br2) or to trap the Br2 and not allow it to escape from the reaction setup. The reaction is shown below but cannot be described using conventional organic curved-arrow pushing.

The resulting mixture is placed in acid waste for the following reasons:

Sodium bisulfite's addition to the reaction mix in the procedure is done to destroy any unreacted bromine (Br2) or to trap the Br2 and prevent it from escaping the reaction setup. Following the reaction, it is necessary to neutralize the mixture with sodium carbonate or another base. After that, the neutralized mixture should be properly disposed of in an acid waste container. Thus, the resulting mixture is placed in acid waste.

Sodium bisulfite is used in excess to the amount of bromine to ensure that all of the bromine is captured or reacted. The resulting mixture is extremely acidic as a result of the reaction. As a result, the mixture must be neutralized before being disposed of.

The most straightforward approach to neutralizing it is to add a basic substance like sodium carbonate, which reacts with the acidic mixture to create water and sodium sulfate (Na2SO4).

As a result, when sodium bisulfite is added in the procedure, the resulting mixture is put into acid waste.

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Related Questions

this portion of the titration curve of a strong acid with a strong base is the same as this region for a weak acid titrated with a strong base.
a. The portion after all of the base has been neutralized
b. The endpoint pH
c. The portion before the endpoint is reached
d. The buffer region

Answers

The portion of the titration curve of a strong acid with a strong base that is the same as the region for a weak acid titrated with a strong base is the buffer region. The correct answer is option: d.

In this region, the pH of the solution changes very slowly as small amounts of base are added to the acid. The buffer region occurs when the amount of base added is roughly equal to the amount of acid in the solution. The other options mentioned, including the portion after all of the base has been neutralized, the endpoint pH are specific to either strong acid or weak acid titration curves and do not apply to both.

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Assume that 0.950g of KHT (potassium hydrogen tartrate) are dissolved in 25.00mL of solution.KHT -> K++ HT-a) calculate the solubility of KHT for these conditions in g KHT / L of solutionb) Calculate the solubility of KHT for these conditions in mol KHT / L of solutionc) Determine [K+] and [HT-] in this solution. If the temperature is Tp, a trace of solid is present and the reaction is at equilibrium. Determine Ksp at this temperature

Answers

a) Solubility (g KHT/L) = (0.950 g KHT) / (0.025 L) = 38 g KHT/L

b) Solubility (mol KHT/L) = (38 g KHT/L) / (188.18 g/mol) = 0.202 mol KHT/L

c) [K+] = [HT-] = 0.202 mol KHT/L

d) Ksp = (0.202)(0.202) = 0.0408

A more detailed explanation of the answer.

a) To calculate the solubility of KHT in g KHT/L of solution, follow these steps:
1. Convert the volume of the solution to liters: 25.00 mL = 0.025 L
2. Calculate the solubility by dividing the mass of KHT by the volume of the solution:
Solubility (g KHT/L) = (0.950 g KHT) / (0.025 L) = 38 g KHT/L

b) To calculate the solubility of KHT in mol KHT/L of solution, follow these steps:
1. Determine the molar mass of KHT (K = 39.10 g/mol, H = 1.01 g/mol, C = 12.01 g/mol, O = 16.00 g/mol):
Molar mass of KHT = K + 2*(C + H + 2*O) = 39.10 + 2*(12.01 + 1.01 + 2*16.00) = 188.18 g/mol
2. Convert the solubility from g KHT/L to mol KHT/L:
Solubility (mol KHT/L) = (38 g KHT/L) / (188.18 g/mol) = 0.202 mol KHT/L

c) To determine [K+] and [HT-] in this solution, follow these steps:
1. Since KHT dissociates into K+ and HT-, the concentrations of K+ and HT- will be equal to the solubility of KHT in mol KHT/L:
[K+] = [HT-] = 0.202 mol KHT/L

As there is a trace of solid present and the reaction is at equilibrium, we can determine the Ksp at this temperature by following these steps:
1. Write the balanced chemical equation for the dissociation of KHT: KHT (s) <-> K+ (aq) + HT- (aq)
2. Write the expression for the Ksp: Ksp = [K+][HT-]
3. Plug in the concentrations calculated earlier: Ksp = (0.202)(0.202) = 0.0408

So, at this temperature (Tp), the Ksp for KHT is 0.0408.

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Why is carbon used to extract metals from their oxides
1-cheap
2- high_____

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Because of its high reactivity, cheap cost, and versatility in the extraction procedure, carbon is used to extract metals from their oxides.

Carbon is commonly used as a reducing agent to extract metals from their oxides because of its high reactivity and low cost. When a metal oxide is heated with carbon, a reduction reaction occurs, with carbon atoms reacting with oxygen atoms to form carbon dioxide (CO2) and the metal atoms being reduced to their elemental form.

One of the main advantages of using carbon for this process is its high reactivity. Carbon has a strong affinity for oxygen, and as such, it readily reacts with metal oxides to reduce them. Carbon is also abundant and relatively cheap, making it a cost-effective reducing agent.

Additionally, the use of carbon as a reducing agent can be carried out in a range of conditions, such as in a blast furnace, making it a versatile method for extracting metals from their ores. This method is widely used in industry for the extraction of metals such as iron, zinc, and lead, among others.

In summary, carbon is used to extract metals from their oxides due to its high reactivity, low cost, and versatility in the extraction process.

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which best describes the reaction, if any, that occurs when aqueous solutions of silver nitrate and sodium phosphate are combined?

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Silver phosphate is created as a precipitate sodium nitrate is formed as a precipitate there is no reaction silver is oxidised silver is reduced.

Does mixing silver I nitrate and sodium chloride aqueous solutions result in a reaction?

The ions of both compounds interchange when silver nitrate (AgNO3) and sodium chloride (NaCl) solution are combined. As a result, white precipitates of silver chloride (AgCl) and sodium nitrate solution (NaNO3) are produced.

What precipitate will result from the reaction between aqueous sodium phosphate and aqueous silver nitrate?

Silver phosphate and sodium nitrate are produced as a result of the interaction between silver nitrate and sodium phosphate. Due to its insoluble in water nature, silver phosphate precipitates out of the solution.

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exchange occurs.
QUESTION 4 DOK 34 ALIGNED STANDARDS
10 points
A chemical reaction between
bromine and sodium occurs.
Bromine has an electronegativity
value of 2.96, while sodium has an
electronegativity value of 0.93,
Will electrons be exchanged between
the two atoms? Explain how you
know.

Answers

With fewer valence electrons than other metals, sodium tends to increase its stability by shedding electrons during action creation.

What happens when bromine and sodium combine?

To create sodium bromide or sodium iodide, hot sodium can also burn in vaporised bromine or iodine. An orange flame and a white solid are the results of each of these reactions.

What number of bonds can bromine form?

In its Lewis structure, bromine contains three lone pairs on each of its atoms and just one Br-Br bond. The three lone pairs and one bond between the bromine atoms are the only connections between them.

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How is potassium-argon dating useful to a paleoanthropologist?

Answers

Answer:

it can be used to date the sedimentary rock where the fossils of ancient humans or their hominid ancestors are found.

Explanation:

hope it helps

What happens over time as sediments settle on land or water?

Answers

When sediments settle on land or water, they can undergo a process known as sedimentation, which can have various effects depending on the type and quantity of sediment involved. Here are some general things that can happen over time as sediments settle:

Deposition: Sediments can accumulate and settle on the bottom of a water body or on land, resulting in the formation of layers of sediment. This process can take thousands or even millions of years, and the resulting sedimentary layers can provide important information about the history of the area.
Compaction: As sediment accumulates, it can become compacted due to the weight of the layers above it. This can result in the compression of the sediment, causing it to become denser and harder over time.
Cementation: Sediments can also become cemented together over time, as minerals in the sediment dissolve and precipitate out, filling the spaces between the grains of sediment and binding them together. This process can result in the formation of sedimentary rocks.
Erosion: Sediments can be eroded away by the action of wind or water, or by human activities such as mining or construction. This can result in the loss of soil and changes to the landscape.
Overall, the process of sedimentation can have a significant impact on the environment over time, as sediments accumulate and are transformed into new forms.

a 64.0 ml portion of a 1.70 m solution is diluted to a total volume of 268 ml. a 134 ml portion of that solution is diluted by adding 149 ml of water. what is the final concentration? assume the volumes are additive.

Answers

The final concentration can be calculated from the dilutions mentioned and it is found to be 0.384 M.

To calculate the final concentration, we need to consider the dilution formula, which states that the initial concentration multiplied by the initial volume is equal to the final concentration multiplied by the final volume.

The first dilution can be calculated as:

C1 × V1 = C2 × V2

1.70 M × 64.0 mL = C2 × 268 mL

108.8 = 268 × C2

C2 = 108.8 ÷ 268

C2 = 0.406 M

This solution has again been diluted. Thus, now the final concentration will be calculated as:

C2 × V2 = C3 × V3

0.406 M × 268 mL = C3 × 283 mL

108.808 = 283 × C3

C3 = 108.808 ÷ 283

C3 = 0.384 M

Therefore, the final concentration of the solution is 0.384 M.

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given that the specific rotation of (r)-2-methoxypentane is −29.6, what is the specific rotation of (s)-2-methoxypentane?

Answers

The specific rotation of (S)-2-methoxypentane is +29.6.

To determine the specific rotation of (S)-2-methoxypentane, given that the specific rotation of (R)-2-methoxypentane is -29.6, follow these steps:

1. Identify the enantiomers: (R)-2-methoxypentane and (S)-2-methoxypentane are enantiomers, which are non-superimposable mirror images of each other.

2. Understand specific rotation: Specific rotation is a property of chiral molecules, and the specific rotation of one enantiomer has the same magnitude but opposite sign as its mirror image enantiomer.

3. Calculate the specific rotation of (S)-2-methoxypentane: Since the specific rotation of (R)-2-methoxypentane is -29.6, the specific rotation of (S)-2-methoxypentane will be the opposite sign with the same magnitude means +29.6.

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What change in volume results if 40 mL of gas is cooled from 33 °C to 5 °C?

Answers

Charles's Law-

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]

Where:-

V₁ = Initial volumeT₁ = Initial temperatureV₂ = Final volumeT₂ = Final temperature

As per question, we are given that -

V₁=40 mLT₁ = 33°CT₂ =5°C

We are given the initial temperature and the final temperature in °C.So, we first have to convert those temperatures in Celsius to kelvin by adding 273-

[tex]\:\:\:\:\:\:\star\sf T_1[/tex] = 33+ 273 = 306K

[tex]\:\:\:\:\:\:\star\sf T_2[/tex] =5+273 = 278K

Now that we have obtained all the required values, so we can put them into the formula and solve for V₂ :-

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2= \dfrac{V_1}{T_1}\times T_2\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2= \dfrac{40}{306}\times 278\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2= 0.13071...........\times 278\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf V_2 = 36.33892...........\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf\underline{ V_2 = 36.34 \:mL}\\[/tex]

Therefore, the volume will become 36.34 mL if 40 mL of gas is cooled from 33 °C to 5 °C.

Make the indicated corrections in the following gas volumes.(show work)

Answers

The required gas volumes obtained at different pressures is a. [tex]279.825cm^3[/tex], b. [tex]0.804m^3[/tex], c. [tex]37.43cm^3[/tex], d. [tex]551.5cm^3[/tex] and e. [tex]200cm^3[/tex].

The ideal gas equation is a mathematical equation used to relate the four main properties of an ideal gas: pressure (P), volume (V), temperature (T), and moles of gas (n). It is expressed as PV = nRT, where R is the ideal gas constant. This equation is used to calculate the pressure, volume, and temperature of an ideal gas given any two of these properties.

a. Given [tex]338cm^3[/tex] at 86.1kPa to 104.0kPa

We can calculate this using the ideal gas law:

P1V1 = P2V2

86.1 * 338 = 104.0 * V2

V2 =[tex]279.825cm^3[/tex]

b. Given [tex]0.873m^3[/tex] at 94.3kPa to 102.3kPa

P1V1 = P2V2

(94.3) * (0.873) = (102.3) * V2

V2 = [tex]0.804m^3[/tex]

c. Given [tex]31.5cm^3[/tex] at 97.8kPa to 82.3kPa

P1V1 = P2V2

(97.8) * 31.5 = 82.3 * V2

V2 = [tex]37.43cm^3[/tex]

d. [tex]524cm^3[/tex] at 110.0kPa to 104.5kPa

P1V1 = P2V2

110.0 * 524 = 104.5 * V2

V2 = [tex]551.5cm^3[/tex]

e. [tex]171cm^3[/tex] at 122.5kPa to 104.3kPa

P1V1 = P2V2

122.5 * 171 = 104.3 * V2

V2 = [tex]200cm^3[/tex]

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write the chemical equation for the ionization of water.

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The chemical equation for the ionization of water is: 2H₂O (l) → 2H+ (aq) + O₂- (aq).

Ionization is the process of breaking apart a molecule into its separate atoms or ions. In the case of water, it is a molecule composed of two hydrogen atoms and one oxygen atom. The ionization of water occurs when these atoms are split apart, creating hydrogen ions (H+) and hydroxide ions (O₂-). This process requires energy, which is usually in the form of heat.

The reaction of water being ionized can be represented by the chemical equation: 2H₂O (l) → 2H+ (aq) + O₂- (aq). This equation shows that two molecules of water (H₂O) are broken apart into two hydrogen ions (H+) and one hydroxide ion (O₂-).

Ionization of water is an important process in many chemical and biological reactions. In the human body, for example, the ionization of water helps to regulate the body's pH level. It is also important for the formation of certain acids and bases, and the solubility of various compounds in water.

In addition, the ionization of water is a necessary step in the formation of electrical currents and is also an important part of the photosynthesis process.

In summary, the chemical equation for the ionization of water is 2H₂O (l) → 2H+ (aq) + O₂- (aq). This process is essential for many chemical and biological reactions and helps to regulate the body's pH level, and the solubility of compounds in water, and is part of the photosynthesis process.

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Ascorbic acid (vitamin C, C6H8O6) is a diprotic acid (K1 =8.0x10^-5 and K2=1.6x10^-12). What is the pH of a 0.270 M solution of ascorbic acid?

Answers

If Ascorbic acid (vitamin C, C6H8O6) is a diprotic acid (K1 =8.0x10^-5 and K2=1.6x10^-12). The pH of a 0.270 M solution of ascorbic acid is 4.10.

What is the pH of a 0.270 M solution of ascorbic acid?

The two dissociation reactions for ascorbic acid are:

H2C6H6O6 ⇌ H+ + HC6H6O6- (K1 = 8.0x10^-5)

HC6H6O6- ⇌ H+ + C6H6O6 2- (K2 = 1.6x10^-12)

To solve the problem, we need to consider the ionization of both H+ ions from ascorbic acid. Let's call the concentration of H+ from the first ionization [H+]1, and the concentration of H+ from the second ionization [H+]2.

K1 = [H+]1 [HC6H6O6-] / [H2C6H6O6]

K2 = [H+]2 [C6H6O6 2-] / [HC6H6O6-]

Since ascorbic acid is a diprotic acid, we need to use the equilibrium expressions for both ionization reactions to determine the concentrations of H+ and the ascorbic acid species.

[H+]1 [HC6H6O6-] / [H2C6H6O6] = 8.0x10^-5

[H+]2 [C6H6O6 2-] / [HC6H6O6-] = 1.6x10^-12

We can assume that the concentration of ascorbic acid that dissociates is much larger than the concentration of H+ formed, so we can use the approximation [H+] << [H2C6H6O6] to simplify the calculations.

[H+]1 = K1 [H2C6H6O6] / [HC6H6O6-] ≈ K1 [H2C6H6O6] / [H2C6H6O6]

[H+]1 ≈ K1 = 8.0x10^-5

[H+]2 = K2 [HC6H6O6-] / [C6H6O6 2-] ≈ K2 [H+]1 [HC6H6O6-] / [C6H6O6 2-]

[H+]2 ≈ K2 [H+]1 = (1.6x10^-12) (8.0x10^-5) = 1.28x10^-16

The total concentration of H+ in the solution is [H+]1 + [H+]2, so the pH of the solution is:

pH = -log([H+]1 + [H+]2)

pH = -log(8.0x10^-5 + 1.28x10^-16)

pH = 4.10

Therefore, the pH of a 0.270 M solution of ascorbic acid is 4.10.

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determine the solubility of kcl at 60 °c in 100g of h2o?

Answers

The solubility of KCl at 60 °C in 100 g of water (H2O) can be determined using experimental data or by using a solubility table. The solubility of a substance refers to the maximum amount of that substance that can dissolve in a given amount of solvent at a particular temperature and pressure.

One possible way to determine the solubility of KCl at 60 °C in 100 g of water is to consult a solubility table, which lists the solubility of various substances in water at different temperatures. According to one such table, the solubility of KCl in water at 60 °C is approximately 47 g per 100 g of water.

This means that 100 g of water at 60 °C can dissolve up to 47 g of KCl before becoming saturated, i.e., no more KCl will dissolve in the water at this temperature.

It is important to note that the solubility of KCl (or any substance) in water can be affected by various factors, such as temperature, pressure, and the presence of other solutes. Therefore, the solubility value obtained from a solubility table is only an approximation and may not be accurate for all conditions.

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When 1 mole of methane (CH4) is burned it releaes 461.9 KJ of heat.
Calculate ΔH for a process in which 8.0 g of methane is burned.

Answers

1 mole of methane (CH₄) is burned it releases 461.9 KJ of heat then the ΔH for the combustion of 8.0 g of methane is 230.1 kJ.

To calculate the ΔH for the combustion of 8.0 g of methane, we need to first convert the mass of methane to moles.

The molar mass of methane (CH₄) is:

C: 12.01 g/mol

H: 1.01 g/mol

4 x H: 4.04 g/mol

Molar mass of CH₄ = 12.01 + 4.04 = 16.05 g/mol

So, 8.0 g of CH₄ is equal to:

n = m/M = 8.0 g / 16.05 g/mol = 0.498 moles of CH₄

Now, we can use the molar heat of combustion to calculate ΔH:

ΔH = n x ΔHcomb

ΔHcomb is the molar heat of combustion, which is given as 461.9 kJ/mol.

ΔH = 0.498 moles x 461.9 kJ/mol = 230.1 kJ

Methane is a chemical compound with the formula CH₄. It is a colorless, odorless, and flammable gas that is the primary component of natural gas. Methane is the simplest hydrocarbon and the main component of biogas and landfill gas. It is also a potent greenhouse gas and a major contributor to climate change. Methane is used as a fuel for heating, cooking, and electricity generation, as well as in industrial processes such as chemical synthesis and metal production.

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Metallic behavior correlates with large atomic size and low ionization energy. Thus, metallic behavior increases down a group and decreases from left to right across a period.true or false

Answers

The statement "Metallic behavior correlates with large atomic size and low ionization energy. Thus, metallic behavior increases down a group and decreases from left to right across a period" is true.

The statement is true because metallic behavior correlates with large atomic size and low ionization energy. So, as you go down a group, the atomic size and ionization energy decrease, resulting in increased metallic behavior. As a result, when going from left to right across a period, the atomic size decreases, and the ionization energy increases, resulting in a decrease in metallic behavior.

As a result, the metallic behavior increases down a group and decreases from left to right across a period.

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if a air mass is rising it must be

Answers

If an air mass is rising, it must be less dense than the surrounding air.

What happens when air mass is rising?

If air mass is rising, it must be less dense than the surrounding air.

This is because when air rises, it is moving into an area of lower pressure than its initial location, which implies that there must be less air above it. Less air above means less weight above, hence resulting in lower density.

The less dense air mass will continue to rise until it reaches an altitude where it is equal in density to that of the surrounding air. This rising motion can lead to cloud formation and potentially precipitation, all depending on the moisture content of the air mass.

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Infant Tylenol contains 0.16 g of acetaminophen (C8H9NO2) in every 5 mL of medicine. What is the Molarity of Tylenol?

Question 6 options:

32 M


0.21 M


0.0002 M


0.32 M

Answers

The molarity of Tylenol is 0.21 M, rounded to two significant figures. Hence, the correct option is (B) i.e. 0.21 M.

To find the molarity of Tylenol, we need to know the number of moles of acetaminophen present in 5 mL of medicine.

First, let's calculate the molecular weight of acetaminophen:

C = 12.011 g/mol x 8 = 96.088 g/mol

H = 1.008 g/mol x 9 = 9.072 g/mol

N = 14.007 g/mol x 1 = 14.007 g/mol

O = 15.999 g/mol x 2 = 31.998 g/mol

Total molecular weight = 96.088 g/mol + 9.072 g/mol + 14.007 g/mol + 31.998 g/mol = 151.165 g/mol

Next, we can use the given mass of acetaminophen in 5 mL of medicine to calculate the number of moles:

0.16 g acetaminophen x (1 mol / 151.165 g) = 0.001058 mol

Finally, we can use the definition of molarity to calculate the molarity of Tylenol:

Molarity = moles of solute / volume of solution in liters

Since we have 0.001058 moles of acetaminophen in 5 mL of medicine, which is equivalent to 0.005 L of solution, we can calculate the molarity as:

Molarity = 0.001058 mol / 0.005 L = 0.2116 M

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Which best explains why sawdust burns more quickly than a block of wood of equal mass under the same conditions?
O The molecules move more quickly in the sawdust than in the block of wood.
O The pressure of oxygen is greater on the sawdust.
O More molecules in the sawdust can collide with oxygen molecules.
O Oxygen is more concentrated near the sawdust than the block of wood.
Mark this and return
Save and Exit
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Answers

On the sawdust, the oxygen pressure is higher. Due to this, pieces of wood burn more quickly than logs of the same mass. A. A log of wood has a larger surface area and requires longer time to burn.

What does sawdust burn more quickly than a chunk of wood?

The surface area of the substance affects how quickly combustion reactions take place. The rate of the combustion reaction increases with surface area. This is due to the large surface area material's frequent exposure to oxygen.

Why burns sawdust more quickly than it should?

The more oxygen molecules that collide per second with the fuel, the faster the combustion reaction is.

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Write a conclusion for Lisa's experiment ​

Answers

From Lisa's experiment, it can be concluded that Tablet C was the best antacid among the four types tested, as it required the least amount of HCl to change the color of the indicator.

How does indigestion tablets work?

Indigestion tablets, also known as antacids, work by neutralizing excess stomach acid. Stomach acid is produced by the body to help digest food, but when there is an excess of acid, it can lead to indigestion, heartburn, and other uncomfortable symptoms.

This indicates that Tablet C was able to neutralize the acid effectively and had the highest buffering capacity compared to the other three tablets. Therefore, it can be recommended as the most effective antacid for treating indigestion.

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Image transcribed:

3. Lisa was investigating which of four different types of indigestion tablet neutralised most acid and was therefore the best 'antacid' of the four. She crushed each tablet to a fine powder, and added the powder to 20 mL of water mixed with two drops of universal indicator solution. Then she added 1 mL of dilute hydrochloric acid at a time until the indicator changed colour.

Lisa's results were:

Tablet A-16 mL

a. Put Lisa's results in a suitable table.

Tablet B-15 mL

Tablet C-8 mL

Tablet D-12 mL

. based on the gc data, what is the ratio of products formed from the reaction with koh in 1-propanol? what are the specific yields of the 2 alkenes? explain what would happen if the solvent is substituted for 2-methyl-2-butanol instead?

Answers

When 1-propanol is used as the solvent instead of 2-methyl-2-butanol, the ratio of the products and the precise yields may vary.

The reaction of KOH with 1-propanol typically results in the formation of two alkenes: propene and 2-propen-1-ol. The ratio of these two products will depend on the reaction conditions, such as temperature, concentration of KOH, and reaction time.

The specific yields of the two alkenes will depend on the efficiency of the reaction, as well as the selectivity of the reaction towards each product. In general, propene is expected to be the major product due to its thermodynamic stability. However, if the reaction conditions favor the formation of 2-propen-1-ol, then the specific yield of this product may be higher.

If the solvent is substituted for 2-methyl-2-butanol, the reaction conditions may be affected due to the differences in physical and chemical properties of the solvent. For example, 2-methyl-2-butanol has a higher boiling point and lower polarity than 1-propanol, which may result in different reaction rates and selectivities. The reaction may also be affected by the steric hindrance of the solvent, which can affect the accessibility of the KOH to the reactant.

Therefore, the ratio of products and specific yields may be different when using 2-methyl-2-butanol as the solvent compared to 1-propanol.

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radioactive decay is a first order kinetic process. radioactive decay is a first order kinetic process. true false g

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The given statement "radioactive decay is a first-order kinetic process" is true because the number of radioactive nuclei that decay per unit time is proportional to the number of radioactive nuclei present.  

Radioactive decay

Radioactive decay is a natural process by which the unstable atomic nucleus loses energy by emitting radiation. This results in a change in the composition of the atomic nucleus, which is accompanied by a release of energy. The three types of radiation that can be emitted during radioactive decay are alpha particles, beta particles, and gamma rays.

Alpha particles are positively charged particles consisting of two protons and two neutrons, beta particles are negatively charged particles emitted by certain radioactive isotopes, and gamma rays are high-energy photons emitted by atomic nuclei during radioactive decay.

First-order kinetics is a type of chemical reaction in which the rate of reaction depends only on the concentration of one reactant. In other words, a first-order reaction is one in which the rate of reaction is proportional to the concentration of the reactant raised to the power of one. This means that the rate of reaction increases linearly with the concentration of the reactant. First-order kinetics is commonly observed in chemical and biochemical systems, as well as in radioactive decay.

In radioactive decay, the number of radioactive nuclei that decay per unit time is proportional to the number of radioactive nuclei present. This property of radioactive decay is called first-order kinetics.

Therefore, the given statement is true.

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calculate the ph of a solution that results from mixing 22.6 ml of 0.23 m dimethylamine ((ch3)2nh) with 17.1 ml of 0.16 m (ch3)2nh2cl. the kb value for (ch3)2nh is 5.4 x 10-4.

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To calculate the pH of a solution that results from mixing 22.6 mL of 0.23 M dimethylamine ((CH3)2NH) with 17.1 mL of 0.16 M (CH3)2NH2Cl, first we need to calculate the initial concentration of dimethylamine and the hydrogen ion. Then, the pH of the solution can be found from the hydrogen ion concentration.

To find the initial concentration of dimethylamine, use the following equation:

CDMA = (22.6 mL x 0.23 M) + (17.1 mL x 0.16 M)

CDMA = 7.868 M

To find the initial concentration of hydrogen ion, use the following equation:

CH+ = CDMA x Kb

CH+ = 7.868 M x 5.4 x 10-4

CH+ = 4.2632 x 10-3 M

To find the pH of the solution, use the following equation:

pH = -log [CH+]

pH = -log (4.2632 x 10-3)

pH = 2.37

Therefore, the pH of the solution that results from mixing 22.6 mL of 0.23 M dimethylamine ((CH3)2NH) with 17.1 mL of 0.16 M (CH3)2NH2Cl is 2.37.

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Which of the following statements is true?

(a) An exothermic reaction will slow down when heated.

(b) The rates of all chemical reactions increase with temperature.

(c) Heating the reactants in an exothermic reaction causes the system to attain a state of equilibrium.

(d) Only exothermic reactions proceed spontaneously at room temperature.

Answers

Answer: A

Explanation: An exothermic reaction generates heat. Unless the reaction is cooled in some way, its temperature increases. If you increase the temperature with an external heater, it slows the reaction down or reverse its direction.

potassium hydrogen phthalate (khc8h4o4) is a weak acid whose ka is 3.91 x 10-6. what will the ph be at the half-equivalence point?

Answers

pH is 5.41 at equivalence point of equivalence point.

Potassium hydrogen phthalate (KHC₈H₄O₄) is a weak acid that has a dissociation constant (Ka) of 3.91 x 10^-6.

To determine the pH at the half-equivalence point of potassium hydrogen phthalate we need to know what is equivalence point is-

The half-equivalence point (pH = pKa) refers to the stage at which half the acid has been converted to the conjugate base, and the pH equals the pKa of the acid. At the half-equivalence point, the number of moles of acid that has been consumed is equal to the number of moles of base that has been consumed.

The formula for the calculation of pH at the half-equivalence point is given below:

pH = pKa + log (cB / cA)

Where,cB is the concentration of the conjugate base, and cA is the concentration of the weak acid.

Since the volume of the titrant is the same at the half-equivalence point, the concentration of the conjugate base and the weak acid will be the same.

So, pH = pKa = -log (3.91 x 10^-6) = 5.41

Therefore, the pH of potassium hydrogen phthalate (KHC₈H₄O₄) at the half-equivalence point is 5.41.

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describe the procedure for the preparation you chose for each ester. make sure the procedure matches the method you selected above and that you include all reagents.

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The procedure for the preparation you chose for each ester are: heat the reagents, add aqueous solution, heat and stir the mixture, cool it down, add sodium bicarbonate, the ester is separated by filtration and lastly crude ester can be purified by recrystallization.

The procedure for the preparation of an ester involves several steps which are in detail below:.

First, the reagents, which can include an acid, an alcohol, and a catalyst, must be combined in a round-bottom flask. Heat is then applied and the mixture is agitated, either manually or with a stirrer.

After the reaction is complete, the mixture is cooled, and an aqueous solution of a base, such as sodium bicarbonate, is added. This causes the ester to precipitate out and is separated from the aqueous layer by filtration.

The crude ester can then be purified, typically by recrystallization. The reagents used will depend on the ester to be prepared. For example, for the preparation of ethyl acetate, acetic acid, ethanol, and sulfuric acid can be used as the reagents.

To complete the reaction, the acid, alcohol, and catalyst are combined in the round-bottom flask, heated and stirred, and cooled. Then, the aqueous solution of sodium bicarbonate is added and the ester is separated by filtration. Finally, the crude ester can be purified by recrystallization.

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Fill in the table. If you could help that would be appreciated.

Answers

Modeling DNA Mutations Key involves:

e) A-T-T-G-T-A-G-A-C-G-C-T-T-A-T-G-A-C.

f) The protein produced from the mutated strand is Protein B.

g) The effect of this mutation on the organism is beneficial.

What are mutation keys?

Mutation keys are a set of rules or guidelines used to represent changes in DNA sequences. They are commonly used in genetics to represent the effects of mutations on the amino acid sequence of a protein.

A mutation key can be a table or a chart that lists the different types of mutations, such as substitution, insertion, or deletion, and the resulting changes in the DNA sequence, the amino acid sequence, and the functional consequences of the mutation.

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Image transcribed:

Modeling DNA Mutations Key

Base Sequence--|--Protein Produced--|--Effect of Mutation

T-T-C-G-T-AGACGCT-T-A-T-GA-C--|--Protein A--|--Neutral

ACC-GT-A-GA-C-G-C-T-T-A-T-G-A-C--|--Protein A--|--Neutral

A-T-GG-T-A-GACGCT-T-A-T-G-A-C--|--Protein A--|--Neutral

GT-CGT-A-GACGCTT-A-T-G-A-C--|--Protein B--|--Beneficial

AAC-GTAGACGC-T-T-A-T-G-A-C--|--Protein B--|--Beneficial

A-T-T-G-T-A-GACGCT-T-A-T-G-A-C--|--Protein B--|--Beneficial

C-T-C-G-T-A-GAC-GC-T-T-A-T-G-A-C--|--Protein C--|--Harmful

AGCGTAGACGCT-TAT-GAC--|--Protein C--|--Harmful

A-T-A-G-T-A-GACGCT-T-A-T-G-A-C--|--Protein C--|--Harmful

e) Find the base sequence from the key that matches the base sequence of the second mutated DNA strand from row C of Table 2.

f) Note the protein produced from this mutated strand and record it in row D of Table 2.

g) Note the effect of this mutation on the organism and record it in row E of Table 2.

Row--|--Description--|--Answers

A--|--Base sequence of original strand--|--A-T-C-G-T-A-G-A-C-G-C-T-T-A-T-G-A-C

B--|--Protein produced from original strand--|--Protein A

C--|--Base sequence of mutated strand--|--________

D--|--Protein produced from mutated strand--|--_______

E--|--Effect of mutation--|--______

any compound that increases the number of hydronium ions when dissolved in water, is called ?

Answers

A compound that increases the number of hydronium ions (H₃O⁺) when dissolved in water is called an acid. Acids are characterized by their ability to donate protons (H⁺) to water molecules, resulting in the formation of hydronium ions. This process is known as acid dissociation.

The strength of an acid is determined by its ability to donate protons. Strong acids, such as hydrochloric acid (HCl) or sulfuric acid (H₂SO₄), completely dissociate in water, resulting in a high concentration of hydronium ions. Weak acids, such as acetic acid (CH₃COOH), only partially dissociate in water, resulting in a lower concentration of hydronium ions.

Acids can have a wide range of applications in industry and everyday life, from the production of fertilizers and cleaning products to the preservation of food and the regulation of pH in the human body.

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Under which set of conditions would H₂ (g) be the most dissolved in H₂O(l)?

101.3 kPa and 75°C
120 kPa and 25°C
101.3 kPa and 25°C
120 kPa and 75°C

Answers

The most dissolved H₂ (g) in H₂O (l) would occur under 101.3 kPa and 75°C.

The attraction between an electronegative atom serving as the hydrogen bond acceptor and a hydrogen atom covalently bonded to a more electronegative "donor" atom or group (Dn) is known as a hydrogen bond, or H-bond (Ac).Under 101.3 kPa and 75 °C, the maximum dissolved H2 (g) in H2O (l) would be present.At higher temperatures, the solvent molecules will have higher kinetic energy, allowing them to break the hydrogen bonds between the molecules and dissolve H₂ (g) more easily. At higher pressures, there will be more molecules of H₂ (g) in a given volume, increasing the chances of it dissolving into the solvent.

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in general, which reaction is favored (forward, reverse, or neither) if the value of keq at a specified temperature is

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When the value of K_eq at a specified temperature is in general, which reaction is favored (forward, reverse, or neither)?When the value of K_eq at a given temperature is greater than 1,

the forward reaction is favored. Conversely, when K_eq is less than 1, the reverse reaction is favored. At equilibrium, when K_eq is equal to 1, the reaction is neither forward nor reverse but is instead stable. Furthermore, it implies that both the forward and reverse reactions occur at the same rate.Thus, in general, the reaction that is favored depends on the value of K_eq. When the value of K_eq is greater than 1, the forward reaction is favored. Conversely, when K_eq is less than 1, the reverse reaction is favored. When K_eq equals 1, the reaction is neither forward nor reverse but is instead at equilibrium.

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