The last sentence in the “Introduction” was: “In this lab you will determine the density (thus characterizing a substance) of a liquid and of a solid-liquid mixture of unknown composition, and then determine the density of a liquid and a solid of known compositions and evaluate how accurate your determinations were.” Give names of those four substances mentioned in the sentence above. A liquid of unknown composition: _________________________________________, a liquid of known composition: ___________________________________________, a solid-liquid mixture of unknown composition: _______________________________, a solid of known composition:

Answer:

Explanation: First, convert the mass of graphite from milligrams (mg) to grams (g).

As 1,000 milligrams in 1 gram

therefore,

300 mg = 300/1000 = 0.3 grams

Now, we can use the molar mass of carbon to calculate the number of moles. We divide the mass of the sample by the molar mass:

Number of moles = Mass (g) / Molar mass (g/mol)

Number of moles = 0.3 g / 12.01 g/mol

Number of moles ≈ 0.02498 moles (rounded to five decimal places)

Therefore, there are approximately 0.02498 moles of carbon in 300 mg of graphite.

Answer::

Explanation::

Answer:Electrical Conductivity,soluble

Explanation:

Salt is a non-magnetic solid and is soluble in water. Sand is a non-magnetic solid and is insoluble in water.

Electrical Conductivity: Salt is an electrolyte and conducts electricity when dissolved in water or in a molten state. This is because salt dissociates into ions (Na+ and Cl-) that can carry electric current. In contrast, sand is a covalent compound and does not conduct electricity, as it does not dissociate into ions in the same way as salt. Sand is considered an insulator in terms of electrical conductivity.

The balanced chemical equation for the decomposition of aspirin (acetylsalicylic acid) is:

[tex]2C_{9}H_{8}O_{4} (aspirin) → 2C_{7}H_{6}O_{3} (salicylic acid) + 2CO_{2} (Carbon dioxide) + H_{2}O (water)[/tex]

In this reaction, the aspirin molecule breaks down into salicylic acid, carbon dioxide, and water. The reaction is typically catalyzed by heat or exposure to acidic or basic conditions.

Aspirin, or acetylsalicylic acid, contains ester functional groups that can undergo hydrolysis. Under suitable conditions, the ester bond in aspirin is cleaved, leading to the formation of salicylic acid, which is the primary decomposition product. Additionally, carbon dioxide and water are released as byproducts of the reaction.

The balanced equation shows that for every two molecules of aspirin, two molecules of salicylic acid, two molecules of carbon dioxide, and one molecule of water are formed. Understanding the decomposition of aspirin is important in pharmaceutical and chemical industries to ensure the stability and shelf-life of the compound, as well as to study its breakdown products and potential side reactions.

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Answer:

To prepare 250 ml of 0.15 M KNO3 solution, you will need to follow these steps:

Calculate the amount of KNO3 needed:

Molarity (M) = moles of solute/liters of solution

Rearranging the formula, moles of solute = M x liters of solution

Moles of KNO3 needed = 0.15 M x 0.25 L = 0.0375 moles

Calculate the mass of KNO3 needed:

Mass = moles x molar mass

The molar mass of KNO3 is 101.1 g/mol

Mass of KNO3 needed = 0.0375 moles x 101.1 g/mol = 3.79 g

Dissolve the calculated amount of KNO3 in distilled water:

Weigh out 3.79 g of KNO3 using a digital balance

Add the KNO3 to a clean and dry 250 ml volumetric flask

Add distilled water to the flask until the volume reaches the 250 ml mark

Cap the flask and shake it well to ensure the KNO3 is completely dissolved

Verify the concentration of the solution:

Use a calibrated pH meter or a spectrophotometer to measure the concentration of the solution

Adjust the volume of distilled water or the mass of KNO3 as needed to achieve the desired concentration

It is important to note that KNO3 is a salt that can be hazardous if ingested or inhaled in large quantities. Therefore, it is recommended to handle it with care and wear appropriate personal protective equipment.

Explanation:

The statement "All organic compounds contain the element carbon, but not all compounds containing the element 'carbon' are organic" can be justified based on the definition and characteristics of organic compounds.

Organic compounds are compounds primarily composed of carbon and hydrogen atoms, often with other elements like oxygen, nitrogen, sulfur, and phosphorus. These compounds are typically associated with living organisms and are known for their unique properties and behavior, including the ability to form complex structures, exhibit covalent bonding, and undergo organic reactions.

On the other hand, there are compounds that contain carbon but are not classified as organic. One notable example is carbon dioxide ([tex]CO_{2}[/tex]), which is a simple inorganic compound composed of carbon and oxygen. Carbon dioxide does not possess the characteristic properties of organic compounds, such as the ability to form long chains or undergo organic reactions.

Additionally, there are inorganic compounds like carbonates (such as calcium carbonate) and carbides (such as calcium carbide) that contain carbon but are not considered organic. These compounds have distinct chemical and physical properties different from those of organic compounds.

In summary, while all organic compounds contain carbon, not all compounds containing carbon are organic. The classification of a compound as organic or inorganic depends on its overall molecular structure, bonding, and characteristic properties.

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To dissolve sugar cubes quickly in a cup of tea, here are two effective methods you can try:Stirring and Crushing, Hot Water Pre-Dissolution.

Stirring and Crushing:

a. Start by placing the sugar cube(s) into the cup of tea. The larger the sugar cubes, the longer they will take to dissolve.

b. Use a spoon or a stirring rod to vigorously stir the tea. The stirring action increases the contact between the sugar cubes and the hot liquid, helping to speed up the dissolution process.

c. While stirring, apply some pressure to the sugar cubes against the walls or base of the cup. This helps to break down the cubes into smaller pieces, exposing more surface area to the tea. Crush the cubes with the back of the spoon or the stirring rod.

d. Continue stirring until all the sugar is dissolved. You can test by observing whether any sugar crystals are visible on the spoon or at the bottom of the cup. If needed, stir a bit longer or crush any remaining sugar crystals.

Hot Water Pre-Dissolution:

a. Fill a separate cup with hot water, ensuring it is hot enough to dissolve the sugar cubes completely.

b. Place the sugar cubes in the hot water and stir until they are fully dissolved. This pre-dissolves the sugar cubes, making it easier and quicker for them to dissolve in the tea.

c. Once the sugar cubes are dissolved in the hot water, pour the sugar solution into your cup of tea.

d. Stir the tea briefly to ensure any remaining undissolved sugar is incorporated.

e. The pre-dissolved sugar solution will mix more readily with the tea, accelerating the overall dissolution process.

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The volume of [tex]H_2S[/tex]collected at 32°C when the atmospheric pressure was 749 torr is approximately 0.0231 liters.

To calculate the volume of [tex]H_2S[/tex]collected, we need to use the ideal gas law equation:

PV = nRT

Where:

P = total pressure (in torr)

V = volume of gas (in liters)

n = number of moles of gas

R = ideal gas constant (0.0821 L·atm/(mol·K))

T = temperature (in Kelvin)

First, let's calculate the number of moles of [tex]H_2S[/tex]produced. From the balanced chemical equation, we see that 1 mole of CuS reacts to produce 1 mole of [tex]H_2S[/tex]. Given the molar mass of CuS (63.5 g/mol) and the mass of CuS (4.22 g), we can calculate the number of moles:

moles of CuS = mass of CuS / molar mass of CuS

moles of CuS = 4.22 g / 63.5 g/mol

moles of CuS ≈ 0.0664 mol

Since the reaction produces 1 mole of [tex]H_2S[/tex]for every mole of CuS, the number of moles of [tex]H_2S[/tex]is also 0.0664 mol.

Next, let's convert the temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15

T(K) = 32°C + 273.15

T(K) ≈ 305.15 K

Now, we can calculate the partial pressure of [tex]H_2S[/tex]using Dalton's law of partial pressures:

Partial pressure of [tex]H_2S[/tex]= Total pressure - Vapor pressure of water

Partial pressure of [tex]H_2S[/tex]= 749 torr - 36 torr

Partial pressure of [tex]H_2S[/tex]≈ 713 torr

Finally, we can rearrange the ideal gas law equation to solve for the volume:

V = (nRT) / P

V = (0.0664 mol * 0.0821 L·atm/(mol·K) * 305.15 K) / 713 torr

V ≈ 0.0231 L

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Answer:

It is commonly used as a fertilizer or feed supplement

The actual result of the calculation is 101.920. Therefore, none of the options provided in the question matches the correct answer.

To calculate the given expression: (43 × 2.310) + (7 × 0.370), we perform the multiplication first and then add the results.

Multiplying the counted numbers:

43 × 2.310 = 99.330

Multiplying the measured numbers:

7 × 0.370 = 2.590

Now, we add the results:

99.330 + 2.590 = 101.920

The correct answer is not provided in the given options: a) 38.35, b) 38.4, c) 38, or d) 40.

The actual result of the calculation is 101.920. Therefore, none of the options provided in the question matches the correct answer.

It's important to note that when performing calculations, it is crucial to accurately follow the order of operations (multiplication before addition) and ensure precision when dealing with decimal numbers.

In this case, the correct answer is not among the options provided, and the accurate result is 101.920.

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Approximately 7.26 g of carbon dioxide is created in the interaction with 8.23 g of oxygen gas.

When octane is burned in air, it undergoes a combustion reaction with oxygen gas ([tex]O_2[/tex]) to produce carbon dioxide ([tex]CO_2[/tex]) and water ([tex]H_2O[/tex]). The balanced chemical equation for this reaction is:

[tex]2 C8H18 + 25 O2 - > 16 CO2 + 18 H2O[/tex]

From the equation, we can see that 25 moles of [tex]O_2[/tex] react to produce 16 moles of [tex]CO_2[/tex]. To find the mass of [tex]CO_2[/tex] produced by the reaction of 8.23 g of [tex]O_2[/tex], we need to convert the mass of [tex]O_2[/tex] to moles and then use the mole ratio from the balanced equation.

The molar mass of [tex]O_2[/tex] is approximately 32 g/mol. Therefore, the number of moles of [tex]O_2[/tex] is calculated as follows:

moles of [tex]O_2[/tex] = mass of [tex]O_2[/tex] / molar mass of [tex]O_2[/tex]

           = 8.23 g / 32 g/mol

           = 0.257 mol

According to the balanced equation, 25 moles of [tex]O_2[/tex] produce 16 moles of [tex]CO_2[/tex]. Using this ratio, we can calculate the number of moles of [tex]CO_2[/tex] produced:

moles of [tex]CO_2[/tex] = (moles of [tex]O_2[/tex]) × (moles of [tex]CO_2[/tex] / moles of [tex]O_2[/tex])

           = 0.257 mol × (16 mol [tex]CO_2[/tex] / 25 mol [tex]O_2[/tex])

           = 0.165 mol

Finally, we can convert the moles of [tex]CO_2[/tex] to grams using the molar mass of [tex]CO_2[/tex], which is approximately 44 g/mol:

mass of [tex]CO_2[/tex] = moles of [tex]CO_2[/tex] × molar mass of [tex]CO_2[/tex]

          = 0.165 mol × 44 g/mol

          = 7.26 g

Therefore, the mass of carbon dioxide produced by the reaction of 8.23 g of oxygen gas is approximately 7.26 g.

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Answer:

Concentration of sodium carbonate solution = Moles of sodium carbonate / (135 mL / 1000) (convert mL to L)

Explanation:

To find the concentration of the sodium carbonate solution, we need to use the concept of stoichiometry and the amount of precipitate recovered.

First, we need to calculate the moles of the precipitate (barium carbonate) using its mass:

Mass of precipitate = 7.13 g

Next, we determine the moles of barium carbonate using its molar mass. The molar mass of barium carbonate (BaCO3) is 197.34 g/mol:

Moles of barium carbonate = Mass of precipitate / Molar mass of barium carbonate

Moles of barium carbonate = 7.13 g / 197.34 g/mol

Now, since the reaction between barium chloride and sodium carbonate is 1:1, the moles of barium carbonate also represent the moles of sodium carbonate present in the solution.

Therefore, the moles of sodium carbonate = Moles of barium carbonate

Now, we need to calculate the volume of the sodium carbonate solution using its concentration. Let's assume the concentration of the sodium carbonate solution is "C" mol/L.

Moles of sodium carbonate = Concentration of sodium carbonate solution * Volume of sodium carbonate solution

Since we have the moles of sodium carbonate and the volume is given as 135 mL, we can rearrange the equation to solve for the concentration:

Concentration of sodium carbonate solution = Moles of sodium carbonate / Volume of sodium carbonate solution

Concentration of sodium carbonate solution = Moles of sodium carbonate / (135 mL / 1000) (convert mL to L)

Substituting the value of moles of sodium carbonate, we can calculate the concentration.

Note: Make sure to perform the necessary unit conversions to ensure consistency in units.

The bullet has a kinetic energy of approximately 344.45 joules (J), while the ocean liner has a kinetic energy of approximately 676,200,000 joules (J). As we can see, the ocean liner has significantly more kinetic energy than the bullet due to its larger mass and velocity.

To calculate the kinetic energy of an object, we use the formula:

Kinetic Energy (Ek) = 0.5 * mass * velocity^2

Let's calculate the kinetic energy for both the bullet and the ocean liner:

For the bullet:

Mass (m) = 0.0020 kg

Velocity (v) = 415 m/s

Ek-bullet = 0.5 * 0.0020 kg * (415 m/s)^2

Ek-bullet = 0.5 * 0.0020 kg * 172225 m^2/s^2

Ek-bullet = 344.45 J

For the ocean liner:

Mass (m) = 6.9 * 10^7 kg

Velocity (v) = 14 m/s

Ek-ocean liner = 0.5 * (6.9 * 10^7 kg) * (14 m/s)^2

Ek-ocean liner = 0.5 * (6.9 * 10^7 kg) * 196 m^2/s^2

Ek-ocean liner = 676200000 J

Therefore, the bullet has a kinetic energy of approximately 344.45 joules (J), while the ocean liner has a kinetic energy of approximately 676,200,000 joules (J). As we can see, the ocean liner has significantly more kinetic energy than the bullet due to its larger mass and velocity.

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For the given reaction K3PO4 → 3K+ + PO4³-, the correct mole ratio would be 3 mol K3PO4 to 1 mol K+ and 3 mol K+ to 1 mol PO4³-, as seen in answer option B.

The mole ratio in a chemical reaction is indeed determined by the coefficients of the balanced chemical equation. In the given reaction:

K3PO4 → 3K+ + PO43-

The correct mole ratio can be established as follows:

1 mole of K3PO4 corresponds to:

- 3 moles of K+

- 1 mole of PO43-

This accurately represents the stoichiometry of the reaction. Therefore, as you correctly pointed out, the mole ratios can be expressed as:

- 3 moles of K3PO4 to 1 mole of K+

- 1 mole of K3PO4 to 1 mole of PO43-

The correct answer choice, corresponding to this mole ratio, is indeed B) 3 mol K3PO4 to 1 mol K+ and 1 mol K3PO4 to 1 mol PO43-. This reflects the relationship between the reactants and products as described by the balanced chemical equation.

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Answer:

Explanation:

The decay of a single nucleus is random. In groups, behavior is predictable (you can predict half-life), but we can't predict when an atom will decay.

Answer: N = +5, O = -2, H = +1

Each H ion has a positive 1 oxidation state when reacting with nonmetals. Each oxygen generally has a -2 (unless in peroxides). The sum of all the states will be 0, so lets set

H + N + O3 = 0

+1 + N - 2(3) = 0

N = +5

so N = +5, O = -2, H = +1

Answer:

I'm not entirely sure what your study is about, but I can tell you that any research or study that contributes new knowledge or insights to a particular field can be valuable. It's important to identify gaps in the existing literature and to approach your research with a clear and focused question or objective. Ultimately, the value of your study will depend on the quality of your research and the significance of your findings.

Answer: The answer is incus

The combustion of 4.7 kg of pure octane ([tex]C_8H_{18[/tex]) produces approximately 15 kg of carbon dioxide ([tex]CO_2[/tex]).

1. Start by writing the balanced equation for the combustion of octane ([tex]C_8H_{18[/tex]):

  [tex]C_8H_{18[/tex] + 12.5O2 → [tex]8CO_2[/tex] + [tex]9H_2O[/tex]

  This equation shows that for every 1 mole of octane burned, 8 moles of carbon dioxide and 9 moles of water are produced.

2. Determine the molar mass of octane ([tex]C_8H_{18[/tex]):

  The molar mass of carbon (C) is approximately 12.01 g/mol.

  The molar mass of hydrogen (H) is approximately 1.008 g/mol.

  Calculating the molar mass of octane: (8 * 12.01 g/mol) + (18 * 1.008 g/mol) ≈ 114.23 g/mol.

3. Calculate the number of moles of octane in 4.7 kg:

  Number of moles = mass (in grams) / molar mass

  Moles of octane = (4.7 kg * 1000 g/kg) / 114.23 g/mol ≈ 41.11 mol

4. Determine the number of moles of carbon dioxide produced:

  From the balanced equation, we know that for every mole of octane burned, 8 moles of carbon dioxide are produced.

  Moles of carbon dioxide = 41.11 mol octane * 8 mol CO2 / 1 mol octane ≈ 328.88 mol

5. Calculate the mass of carbon dioxide produced:

  Mass = moles * molar mass

  Mass of carbon dioxide = 328.88 mol * (12.01 g/mol + 2 * 16.00 g/mol) ≈ 7,883.51 g ≈ 7.88 kg

6. Express the answer using two significant figures:

  The mass of carbon dioxide produced is approximately 7.88 kg when 4.7 kg of octane is burned.

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Based on the given bright-line spectra and the observed wavelengths in the mixture's spectrum, the elements G and J are the ones present in the mixture.

From the given bright-line spectra and the spectrum of the mixture, we can determine the elements present in the mixture by comparing the specific wavelengths observed. Examining the bright-line spectra, we can identify that G has a distinct wavelength at 650 nm, J at 600 nm, L at 550 nm, and M at 500 nm.

Looking at the spectrum of the mixture, we can observe two prominent wavelengths, 650 nm and 600 nm. These correspond to the wavelengths of G and J, respectively. Since the spectrum of the mixture does not exhibit the wavelengths specific to L (550 nm) or M (500 nm), we can conclude that only G and J are present in the mixture.

Therefore, based on the given bright-line spectra and the observed wavelengths in the mixture's spectrum, the elements G and J are the ones present in the mixture.

This analysis relies on the principle that each element has characteristic wavelengths at which they emit light. By comparing the observed wavelengths in the mixture's spectrum with those of the individual elements, we can determine the elements present in the mixture.

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Answer:

B. nitrogen-fixing bacteria

Explanation:

Nitrogen is converted from atmospheric nitrogen (N2) into usable forms such as NO2-,

In a process known as fixation. The majority of nitrogen is fixed by bacteria, most of which are symbiotic with plants.

Recently fixed ammonia is then converted to biologically useful forms by specialized bacteria.

Explanation:

When an unopened bottle of carbonated water appears to contain no gases, it is actually because the gas is dissolved in the water under pressure. This large-scale behavior can be explained by understanding the relationship between pressure, solubility, and the behavior of particles at a small scale.

Carbonated water is typically created by dissolving carbon dioxide (CO2) gas in water under pressure. At a small scale, water molecules form a network of hydrogen bonds, creating spaces where gas molecules can fit. When CO2 is dissolved in water, it forms carbonic acid (H2CO3), which contributes to the slightly acidic taste of carbonated water. The solubility of CO2 in water increases with increasing pressure.

Henry's Law describes the relationship between the solubility of a gas in a liquid and the partial pressure of the gas above the liquid. According to Henry's Law, at a constant temperature, the amount of dissolved gas is proportional to the partial pressure of that gas in equilibrium with the liquid. In the case of carbonated water, when the bottle is sealed, the pressure inside the bottle is higher than atmospheric pressure, and a larger amount of CO2 can dissolve in the water.

When you open the bottle, the pressure inside the bottle rapidly decreases to match the atmospheric pressure. As a result, the solubility of CO2 in the water decreases, and the excess CO2 comes out of the solution in the form of bubbles. This is the fizzing you observe when opening a bottle of carbonated water. At a small scale, the CO2 molecules that were once dissolved in the water now form bubbles, which grow and rise to the surface, eventually escaping into the air.

Glycolysis and photosynthesis are necessary processes: glycolysis produces ATP for energy, while photosynthesis converts sunlight into glucose and oxygen. They are similar in energy transformation and enzymatic reactions but differ in organisms, oxygen/light dependence, and cellular location.

Glycolysis and photosynthesis are both necessary fundamental processes due to their vital roles in energy production and carbon fixation, respectively. Glycolysis is a central pathway in cellular respiration that breaks down glucose to produce ATP, the main energy currency of cells.

It occurs in the cytoplasm of all living organisms and is essential for the generation of energy required for various cellular activities. On the other hand, photosynthesis is the process by which plants, algae, and some bacteria convert sunlight, water, and carbon dioxide into glucose and oxygen. It takes place in the chloroplasts of plants and is responsible for oxygen production and the primary source of organic carbon in ecosystems.

In terms of similarities, both glycolysis and photosynthesis involve the transformation of energy. Glycolysis converts the chemical energy stored in glucose molecules into ATP, while photosynthesis converts solar energy into chemical energy in the form of glucose.

Both processes also involve multiple enzymatic reactions and occur in different cellular compartments (cytoplasm for glycolysis and chloroplasts for photosynthesis). Additionally, they are essential for the survival and functioning of organisms, as glycolysis provides the energy needed for cellular processes, and photosynthesis is responsible for maintaining oxygen levels and providing organic carbon for food chains.

However, there are significant differences between the two processes. Glycolysis occurs in all living organisms, including plants, animals, and microorganisms, while photosynthesis is primarily limited to plants, algae, and some bacteria.

Glycolysis is an anaerobic process that does not require oxygen, whereas photosynthesis is an aerobic process that relies on the presence of light and produces oxygen as a byproduct. Furthermore, glycolysis occurs in the cytoplasm, which is present in all cells, while photosynthesis occurs in specialized organelles called chloroplasts, which are only found in plant cells.

In summary, both glycolysis and photosynthesis are crucial fundamental processes. Glycolysis generates ATP for cellular energy, while photosynthesis converts solar energy into glucose and oxygen. They share similarities in energy transformation and enzymatic reactions but differ in their occurrence across organisms, dependence on oxygen and light, and cellular location.

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Answer:Make sure vyour formatting is clear and easy to understand. Remember, it’s all about helping others understand the answer.

Explanation:

Make sure your formatting is clear and easy to understand. Remember, it’s all about helping others understand the answer.

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The steps I performed:

Answer:

The correct option is: carbon dioxide, methane, water vapor.

Explanation:

Greenhouse gases, such as carbon dioxide (CO2), methane (CH4), and water vapor (H2O), trap heat in the Earth's atmosphere, leading to the greenhouse effect and global warming. Water vapor is the most abundant greenhouse gas, while carbon dioxide and methane also play significant roles. Carbon dioxide is released through activities like burning fossil fuels, deforestation, and industrial processes. Methane is produced by sources such as agriculture, natural gas production, and organic waste decay. Other gases like oxygen, argon, nitrogen, and helium do not significantly contribute to the greenhouse effect.

These substances have dispersed particles that are large enough to scatter light, making the beam visible. Therefore, out of the options provided, a mixture that scatters light is an example of a colloid. Option B)

A colloid is a type of mixture in which particles are dispersed throughout a medium, creating a homogeneous appearance. Unlike solutions, where the particles are completely dissolved, and suspensions, where the particles settle out, colloids have particles that are larger than those in solutions but smaller than those in suspensions. One characteristic of colloids is that they can scatter light due to the size of the particles. This scattering of light is known as the Tyndall effect. Examples of colloids include milk, fog, and aerosol sprays. These substances have dispersed particles that are large enough to scatter light, making the beam visible. Therefore, out of the options provided, a mixture that scatters light is an example of a colloid. Therefore option B) is correct

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Note Complete Question

which is an example of a colloid?

a mixture that settles out,

b mixture that scatters light,

c mixture that is separated by filtration,  

d salt and water mixture?

Answer:

Tcs foods are foods that pose a greater risk of causing foodborne illness if not prepared.

Non Tcs foods on the other hand, are foods that are less likely to support the growth of bacteria and have a lower risk of causing foodborne illness.

Answer:

BaCrO₄

Explanation:

Bariu(Ba) + Chromium(Cr) + 4 Oxygen( O₄)

Answer:

BaCrO4.

Explanation:

Barium chromate is a yellow, crystalline compound, BaCrO4, used as a pigment (barium yellow).

- Separate elements/compounds;

  Barium is a whitish, malleable, active, divalent, metallic element, occurring in combination chiefly as barite or as witherite. Symbol: Ba; atomic weight: 137.34, atomic number: 56; specific gravity 3.5 at 20°C.

Chromate is a salt of chromic acid, as potassium chromate, K2CrO4.

Chromic acid is a hypothetical acid, H2CrO4, known only in the solution or in the form of salts.