which of the following is a true statement? for a saturated solution q > ksp for an supersaturated solution q > ksp. for an unsaturated solution q > ksp. for a saturated solution q < ksp. precipitation is expected for a solution where q < ksp.

Answer:

0.872

Explanation:

The correct answer is B) zymomonas. Ethanolic fermentation, which is the process of converting sugars into ethanol (alcohol) and carbon dioxide by yeast, is commonly used in the production of beverages like beer, wine, and pulque.

While there are various microorganisms that can carry out fermentation, yeast is the primary organism involved in ethanolic fermentation.

Zymomonas is a type of bacteria that is known for its ability to perform ethanolic fermentation. It can efficiently convert sugars into ethanol and carbon dioxide, making it suitable for industrial applications such as the production of alcoholic beverages.

Clostridium, Leuconostoc, Lactobacillus, and Propionibacterium are other types of microorganisms, but they are not primarily involved in ethanolic fermentation. They may have different metabolic pathways or be associated with other fermentation processes, but they are not the main organisms responsible for the production of ethanol in the context of beverage production.

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Answer:

B.

Explanation:

Minerals can be located using geochemical surveys and remote sensors that analyze satellite images. Following that, mining or quarrying is used to remove many minerals.

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Answer:

Lose 2 electrons

Explanation:

The positive signs of ∆H° and ∆S° indicate that the vaporization of water at 50 degrees Celsius requires an input of heat and results in an increase in randomness. However, the negative sign of ∆G° indicates that despite the initial energy input required, the process is still favorable and will release energy in the end.

The signs of ∆H°, ∆S°, ∆G° for the vaporization of water at 50 degrees Celsius are as follows:
1. ∆H° (enthalpy change): The enthalpy change for the vaporization of water at 50 degrees Celsius is positive (+). This means that heat is absorbed from the surroundings during the process. The positive sign indicates that the reaction is endothermic. 2. ∆S° (entropy change): The entropy change for the vaporization of water at 50 degrees Celsius is also positive (+). This means that the randomness or disorder of the system increases during the process.

3. ∆G° (Gibbs free energy change): The Gibbs free energy change for the vaporization of water at 50 degrees Celsius is negative (-). This means that the process is thermodynamically favorable and spontaneous. The negative sign indicates that the reaction releases energy.

.

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3 moles⋅22.4 L/1 mole=67.2 L Likewise, 0.5 moles of gas will occupy half the volume 1 mole occupies 0.5 moles⋅22.4 L/1 mole = 11.2 L

Standard Temperature and Pressure conditions imply a temperature of 273.15 K and a pressure of 1 atm. When these conditions are met, 1 mole of any ideal gas occupies exactly 22.4 L. Keep in mind that ideal gas particles are assumed to have no volume of their own, that's why, for example, at STP 1 mole of helium gas will occupy the same volume as, say, 1 mole of chlorine gas.

However, as you can see in the picture, the two gases will have different masses because of the difference in their molar mass. The balloon filled with helium will weigh less than the one filled with chlorine, despite the fact that they occupy the same volume.

So, if 1 mole occupies 22.4 L, the immediate conclusion is that a bigger number of moles will occupy more than 22.4 L, and a smaller number of moles will occupy less than 22.4 L.

In your case, 3 moles of gas will occupy 3 times more volume than 1 mole of gas.

3 moles⋅ 22.4 L/1 mole=67.2 L

Likewise, 0.5 moles will occupy half the volume 1 mole occupies

0.5 moles⋅ 22.4 L/1 mole=11.2 L

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C/ If Fe is greater than Fg, the negatively charged oil droplet will move freely toward the negatively charged plate.

In the Millikan oil droplet experiment, the negatively charged oil droplets are subjected to an electric field created by the charged plates. The electric force (Fe) acts on the oil droplet in a direction opposite to the gravitational force (Fg). When Fe is greater than Fg, the electric force overcomes the gravitational force, causing the negatively charged oil droplet to experience an upward force. As a result, the oil droplet moves freely upward toward the negatively charged plate.

Option B is incorrect because if Fe is equal to Fg, the forces balance each other, resulting in a stationary droplet. However, the question states that Fe is increased until it is greater than Fg, implying that the droplet is no longer stationary but moves in response to the electric force.

Therefore, option C is the correct answer, as it describes the effect of an electric field on the motion of a negatively charged oil droplet in the Millikan oil droplet experiment.

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If 16% of the nucleotides in one strand of a DNA molecule contain the base g, what percent of the nucleotides on the complementary strand will also contain the base g but it is impossible to determine from the information given.

DNA double helix. The hydrogen bonds are damaged and the helix is ​​opened. Each strand of DNA acts as a template for synthesizing a brand new complementary strand. During replication, equal DNA double helices are formed, every with a brand new strand and a vintage strand.

Also, cytosine (C) enhances guanine (G) and adenine (A) enhances thymine (T). Therefore, the complementary strand for 5'-TAGAC 3' is 3'-ATTCG 5', similar to alternative c.

His Chargaff's discovery of A = T and C = G, combined with a few very essential X-ray crystallographic research through British researchers Rosalind Franklin and Maurice Wilkins, Watson Contributed to Crick's derivation of his 3-D double helix model. DNA structure.

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Answer:

79.8g/dm³

Explanation:

As you can see, the solution in the problem contains 0.5 moles of copper sulfate per dm³. To solve this question we must convert these moles to grams using its molar mass (Molar mass CuSO4 = 159.609g/mol) as follows:

0.5mol CuSO4/dm³ * (159.609g/mol) =

The solubility of the potassium nitrate will increase as we increase the temperature because increases the kinetic energy of solute particles. More KNO₃ will dissolve in the solution and makes it unsaturated.

Solubility can be described as the maximum amount of Solute that can dissolve in a known quantity of solvent at a particular temperature. By changing the temperature we can increase the solubility of a solute. Sparingly soluble solids or liquids can be completely liquified by raising the temperature.

Solid solubility depends on the nature of the solute and the solvent. Substances such as sugar, common salt (NaCl), etc dissolve in water while substances such as naphthalene do not dissolve in water.

The solubility of potassium nitrate KNO₃ increases with an increase the temperature but the increasing temperature of NaCl solution does not show much change in the solubility of NaCl.

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The following elements have an impact on reaction rate: a) temperature; e) concentration; and h) catalysts. the appropriate choices are a, e, and h.

A reaction's pace can be greatly impacted by variations in temperature and concentration. More particles have the energy to react as a result of a rise in temperature, hence the pace of reactions typically rises as well. The frequency of collisions also rises with an increase in reactant concentration, which results in a higher reaction rate. The collision theory of chemical kinetics can be used to explain these outcomes. However temperature and concentration variations can also impact the equilibrium constant and change the reaction's equilibrium position. Changes in enthalpy, entropy, and free energy can all be influenced by the presence of a catalyst, lowering the activation energy and speeding up the rate of a process.

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1.00 pint of milk is equal to 473.18 milliliters, based on the conversion factor of 1 pint = 473.18 milliliters.

To convert pints to milliliters, we can use the conversion factor of 1 pint = 473.18 milliliters.

Since we have 1.00 pints of milk, we can multiply it by the conversion factor to find the volume in milliliters:

1.00 pint * 473.18 milliliters/pint = 473.18 milliliters.

Therefore, 1.00 pint of milk is equivalent to 473.18 milliliters. It's important to note that this conversion factor is based on the standard definition of a pint, which is equal to 473.18 milliliters. In some countries, the pint may have a different value, so it's essential to use the appropriate conversion factor based on the specific context or region.

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The total time required for the decay is approximately 1416.0 hours, which is closest to the given option C. 1006.9 hours.

Option (C) is correct.

To calculate the number of half-lives required for the decay. The half-life of 140 Ba is given as 283.2 hours.

First, we calculate the fraction of the original amount remaining after each half-life. Since the half-life represents the time it takes for half of the substance to decay, the fraction remaining after each half-life is 1/2.

To find the number of half-lives required to decay from 35 mg to 1.0 mg, we can set up the following equation:

(35 mg) * (1/2)ⁿ = 1.0 mg

Where 'n' is the number of half-lives.

Now, let's solve for 'n':

(1/2)ⁿ = 1.0 mg / 35 mg

(1/2)ⁿ = 0.02857

To find 'n', we can take the logarithm (base 1/2) of both sides:

n = log base 1/2 (0.02857)

Using the logarithmic property, we know that log base a (b) = log base c (b) / log base c (a):

n = log (0.02857) / log (1/2)

Using a calculator, we can find:

n ≈ 4.243

Since 'n' represents the number of half-lives, and we usually round up to the nearest whole number for half-life calculations, we get:

n ≈ 5

Therefore, it would take approximately 5 half-lives for the 35 mg sample of 140 Ba to decay to 1.0 mg.

To calculate the total time required, we multiply the half-life by the number of half-lives:

Total time = 283.2 hours * 5

Total time ≈ 1416 hours

Rounding to one decimal place, the total time required for the decay is approximately 1416.0 hours, which is closest to the given option C. 1006.9 hours.

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Answer:

3

Explanation:

begining 0's are never significant

all numbers 1-9 are always significant

ending 0's are only significant when there is

a decimal.

Answer:

Lower, Higher

Explanation:

The density of gold in SI units is 19,000 kg/m³. Gold's relative density of 19 means that it is 19 times denser than water at 4°C, which has a density of 1000 kg/m³. Therefore, by multiplying the relative density of gold by the density of water, we can find the density of gold in SI units. In this case, 19 multiplied by 1000 kg/m³ equals 19,000 kg/m³, which is the density of gold in SI units.

In the context of this problem, the relative density of a substance is defined as the ratio of its density to the density of a reference substance, in this case, water at 4°C. By multiplying the relative density by the density of the reference substance, we can calculate the density of the substance in SI units.

In this example, the relative density of gold is 19, indicating that gold is 19 times denser than water at 4°C. Multiplying the relative density by the density of water (1000 kg/m³) yields the density of gold as 19,000 kg/m³ in SI units.

To summarize, the density of gold in SI units is 19,000 kg/m³. The relative density of gold is 19, indicating that it is 19 times denser than water at 4°C, which has a density of 1000 kg/m³.

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Answer:

Gregor Mendel, through his work on pea plants, discovered the fundamental laws of inheritance. He deduced that genes come in pairs and are inherited as distinct units, one from each parent. Mendel tracked the segregation of parental genes and their appearance in the offspring as dominant or recessive traits.

Through his careful breeding of garden peas, Gregor Mendel discovered the basic principles of heredity and laid the mathematical foundation of the science of genetics.

Explanation:

The substance used as a reference standard in both the freezing-point and vapor-pressure models is A. Deionized water.

In osmometry, a reference standard is used to calibrate the instrument and establish a baseline for measurements. Both freezing-point and vapor-pressure osmometers require a known substance with well-defined properties for accurate calibration.

Deionized water (option A) is commonly used as the reference standard in osmometers because its freezing-point and vapor-pressure properties are well-established and easily reproducible.

The freezing-point of pure water is 0°C (32°F) at standard atmospheric pressure, and its vapor pressure is also well-defined.

Other substances like NaCl (option B) and KCl (option C) are used as calibration standards in specific contexts, but they are not universally applicable to both freezing-point and vapor-pressure osmometers.

Based on the given information, the substance used as a reference standard in both the freezing-point and vapor-pressure models is deionized water (option A). It serves as a reliable and widely accepted calibration standard for osmometers due to its well-defined freezing-point and vapor-pressure properties.

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To calculate the mass of lead deposited on the cathode of a lead-acid car battery when a current of 26A is fed for 71 seconds, we need to consider the Faraday's law of electrolysis.

By determining the number of moles of electrons transferred during the reduction reaction, we can calculate the corresponding mass of lead deposited.

According to Faraday's law of electrolysis, the mass of a substance deposited during an electrolytic process is directly proportional to the number of moles of electrons transferred. To calculate the mass of lead deposited on the cathode, we need to determine the number of moles of electrons transferred.

Given that the current is 26A and the time is 71 seconds, we can calculate the total charge transferred using the formula Q = I * t, where Q is the charge in coulombs, I is the current in amperes, and t is the time in seconds. Substituting the values, we have Q = 26A * 71s = 1846C.

Since one mole of electrons corresponds to 96,485 coulombs, we can calculate the number of moles of electrons transferred by dividing the total charge by the Faraday constant: 1846C / 96,485 C/mol = 0.0191 mol.

The balanced reduction half-reaction for the deposition of lead is Pb^2+(aq) + 2e^− -> Pb(s). From the balanced reaction, we see that 2 moles of electrons are required to deposit 1 mole of lead.

Therefore, the number of moles of lead deposited on the cathode is 0.0191 mol / 2 = 0.00955 mol.

To calculate the mass of lead, we need to multiply the number of moles by the molar mass of lead, which is 207.2 g/mol: 0.00955 mol * 207.2 g/mol = 1.98 g.

Thus, the mass of lead deposited on the cathode of the car battery is approximately 1.98 grams.

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The atomic number and mass number of sulfur are 16 and 32. So it has 16 electrons, 16 protons, 16 neutrons. The metals with 2 valence electrons are elements of group 2 and have high ionization energy & low electron affinity.

Valence electrons can be defined as the electrons present in the outermost shell while the electrons filled the inner shell are called core electrons. Lewis structures of the elements can be helpful to find the number of valence electrons.

Only electrons present in the outermost shell i.e. valence electrons can contribute to the formation of a bond and decide the reactivity of the specific element.

Valence electrons are occupying different electron shells and these electrons are responsible for the interaction between atoms and cause the formation of chemical bonds.

The elements of group 2 have electronic configuration ns². Therefore, they have high ionization energy to knock out the electrons and low electron affinity to add more electrons.

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The balanced equation is 2Na + 2H\(_2\)O \(\rightarrow\) 2NaOH + H\(_2\) and the coefficient of H\(_2\)O is 2. Stoichiometry represents the number of coefficients.

An equation for just a chemical reaction is said to be balanced if both the reactants as well as the products have the same number of atoms and total charge for each component of the reaction. In other words, each sides of both the reaction have an equal balance of mass and charge.

It is typical that balancing chemical formulas for both charge and mass in aqueous solutions. Equal numbers and types of atoms are produced on both aspects of the equation when balancing for mass. When the charge is balanced, there is no net charge on either side of the equation.

The balanced equation can be written as

2Na + 2H\(_2\)O = 2NaOH + H\(_2\)

The coefficient of H\(_2\)O is 2.

Therefore, the balanced equation is 2Na + 2H\(_2\)O \(\rightarrow\) 2NaOH + H\(_2\) and the coefficient of H\(_2\)O is 2.

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Answer:

Explanation:

g

Liquids have characteristics that fall somewhere in between those of solids and gases, yet they resemble solids more. Intermolecular forces, as opposed to intramolecular forces, such as the covalent bonds that bind atoms in molecules and polyatomic ions, keep molecules in a liquid or solid together. In comparison to covalent bonds, intermolecular forces are often significantly weaker. For instance, in 1 mol of water, the intramolecular forces must be overcome in order to break both O-H bonds, while the intermolecular forces must be overcome in order to turn 1 mol of liquid water into water vapor at 100 °C, which needs just around 41 kJ. (Despite this figure appearing to be modest, the intermolecular forces in liquid water are some of the highest intermolecular forces ever measured.