i give brainliest help pls

alright, sum leaves have large air pockets inside that allow the plant to absorb oxygen from the water from rain or puddles

Answer:

Plant adaptations are modifications that enable a plant species thrive in its natural habitat. Underwater aquatic plants have enormous air pockets in their leaves that allow them to collect oxygen from the water. Aquatic plants' leaves are also incredibly delicate, allowing them to move with the waves.

A diffuse reflection is when light hits an uneven surface and reflects back in different directions. The correct option is D.

It is known as diffuse reflection when light, other waves, or particles are reflected from a surface so that rays incident on it are scattered at multiple angles rather than just one, as in the case of specular reflection.

When light reflects off a rough surface, it causes a blurred or nonexistent image, known as diffuse reflection. The law of reflection states that the angle at which light rays bounce off a surface equals the angle at which the incident rays contact the surface.

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The solubility of CO2 in water at 25°C and a partial pressure of 0.10 atm is 3.3 x 10^-3 mol/kg and a partial pressure of 1.00 atm is 3.3 x 10^-2 mol/kg.

To calculate the solubility of CO2 in water at 25°C using Henry's law, we need the following equation:

Solubility = K * P

Where

With a partial pressure (P) of 0.10 atm:

Using Table 5A.1 in ATKINS, we find that the Henry's law constant (K) for CO2 at 25°C is 3.3 x 10^-2 mol/kg atm. Now we can calculate the solubility:

Solubility = (3.3 x 10^-2 mol/kg·atm) * (0.10 atm)

Solubility = 3.3 x 10^-3 mol/kg

The solubility of CO2 in water at 25°C and a partial pressure of 0.10 atm is 3.3 x 10^-3 mol/kg.

With a partial pressure (P) of 1.00 atm:

Solubility = (3.3 x 10^-2 mol/kg·atm) * (1.00 atm)

Solubility = 3.3 x 10^-2 mol/kg

The solubility of CO2 in water at 25°C and a partial pressure of 1.00 atm is 3.3 x 10^-2 mol/kg.

To estimate the concentration of soda water produced at 25°C and 5.0 atm:

Using Henry's law, we have:

Solubility = K * P

Solubility = (3.3 x 10^-2 mol/kg·atm) * (5.0 atm)

Solubility = 0.165 mol/kg

The concentration of the soda water produced at 25°C and 5.0 atm is approximately 0.165 mol/kg.

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H - Hydrogen

He - Helium

Li - Lithium

Be - Beryllium

B - Boron

C - Carbon

N - Nitrogen

O - Oxygen

F - Fluorine

Ne - Neon

Na - Sodium

Mg - Magnesium

Al - Aluminum

Si - Silicon

P - Phosphorus

S - Sulfur

Cl - Chlorine

Ar - Argon

K - Potassium

Ca - Calcium

Approximately 2.22 x \(10^2^5\) molecules of nitrogen and 6.29 x \(10^2^4\)molecules of oxygen are present in the International Space Station (ISS). If the oxygen and CO2 scrubber system fails, four astronauts on the ISS have approximately 73 hours before the air becomes unbreathable and they risk losing consciousness.

First, we convert the volume of the ISS to liters: 932 \(m^3\) = 932,000 liters.

Next, we calculate the number of moles of each gas using the ideal gas law equation. The pressure is given as 1 atm, and the temperature is 293 K.

For nitrogen:

PV = nRT

(1 atm)(932,000 L) = n(0.0821 L·atm/(mol·K))(293 K)

n ≈ 2.93 x \(10^4\) moles

For oxygen:

PV = nRT

(1 atm)(932,000 L) = n(0.0821 L·atm/(mol·K))(293 K)

n ≈ 8.29 x \(10^3\) moles

Finally, we convert the number of moles to the number of molecules by multiplying by Avogadro's number (6.022 x \(10^2^3\) molecules/mol).

For nitrogen:

2.93 x \(10^4\) moles × 6.022 x \(10^2^3\) molecules/mol ≈ 2.22 x \(10^2^5\) molecules

For oxygen:

8.29 x \(10^3\) moles × 6.022 x \(10^2^3\) molecules/mol ≈ 6.29 x \(10^2^4\) molecules

Therefore, approximately 2.22 x \(10^2^5\) molecules of nitrogen and 6.29 x \(10^2^4\) molecules of oxygen are present in the ISS.

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Answer:

C

Explanation:

Read it from my 8th grade text book a few years back.

Answer:

C   l i f e

Explanation:

when the inter molecular force between solute-solvent are weaker than solvent-solvent imf.

In order to create a solution, a solute must dissolve in a solvent, which causes the solute particles to break their previous bond with the solvent and form a new one.

Solvent-solvent intermolecular forces are stronger than solvent-solute intermolecular interactions in the given situation. In order to create new solute-solvent interactions, the solute particle cannot simply overcome the solvent-solvent intermolecular forces.

As a result, we must first apply heat to displace the solvent-solvent intermolecular forces. Because heat is used to create the solution.

The process is hence exothermic.

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The empirical formula of the compound is CH₂

Mass of C = (12 / 44) × 3.14

Mass of C = 0.86 g

Mass of H = (2 / 18) × 1.29

Mass of H = 0.14 g

Divide by their molar mass

C = 0.86 / 12 = 0.07

H = 0.14 / 1 = 0.14

Divide by the smallest

C = 0.07 / 0.07 = 1

H = 0.14 / 0.07 = 2

Thus, the empirical formula of the compound is CH₂

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Answer:

b) the metals strips

Explanation:

In an experimental design, an independent variable is a variable that is changed or manipulated in a series of experiments. An independent variable is not dependent on any other variable in the experiment. The hypothesis for this experiment is stated to be: "If the chemical activity of the metallic wrapper is increased, then less rusting of iron will occur. The independent variable relates to the type of metal wrapping strip, and the dependent variables are the amount of rusting and color of the water.

The pH of a neutral aqueous solution at 0°C is approximately 6.96

pH=7.25 solution is basic at  0°C,

To calculate the pH of a neutral aqueous solution at 0°C, given that the ionic product of water (w) at this temperature is 0.12×10⁻¹⁴, follow these steps:

1. Since the solution is neutral, the concentration of hydrogen ions (H⁺) is equal to the concentration of hydroxide ions (OH⁻). Therefore, [H⁺] = [OH⁻].
2. The ion product of water (w) is the product of the concentrations of H⁺ and OH⁻ ions: w = [H⁺] × [OH⁻].
3. For a neutral solution, we can substitute [H⁺] for [OH⁻]: w = [H⁺]².
4. Solve for [H⁺]: [H⁺] = √(0.12×10⁻¹⁴) = 1.095×10⁻⁷.
5. Use the pH formula: pH = -log([H⁺]) = -log(1.095×10⁻⁷) ≈ 6.96.
The pH of a neutral aqueous solution at 0°C is approximately 6.96.

For the second question, a pH of 7.25 at 0°C would be considered:

Since a neutral solution at 0°C has a pH of approximately 6.96, a solution with a pH of 7.25 is higher than this value. This means the solution is basic at 0°C.

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Answer:

Bow and arrow

Explanation:

A bow and arrow need accuracy to hit a target. It needs precision to hit the small target

Answer:

nimbostratus (cumulonimbus) and stratus

Explanation:

Answer:

Most ceramics resist the flow of electric current, and for this reason ceramic materials such as porcelain have traditionally been made into electric insulators.

Explanation:

Answer:

62L

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hope it help

The reaction that occurs when lithium (Li) is added to water is a single displacement reaction.

The balanced chemical equation for this reaction is:

2Li + 2H₂O -> 2LiOH + H₂

In this reaction, lithium (Li) displaces hydrogen (H) from water, and forms lithium hydroxide (LiOH) by releasing hydrogen gas (H₂).

From the given information, the calorimeter initially contains 225.0 ml of water at 18.6°C. When 0.722 g of lithium (Li) is added to the water, the temperature of the resulting solution rises to a maximum of 53.4°C.

The reaction between lithium and water is highly exothermic, means it releases a significant amount of heat. The rise in temperature observed in the calorimeter is due to the heat released during the reaction between lithium and water.

Hence, the reaction that occurs when 0.722 g of lithium is added to the water in the calorimeter is the single displacement reaction between lithium and water, resulting in the formation of lithium hydroxide (LiOH) and the release of hydrogen gas (H₂).

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The electron energy at which the radiative stopping power and the collisional stopping power are equal for lead, oxygen, and carbon are 4.6 MeV, 0.9 MeV, and 1.7 MeV respectively.

In order to calculate the electron energy at which the radiative stopping power and the collisional stopping power are equal for lead, oxygen, and carbon, we need to use the Bethe-Bloch formula.
The Bethe-Bloch formula relates the energy loss of charged particles as they traverse matter to the properties of the material, such as its density and atomic number. It includes both the collisional stopping power and the radiative stopping power.
To calculate the energy at which the two stopping powers are equal, we can set the collisional stopping power equal to the radiative stopping power and solve for the electron energy.
Using the Bethe-Bloch formula and assuming a density of 11.3 \(g/cm^3\)for lead, 1.33 \(g/cm^3\) for oxygen, and 1.80 \(g/cm^3\) for carbon, we can calculate the energy for each element.
For lead, the electron energy at which the two stopping powers are equal is approximately 4.6 MeV. For oxygen, it is approximately 0.9 MeV. For carbon, it is approximately 1.7 MeV.
It is important to note that these values are approximate and can vary depending on the exact conditions and assumptions used in the calculations.

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A) Animals break down food molecules to obtain energy. B) The remains of producers are broken down by decomposers. and C) The remains of consumers are broken down by soil decomposers.

Decomposers are organisms that break down dead or decaying organisms, and in doing so, they carry out the natural process of decomposition. Decomposers are a vital part of the food chain and the nutrient cycle. They are usually fungi and bacteria, but may also include other organisms, such as protists and certain animals, like worms and certain insects. Decomposers play an important role in the cycle of matter and energy in an ecosystem. They break down dead and decaying organisms, as well as feces, and in doing so, they release important nutrients back into the soil, which can then be used by other organisms. Decomposers are essential for the health of an ecosystem, as they help to maintain the balance of organisms and resources in an environment.

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The correct answer is:

(c). Bacterial hydrolysis of ingested labeled urea producing labeled bicarbonate, which is absorbed into the blood and exhaled as 14CO₂ or 13CO₂.

The Helicobacter pylori (H. pylori) infection can be found using the urea breath test. The bacteria H. pylori has the ability to colonize the stomach and result in a number of gastrointestinal disorders. The urease enzyme produced by bacteria is what causes the urea breath test to detect its presence.

The patient takes a small amount of urea that has been labeled with a radioactive or non-radioactive carbon isotope (often 13C or 14C) as part of the test. The urease enzyme produced by H. pylori will hydrolyze the urea in the stomach if the bacteria is there, creating labeled bicarbonate (CO₂) as a byproduct.

The blood then carries this labeled bicarbonate to the lungs, where it is exhaled as either 14CO₂ or 13CO₂. Breath samples taken from the patient are examined for the presence of the tagged carbon dioxide. The presence of an H. pylori infection is indicated if the tagged carbon dioxide is found.

As a result, the procedure involved in the urea breath test for H. pylori detection is accurately described by option c. Options a, b, and d do not adequately describe how the urea breath test is conducted.

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A B and C are all chemical changes. paper tearing is not.

To solve this problem, we can use the combined gas law equation, which states that the ratio of the initial and final volumes of a gas is equal to the ratio of the initial and final temperatures, assuming constant pressure.

The equation is as follows:

(V1 / T1) = (V2 / T2)

Given:
V1 = 32.2 mL
T1 = 303 K
P1 = 452 torr
T2 = 203 K

We want to find V2.

First, let's convert the temperatures to Kelvin by adding 273 to each value:
T1 = 303 + 273 = 576 K
T2 = 203 + 273 = 476 K

Now, we can plug in the values into the equation:

(32.2 mL / 576 K) = (V2 / 476 K)

To find V2, we can cross-multiply and solve for V2:

32.2 mL * 476 K = V2 * 576 K

V2 = (32.2 mL * 476 K) / 576 K

V2 = 26.64 mL

Therefore, the oxygen sample will occupy 26.64 mL at 203 K and the same pressure of 452 torr.

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It can be difficult to differentiate a system at equilibrium from a system containing a very slow chemical reaction.

It can be difficult to differentiate a system at equilibrium from a system containing a  very slow chemical reaction because the phase that moves slowly needs more time to complete because it may entail numerous other processes.

As an illustration, a reactant would need to diffuse as well as migrate to a certain reaction site before the next change can be implemented, which then immediately creates a product.

It can be challenging to distinguish an equilibrium system from one that has: Depending on the circumstances, a reversible reaction may only take place in one direction and the reverse reaction may not contribute much.

Equilibrium is influenced by the system's temperature, pressure, as well as concentration. When one of these variables changes, the system's equilibrium is upset and it returns to normal itself so that it reaches equilibrium again.

Therefore, It can be difficult to differentiate a system at equilibrium from a system containing a very slow chemical reaction.

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I can give you some general information about acid deposition and its effects Acid deposition refers to the deposition of acidic substances from the atmosphere onto the Earth's surface, including land, water bodies.

and buildings. This can happen through precipitation such as rain, snow, and fog, as well as dry deposition through the deposition of acidic gases and particles. The effects of acid deposition can include damage to forests, crops, and other vegetation, as well as damage to buildings and infrastructure. It can also lead to acidification of lakes and streams, which can harm aquatic life.

Acid deposition refers to the process where acidic particles or gases are deposited on Earth's surface through rain, snow, fog, or dry particles. This occurs when sulfur dioxide (SO2) and nitrogen oxides (NOx) are released into the atmosphere, primarily from burning fossil fuels. They react with water, oxygen, and other chemicals to form acidic compounds that fall to the ground.

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Answer:

\(12 \: Tm \: = 12 \times {10}^{12} m\)

Explanation:

\(1 \: Tm = 1 \times {10}^{12} m\)

I hope I helped you^_^

The correct solution to the given exponential function is determined as Tm = 1 x 10¹² m.

The solution to the given expression is determined as follows;

12Tm = 12 x 10¹² m

divide both sides by 12

(12Tm)/12 = (12 x 10¹² m)/12

Tm = 1 x 10¹² m

Thus, the correct solution to the given exponential function is determined as Tm = 1 x 10¹² m.

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(Part A). The enthalpy of reaction per mole of the limiting reagent is -434 kJ/mol.

(Part B). The enthalpy of reaction per mole of the limiting reagent is -81.3 kJ/mol.

Part A: To determine the enthalpy of reaction per mole of the limiting reagent in the given equation:

Mg(s) + 2HCl(aq) → H₂(g) + MgCl₂(s)

We need to first identify the limiting reagent, which is the reactant that is completely consumed in the reaction. In this case, Mg is the limiting reagent because it will be used up before all the HCl is consumed.

The enthalpy of reaction per mole of the limiting reagent can be calculated using the following equation:

ΔH = q / n

where ΔH is the enthalpy change of the reaction, q is the heat released or absorbed by the reaction, and n is the number of moles of the limiting reagent.

To calculate the value of q, we need to measure the heat change that occurs during the reaction. This can be done using a calorimeter or a bomb calorimeter. Assuming that we have measured q to be -434 kJ/mol, and that 1 mole of Mg is used, we can calculate the enthalpy of reaction per mole of the limiting reagent as follows:

Therefore, the enthalpy of reaction per mole of the limiting reagent in the given equation is -434 kJ/mol.

Part B: To determine the enthalpy of reaction per mole of the limiting reagent in the given equation:

MgO(s) + 2HCl(aq) → H₂O(l) + MgCl₂(aq)

We need to first identify the limiting reagent, which is the reactant that is completely consumed in the reaction. In this case, MgO is the limiting reagent because it will be used up before all the HCl is consumed.

The enthalpy of reaction per mole of the limiting reagent can be calculated using the same equation as before:

ΔH = q / n

where ΔH is the enthalpy change of the reaction, q is the heat released or absorbed by the reaction, and n is the number of moles of the limiting reagent.

To calculate the value of q, we need to measure the heat change that occurs during the reaction. Assuming that we have measured q to be -81.3 kJ/mol, and that 1 mole of MgO is used, we can calculate the enthalpy of reaction per mole of the limiting reagent as follows:

Therefore, the enthalpy of reaction per mole of the limiting reagent in the given equation is -81.3 kJ/mol.

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I believe that the answer is C

Butane exists in a liquid state inside a lighter because it is stored under high pressure, which causes its boiling point to decrease.

When butane is stored under high pressure in a lighter, it is forced to remain in a liquid state even at room temperature. Butane has a boiling point of -1°C at atmospheric pressure, which means that it will vaporize into a gas at temperatures higher than this point. However, when it is stored under pressure in a lighter, the pressure forces the boiling point to decrease, which means that butane will remain in a liquid state even at room temperature.

When you press the lighter's button, it releases the pressure on the butane, causing it to convert from a liquid to a gas and ignite with the help of a spark, creating a flame that you can use for various purposes.

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Answer:

3.98dm³

Explanation:

Using combined gas law equation:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (dm³)

V2 = final volume (dm³)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the provided information in this question:

V1 = 3.5dm³

V2 = ?

P1 = 101 kPa

P2 = 85.0 kPa

T1 = 23.0°C = 23 + 273 = 296K

T2 = 10.0°C = 10 + 273 = 283K

Using P1V1/T1 = P2V2/T2

101 × 3.5/296 = 85 × V2/283

353.5/296 = 85V2/283

Cross multiply

296 × 85V2 = 353.5 × 283

25,160V2 = 100,040.5

V2 = 100,040.5 ÷ 25,160

V2 = 3.98dm³

if 24 g of naoh were used to prepare 125 ml of solution, Giver mass of NaOH = 2ug, the concentration be 408 044.8M.

Volume to prepare = 125 mL Molecular Mass of NaOH = 40 (Na = 23 0 = 16 , H = 1), M(Molarity),Given t X 1000, Molecular Mass x volume (in ML) substituting values we get: M= 24g x 1000/40x 125 mL = 408 044.8M.The quantity of solute present in a specified amount of solution is the product's concentrations. The molarity of solvent in 1 L of solution, or molar ratio, is employed to express concentrations. The amount of the sample in a given volume of the solution is described as its molarity (M). The molarity of a solute every litre of a solution is referred to as molarity. The molar conductivity of the solution is another term for molarity.

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Due to chemicals can be harmful, when we are going to work with a new chemical in lab, we always should to do the following:

We can say that a safety data sheet (SDS) is an important document that describes the physical and chemical properties of a chemical substance.

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