What is volume? a. How close particles are stuck together b. The amount of space occupied by matter c. The amount of particles in any given object d. The weight of an object Please select the best answer from the choices provided

The wave property that should be tested in this experiment is absorption, which refers to the extent to which a material can absorb sound waves. The correct answer is option a.

By testing how different materials interact with sound waves from the straw horn, the student can gain insight into the properties of those materials and their ability to absorb sound.

A. Wave property: absorption

Method: playing the straw horn in a room with hard surfaces and a room with soft surfaces

To test this property, the student should play the straw horn in a room with hard surfaces, such as walls and floors made of concrete or tile, and a room with soft surfaces, such as walls and floors made of carpet or drapes.

By comparing the sound produced in each room, the student can observe how the sound waves interact with different materials and how effectively each material absorbs the sound.

This method allows the student to investigate how different materials absorb sound waves and how this affects the sound produced by the straw horn. This information can be valuable in understanding how sound travels in different environments and how to optimize sound quality in different settings.

The correct answer is option a.

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The Tyndall effect is the scattering of light by colloidal particles or suspensions, causing the particles to become visible.

In fog, water droplets act as colloidal particles and scatter light, making it difficult to see clearly. High beams produce a greater amount of light, which causes more scattering and reflection in the fog, resulting in decreased visibility. This is because the water droplets in the fog are closer together and more concentrated in the path of the high beams, causing more light to be reflected back towards the driver's eyes.

Using low beams, on the other hand, produces less light and reduces the amount of scattering and reflection in the fog, resulting in better visibility. Therefore, it is recommended to use low beams when driving in foggy conditions to avoid glare and improve visibility.

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The volume of the same gas is 320 L.

Use the combined gas law to solve for the final volume of the gas:

(P1V1/T1) = (P2V2/T2)

Substituting the given values, we get:

Solving for V2, we get:

Therefore, the volume of the gas at the new conditions is 320 L.

The combined gas law relates the pressure, volume, and temperature of a gas in a closed system. It states that the product of pressure and volume divided by the temperature is a constant for a given mass of gas in a closed system undergoing changes in pressure, volume, and temperature. Mathematically, the combined gas law can be represented as:

Where P₁ and V₁ are the initial pressure and volume, T₁ is the initial temperature, P₂ and V₂ are the final pressure and volume, and T₂ is the final temperature. This equation is useful in predicting the behavior of gases when the conditions of pressure, volume, and temperature are changed. The combined gas law is a combination of Boyle's law, Charles's law, and Gay-Lussac's law, and it can be derived from the ideal gas law.

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Sound is not always an indicator of an exothermic reaction. While some exothermic reactions may produce sound, others may not.

However, certain exothermic reactions that produce a lot of heat can cause nearby air molecules to rapidly expand and create pressure waves, which we hear as a sound.

For example, combustion reactions that involve burning fuels such as gasoline, natural gas, or propane can produce a loud, explosive sound as the fuel rapidly oxidizes and releases a large amount of energy in the form of heat and light.

Additionally, some exothermic reactions can cause materials to break or shatter, producing a loud cracking or popping sound. For example, the reaction between baking soda and vinegar produces carbon dioxide gas, which can cause a balloon filled with the mixture to pop with a loud sound.

So while sound alone is not conclusive evidence of an exothermic reaction, it can be a possible indicator in certain cases where the reaction produces a significant amount of heat or causes physical changes in the surrounding materials.

Other factors such as changes in temperature, light emission, or gas production may also be used as evidence to confirm an exothermic reaction.

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In the equation 2h2 + o2 + 2h2o, the two hydrogen molecules (H2) react with one oxygen molecule (O2) to form two molecules of water (H2O). This reaction is known as combustion and it requires a certain ratio of hydrogen to oxygen in order for the reaction to take place.

Here correct answer is D)

In this equation, the ratio of hydrogen to oxygen is 2:1. This means that for every one liter of hydrogen, two liters of oxygen are needed in order for the reaction to take place.

In answer to the questions, a) one liter of hydrogen would react with two liters of oxygen, b) one liter of hydrogen would react with 22.4 liters of oxygen, c) 22.4 liters of hydrogen would react with one liter of oxygen, and d) two liters of hydrogen would react with one liter of oxygen.

This equation is a great example of the law of conservation of mass, as the total number of atoms on each side of the equation remain the same

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The pH of the NaOH solution prepared to react with the pennies is 14.550.

To determine the molarity of the NaOH solution prepared to react with the pennies, follow these steps:

1. Calculate the moles of NaOH: Divide the mass of NaOH by its molar mass (142.1 g / 39.997 g/mol) = 3.553 moles of NaOH.

2. Calculate the volume of the solution: Convert the volume from mL to L (1000.0 mL * (1 L / 1000 mL)) = 1.000 L.

3. Calculate the molarity: Divide the moles of NaOH by the volume of the solution (3.553 moles / 1.000 L) = 3.553 M.

The molarity of the NaOH solution prepared to react with the pennies is 3.553 M.

To determine the pH of the solution:

1. Use the formula: pH = -log[H+], where [H+] represents the concentration of hydrogen ions in the solution.

2. Since NaOH is a strong base, it dissociates completely in water. The concentration of OH- ions is equal to the molarity of NaOH (3.553 M).

3. Calculate the pOH: pOH = -log[OH-] = -log(3.553) = -0.550.

4. Convert pOH to pH: pH = 14 - pOH = 14 - (-0.550) = 14.550.

The pH of the NaOH solution prepared to react with the pennies is 14.550.

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The pH of the resulting solution after adding 0.0248 moles of hydrochloric acid to the buffer containing 0.348 M ammonium chloride and 0.339 M ammonia is approximately 7.967.

To calculate the pH of the resulting solution after adding hydrochloric acid to a buffer containing 0.348 M ammonium chloride and 0.339 M ammonia, follow these steps:

1. Determine the initial moles of ammonium chloride (NH₄Cl) and ammonia (NH₃) in the solution:
- Moles of NH₄Cl = (0.348 M) x (0.125 L) = 0.0435 moles
- Moles of NH₃ = (0.339 M) x (0.125 L) = 0.042375 moles

2. Calculate the moles of NH₄Cl and NH₃ after the reaction with HCl:
- Moles of HCl added = 0.0248 moles
- The reaction between NH₃ and HCl produces NH₄Cl: NH₃ + HCl → NH₄Cl
- Moles of NH₄Cl after reaction = 0.0435 moles (initial) + 0.0248 moles (from HCl) = 0.0683 moles
- Moles of NH₃ after reaction = 0.042375 moles (initial) - 0.0248 moles (reacted with HCl) = 0.017575 moles

3. Calculate the new concentrations of NH₄Cl and NH₃:
- [NH₄Cl] = 0.0683 moles / 0.125 L = 0.5464 M
- [NH₃] = 0.017575 moles / 0.125 L = 0.1406 M

4. Use the Henderson-Hasselbalch equation to find the pH:
- pH = pKₐ + log ([NH₃] / [NH₄⁺])
- The pKₐ of ammonia (NH₃) is 9.25
- pH = 9.25 + log (0.1406 / 0.5464) = 9.25 - 1.283 = 7.967

The pH of the resulting solution after adding 0.0248 moles of hydrochloric acid to the buffer containing 0.348 M ammonium chloride and 0.339 M ammonia is approximately 7.967.

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The final temperature will be approximately -55.6 degrees Celsius when the volume is reduced to 5.25 mL.

According to Charles' Law, when pressure is constant, the volume of a gas is directly proportional to its temperature (in Kelvin). The formula for Charles' Law is V1/T1 = V2/T2.

First, convert the initial temperature from Celsius to Kelvin (68 + 273.15 = 341.15 K). Then, plug in the values: (12.75 mL / 341.15 K) = (5.25 mL / T2).

To solve for T2, multiply both sides by T2 and divide by 5.25 mL: T2 = (341.15 K * 5.25 mL) / 12.75 mL ≈ 139.6 K. Finally, convert back to Celsius: 139.6 K - 273.15 ≈ -55.6 degrees Celsius.

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There are 1.142 grams of C₃H₆ in the 652 mL of sample of the gas at STP.

Using ideal gas equation,

PV = nRT,  pressure is P, volume is V, number of moles in n, gas constant is R, the temperature is T. At STP, the pressure is 1 atm, the temperature is 273 K, and the molar volume is 22.4 L.

We can use the following steps to calculate the number of moles of C₃H₆ present in 652 mL of the gas at STP:

Convert the volume to liters:

652 mL = 0.652 L

Calculate the number of moles using the ideal gas law:

PV = nRT

(1 atm) (0.652 L) = n (0.0821 L·atm/mol·K) (273 K)

n = 0.0272 mol

Calculate the mass of C₃H₆ using its molar mass:

m = n × M

M(C₃H₆) = 42.08 g/mol

m = 0.0272 mol × 42.08 g/mol

m = 1.142 g

It is nearest to option D, hence the mass is 1.22 grams.

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The ratio of NH₃ to O₂ is less than 4:7, it means that NH₃ is the limiting reagent. Therefore, NH₃ will be completely consumed before O₂ and the amount of product formed will be determined by the amount of NH₃ available.

To determine the limiting reagent, we need to compare the amount of each reactant with their respective stoichiometric coefficients in the balanced chemical equation.

First, we convert the given masses of NH₃ and O₂ to moles using their molar masses:

58.3 g NH₃ × (1 mol NH₃ ÷ 17.03 g NH₃) = 3.42 mol NH₃

126 g O₂ × (1 mol O₂ ÷ 32 g O₂) = 3.94 mol O₂

Next, we compare the number of moles of NH₃ and O₂ to the stoichiometric coefficients in the balanced equation:

NH₃ : O₂ ratio = 4:7

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C. No Reaction will be the products when CuF2 reacts with Li

The positive and negative ions of two ionic compounds switch positions to generate two new compounds in a process known as a double replacement reaction. In aqueous solution, double-replacement reactions often take place between compounds.

In conclusion, you cannot balance a reaction by modifying or adding new components. To ensure that mass is preserved, the only thing you can do is alter the quantity of particles, or moles of particles, involved.

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The difference in the rate of sugar and acid reaction compared to the reaction between KI(aq) and Pb(NO₃)₂(aq) is due to the type of chemical bonding involved.

The reaction between sugar and acid involves covalent bonding, which is a strong bond that requires significant energy input to break. This type of bonding is responsible for the slow rate of the sugar and acid reaction.

In contrast, the reaction between KI(aq) and Pb(NO₃)₂(aq) involves ionic bonding, which is a much weaker bond than covalent bonding. As a result, the ions in the reactants are more easily separated, leading to a faster reaction rate.

Ionic bonding involves the transfer of electrons from one atom to another, whereas covalent bonding involves the sharing of electrons between atoms. This difference in electron sharing or transfer contributes to the different reaction rates observed between covalent and ionic bond containing compounds.

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When blackworms are collected from an environment with an acidic pH, it is expected that (B) the pulse rate of the blackworms would increase to minimize the effects of acidosis.

Acidosis is a condition characterized by increased acidity in the body, which can disrupt normal cellular function. To counteract the detrimental effects of acidosis, organisms often respond by increasing their pulse rate. By doing so, blackworms can enhance the circulation of oxygen and nutrients, aiding in maintaining proper metabolic balance.

Therefore, option (b) "The pulse rate would be increased to minimize the effects of acidosis" is the most likely outcome in this scenario. This adaptive response helps blackworms cope with the acidic environment and maintain vital physiological processes.

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To determine the number of moles of O2 needed to burn 9.45 moles of C2H2, we first need to write down the balanced chemical equation for the combustion of acetylene (C2H2):

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

From this equation, we can see that 5 moles of O2 are required to burn 2 moles of C2H2. To find out how many moles of O2 are needed for 9.45 moles of C2H2, we can use a simple proportion:

(5 moles O2 / 2 moles C2H2) = (x moles O2 / 9.45 moles C2H2)

To solve for x (moles of O2 needed), simply cross-multiply and divide:

x = (5 moles O2 * 9.45 moles C2H2) / 2 moles C2H2

x ≈ 23.63 moles O2

Therefore, approximately 23.63 moles of O2 are needed to burn 9.45 moles of C2H2.

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In the first reaction, the starting material (1-bromo-3-methylbutane) reacts with KCN, which acts as a nucleophile.

The cyanide ion (CN-) attacks the carbon with the bromo substituent, leading to a substitution reaction (SN2). As a result, product 1 is formed: 3-methylbutanenitrile.

In the second reaction, product 1 (3-methylbutanenitrile) reacts with hydroxide (OH-) and water (H2O), followed by the addition of H3O+ (hydronium ion).

This involves a two-step process: nucleophilic addition and hydrolysis. The hydroxide ion attacks the nitrile group, creating an intermediate which subsequently undergoes hydrolysis in the presence of H3O+ to form product 2: 3-methylbutanoic acid.

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Oxidation is the process in which an atom, ion, or molecule loses one or more electrons, resulting in an increase in its oxidation state. Reduction, on the other hand, is the process in which an atom, an ion, results in a decrease in its oxidation state. And only [tex]KBr[/tex]  [tex]CaF_2[/tex] would result in precipitate

These two processes occur simultaneously in a chemical reaction and are referred to as redox reactions. When a halide ion is mixed with a silver nitrate solution, a precipitation reaction may occur if the resulting compound is insoluble in water. [tex]KBr[/tex]  [tex]CaF_2[/tex] would result in a precipitate, as they form insoluble compounds with silver ions. [tex]MgCl_2[/tex], [tex]LiI[/tex] and [tex]NaF[/tex] would not result in a precipitate as they form soluble compounds with silver ions.

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--The complete Question is, What is the difference between oxidation and reduction in a chemical reaction?

Which of the following halide-containing salts, when mixed with a silver nitrate solution, would result in a precipitate: CaF2, MgCl2, LiI, NaF, or KBr? --

Answer:

12. Balanced chemical equation: Cu + Cl2 → CuCl2

Type of reaction: Combination or synthesis reaction

13. Balanced chemical equation: 2Ca(ClO3)2 → 2CaCl2 + 3O2

Type of reaction: Decomposition reaction

14. Balanced chemical equation: 2Li + 2H2O → 2LiOH + H2

Type of reaction: Single displacement or substitution reaction

Explanation:

Hope it helps^^

The mass of BSA in the solution is 24.96 μg, calculated by diluting 25 μL of 1.0 mg/mL BSA solution with 25 μL of distilled water and finding a final concentration of 0.0104 mg/mL.

To calculate the mass of BSA in the solution, we first need to find out how much BSA is present in the 25 μL of 1.0 mg/mL BSA solution.

1.0 mg/mL means that there is 1.0 mg of BSA per 1 mL of solution. Therefore, in 25 μL of solution (0.025 mL), there will be:

1.0 mg/mL x 0.025 mL = 0.025 mg of BSA

Next, we need to find out the concentration of BSA in the final solution after mixing. Since we are adding 25 μL of distilled water to the BSA solution, the volume of the BSA solution is now 50 μL (0.050 mL).

To calculate the concentration of BSA in the final solution, we can use the following formula:

C1V1 = C2V2

Where C1 is the initial concentration of BSA, V1 is the initial volume of the BSA solution, C2 is the final concentration of BSA, and V2 is the final volume of the solution.

We know that C1 = 1.0 mg/mL, V1 = 0.025 mL, V2 = 2.4 mL, and we want to find C2.

C2 = (C1V1)/V2 = (1.0 mg/mL x 0.025 mL)/2.4 mL = 0.0104 mg/mL

Now that we know the concentration of BSA in the final solution, we can calculate the mass of BSA in the solution by using the following formula:

mass = concentration x volume

The volume of the final solution is 2.4 mL. To convert this to μL, we need to multiply by 1000:

2.4 mL x 1000 μL/mL = 2400 μL

Now we can calculate the mass of BSA:

mass = 0.0104 mg/mL x 2400 μL = 24.96 μg

Therefore, the mass of BSA in the solution is 24.96 μg.

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To answer this question, we need to use the ideal gas law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. We also need to use the concept of molar volume, which is the volume occupied by one mole of a gas at a specific temperature and pressure.

First, we need to find the number of moles of gas that occupies 350.0L at standard pressure (1 atm) and temperature (273 K). This can be calculated using the formula n = PV/RT, where P = 1 atm, V = 350.0L, R = 0.08206 L atm/mol K, and T = 273 K. Substituting these values, we get n = (1 atm x 350.0L)/(0.08206 L atm/mol K x 273 K) = 14.15 mol.

Next, we need to find the molar volume of the gas at the pressure and volume we want it to occupy. Using the same formula, but with the new pressure (P') and volume (V') values, we get V' = nRT/P'. Since we want the gas to occupy 3.00L, we have V' = 3.00L. We also know that the number of moles (n) and temperature (T) are constant, so we can rearrange the formula to solve for the new pressure (P'). Thus, P' = nRT/V' = (14.15 mol x 0.08206 L atm/mol K x 273 K)/3.00L = 2,062.58 atm.

Therefore, the gas must be compressed to a pressure of 2,062.58 atm in order to occupy a volume of 3.00L, assuming constant temperature and number of moles. This is a very high pressure, and it highlights the importance of understanding the properties of gases and how they behave under different conditions.

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Answer: 1.20x10^24 atoms O

Explanation:

Oxygen is a diatomic element and is O2

Each molecule of oxygen, O2, has 2 atoms of O.

Each mole has 6.022 x 10^23 molecules of O2.

So our equation is

(6.022x10^23) x 2 = 1.2044x10^24 atoms of O2.

and because our initial problem uses 3 sig figs we round that to

1.20 x 10^24 atoms of O.

To solve this problem, we need to use the law of conservation of mass which states that the total mass of reactants equals the total mass of products in a chemical reaction. In this case, we know the mass of the element m and the mass of the compound m3n2 that is produced.

(i) To find the molar mass of the element, we need to first determine the number of moles of the compound produced. We can do this by dividing the mass of the compound by its molar mass.

molar mass of m3n2 = (molar mass of m x 3) + (molar mass of n x 2)
We can find the molar mass of the compound m3n2 by adding the molar mass of three atoms of element m and two atoms of nitrogen. The molar mass of nitrogen is 14 g/mol, and we can use the mass of the compound (43.5 g) to find its molar mass:

molar mass of m3n2 = (molar mass of m x 3) + (molar mass of n x 2)
43.5 g/mol = (3x molar mass of m) + (2x 14 g/mol)
43.5 g/mol - (2x14 g/mol) = 3x molar mass of m
15.5 g/mol = 3x molar mass of m
molar mass of m = 15.5 g/mol / 3 = 5.17 g/mol

So, the molar mass of element m is 5.17 g/mol.

(ii) To find the name of the element, we need to look at the periodic table and find an element with a molar mass close to 5.17 g/mol. From the periodic table, we see that the closest element is boron (B), which has a molar mass of 10.81 g/mol.

Therefore, the element m in this reaction is boron (B).

In summary, we can use the law of conservation of mass and the molar mass of the compound produced to determine the molar mass and name of the element reacted with nitrogen. In this case, we found that the element is boron with a molar mass of 5.17 g/mol.

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258.72 grams of CaBr2 is consumed when 96 g of Ca(OH)2 is produced in the given reaction.

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

CaBr2 + 2KOH → Ca(OH)2 + 2KBr

From the equation, we can see that 1 mole of CaBr2 reacts with 2 moles of KOH to produce 1 mole of Ca(OH)2 and 2 moles of KBr.

We need to first determine the number of moles of Ca(OH)2 produced from 96 g of Ca(OH)2. The molar mass of Ca(OH)2 is:

Ca(OH)2 = 1 x 40.08 (molar mass of Ca) + 2 x 16.00 (molar mass of O) + 2 x 1.01 (molar mass of H)

= 74.10 g/mol

Number of moles of Ca(OH)2 produced = Mass of Ca(OH)2 / Molar mass of Ca(OH)2

= 96 g / 74.10 g/mol

= 1.295 moles

From the balanced equation, we know that 1 mole of CaBr2 reacts with 1 mole of Ca(OH)2. Therefore, the number of moles of CaBr2 consumed in the reaction is also 1.295 moles.

Now, we can calculate the mass of CaBr2 consumed using its molar mass. The molar mass of CaBr2 is:

CaBr2 = 1 x 40.08 (molar mass of Ca) + 2 x 79.90 (molar mass of Br)

= 199.88 g/mol

Mass of CaBr2 consumed = Number of moles of CaBr2 consumed x Molar mass of CaBr2

= 1.295 moles x 199.88 g/mol

= 258.72 g

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The force applied to the 10 kg computer was 50 Newtons.

An electrical device with the capability to accept, store, process, and output data is known as a computer.

The following formula can be used to determine the force exerted on a 10 kilogram computer that is accelerating at a rate of 5 m/s2:

Force = mass x acceleration

Where

Plugging in these values, we get:

Force = 10 kg x 5 m/s²

Force = 50 N

Therefore, the force applied to the 10 kg computer was 50 Newtons.

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ans.

blank 1 = 1

blank 2 = 5

blank 3 = 3

blank 4 = 4

The gas laws are a set of fundamental principles that describe the behavior of gases under different conditions of pressure, volume, and temperature. We can use the ideal gas law to solve this problem:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin by adding 273.15:

T = 0.00 C + 273.15 = 273.15 K

Next, we can plug in the values we have:

P(22.4 L) = (1.00 mol)(0.0821 L·atm/mol·K)(273.15 K)

Simplifying:

P = (1.00 mol)(0.0821 L·atm/mol·K)(273.15 K)/(22.4 L)

P = 1.01 atm

Therefore, the helium will exert a pressure of 1.01 atm on its container.

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The new boiling point of the solution is 100°C + 0.199°C = 100.199°C.

The boiling point of a solution is dependent on the concentration of solute particles in the solvent. This can be calculated using the formula

ΔTb = Kbm

where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant, and m is the molality of the solution (moles of solute per kilogram of solvent).

The molar mass of CaS is 72.14 g/mol, so we can calculate the number of moles of CaS in the solution:

35 g / 72.14 g/mol = 0.4858 mol

The molality of the solution is then:

m = 0.4858 mol ÷ 1.25 kg

m = 0.3886 mol/kg

Next, we need to find the boiling point elevation constant Kb for water. Kb for water is 0.512 °C/m.

Finally, we can calculate the boiling point elevation:

ΔTb = Kb x m

ΔTb = 0.512 °C/m x 0.3886 mol/kg

ΔTb = 0.199 °C

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The substrates that will react are CH₃CH₂CH₂Br and CH₃CH₂CH₂CH₂Br  and (CH₃)₃CNH₂ and CH₃CH₂OH will not react with naome in an sn2 reaction to form an appreciable amount of product.

Based on the Sn2 reaction mechanism, substrates with good leaving groups and low steric hindrance are more likely to react with nucleophiles like NaOMe.

Therefore, the substrates CH₃CH₂Br, (CH₃)₂CHBr, CH₃CH₂I, and (CH₃)₃CBr are expected to react with NaOMe to form appreciable amounts of product. On the other hand, substrates with poor leaving groups or high steric hindrance are less likely to undergo Sn2 reactions.

Therefore, the substrates (CH₃)₃CNH₂ and CH₃CH₂OH are not expected to react with NaOMe to form appreciable amounts of product. Finally, CH₃CH₂CH₂Br and CH₃CH₂CH₂CH₂Br may react with NaOMe, but to a lesser extent due to their higher steric hindrance.

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Complete question :

Determine which of the substrates will and will not react with NaOMe in an Sy2 reaction to form an appreciable amount of product. Substrate will react Substrate will NOT react Answer Bank CH,CH.CH,BE (CH),CBE (CH), CHRE CH, CH,CH,NH, (CH),CCH,BE CH,CH.CH, OH

Estimating temperature to the nearest tenth of a degree

Estimating the mass of the sample to the nearest tenth of a gram

The brand of light bulb used for the heat lamp

Numerous things can go wrong, including human error, ambient variables, measuring instrument limits, and systematic or random deviations in the experimental technique.

The errors that would occur in this experiment can be seen to stem more from the nature of the estimation and are essentially errors that occur due to the computation of the results.

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The order of increasing atomic radii for the given elements is: Br < Sb < Se < Te.

When we talk about atomic radii, we are referring to the size of an atom. The atomic radius increases as we move down a group in the periodic table, and it decreases as we move across a period. This is because as we move down a group, the number of electron shells increases, leading to a larger atomic radius.

Conversely, as we move across a period, the number of protons in the nucleus increases, leading to a stronger attractive force on the electrons, resulting in a smaller atomic radius.

In the case of the four elements given - selenium (Se), antimony (Sb), bromine (Br), and tellurium (Te) - we need to determine their position in the periodic table to determine the order of increasing atomic radii.

Starting from the top, we have selenium (Se) and tellurium (Te) in the same group, but Te has a larger atomic number, so it has more electron shells, resulting in a larger atomic radius. Next, we have antimony (Sb), which is in the same period as Te, but with a smaller atomic number, meaning it has a smaller atomic radius.

Finally, we have bromine (Br), which has the smallest atomic number and is also in the same period as Sb, so it has the smallest atomic radius.

Therefore, the order of increasing atomic radii for the given elements is: Br < Sb < Se < Te.

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Answer:

For 25.6 grams of oxygen you will need 2*25.6 grams of hydrogen because water has two molecules of hydrogen and one molecule of oxygen the final mass of water is 76.8 grams