A student sets a 0. 4 kilogram (kg) soccer ball on the ground and gives it a hard kick. The ball traveled a distance of 37 meters (m). The acceleration of the ball was 2,550 meters per second squared (m/s). Based on the data, how much force in newtons (N) did the student apply to the ball?
A

Answer:

3-chloro propan-1-ol

Explanation:

As long as you're writing IUPAC name of a compound, you must mark that the sequence of carbon atoms in the chain from alpha-carbon, which is the carbon right next to the -OH group in this case

Answer:

displacement =1025km

time=6hours

Explanation:

velocity=displacement /time=1025/6=170.8km/h

The concentration is tested by titration with potassium permanganate, and each sample is tested twice Replication- 10/20/30 hours.

Hydrogen peroxide is a chemical compound that is gaseous in nature.

Since it is unstable it slowly decomposes in the presence of light to form water and oxygen. The control experiment is necessary for the determination of reliability and statistical significance of results obtained.

Testing the concentration of the species after exposure to air for 10,20 and 30 hours is randomization. Replication refers to the number of times the same experiment is carried out. It enhances the reliability of the result of an experiment.

Testing each sample twice is replication.

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Answer:

its a.

Explanation:

Answer:

so since the model atom was changed as new evidence, and the plum pudding model was a ball of positive charge, the alpha particle scattering which broke up the atom

Answer:

1.04 L.

Explanation:

What is given?

Temperature 1 (T1) = 240 K.

Pressure 1 (P1) = 670 mmHg.

Volume 1 (V1) = 1.28 L.

Temperature 2 (T2) = 198 K.

Pressure 2 (P2) = 680 mmHg.

What do we need? Volume 2 (V2).

Step-by-step solution:

To solve this problem, we have to use the combined gas law. The combined gas law expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas. For a combined gas law problem, only the amount of gas is held constant. The formula of combined gas law is:

We need to find the volume 2 (V2), so let's solve for this unknown value and let's replace the given data that we have:

The final volume for this case would be 1.04 L, the volume is being reduced.

The binding energy of silver-109 (Ag-109) in kJ/mol nucleons is not a well-defined concept, as binding energy is typically calculated for atomic nuclei rather than individual isotopes.

The binding energy of an atomic nucleus is the energy required to completely separate all of its constituent protons and neutrons into individual particles. It is usually expressed in units of energy per nucleon, which is the energy required to separate one proton or neutron from the nucleus.

The average binding energy per nucleon for an atomic nucleus is typically highest for medium-mass nuclei, such as those found in the region of the so-called "valley of stability" on the nuclear chart.

The binding energy per nucleon for silver-109 is not likely to be particularly high, as silver is a relatively heavy element and the binding energy per nucleon tends to decrease with increasing atomic number (Z).

Without more information about the specific calculation being used to determine the binding energy of Ag-109, it is not possible to accurately provide a value for the binding energy in kJ/mol nucleons.

The required masses (g/mol) are also not specified in the question, so it is not clear what context these values might be used in.

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Answer:

The thermometer in the [black] paper shows the greatest increase in temperature because black [absorbs] all wavelengths of visible light. The thermometer in the [white] paper shows the smallest increase in temperature because white [reflects] all colors of visible light. The thermometers in the green, red, and blue paper all show an increase in temperature. However, they don't increase as much as the thermometer in the black paper because these colors [absorb all other colors and reflect their own color.]

Explanation:

Hope this helps you understand!

The thermometer in the black paper shows the greatest increase in temperature because black absorbs all wavelengths of visible light. The thermometer in the white paper shows the smallest increase in temperature because white reflects all colors of visible light. The thermometers in the green, red, and blue paper all show an increase in temperature. However, they don't increase as much as the thermometer in the black paper because these colors absorb all other colors and reflect their own color.

The temperature of the water will increase by 0.177°C when 0.735 J of heat is added to 0.9916 g of water.

The specific heat capacity of water is 4.184 J/g °C, which means that 4.184 J of heat is needed to raise the temperature of 1 g of water by 1°C. The amount of heat required to raise the temperature of a substance is determined by its mass and specific heat capacity. The formula is:

Q = mcΔT

Where:

Q = Heat capacity

m = Mass of the substance

c = Specific heat capacity of the substance

ΔT = Change in temperature

To calculate the change in temperature when 0.735 J of heat is added to 0.9916 g of water, the formula can be rearranged as:

ΔT = Q / (mc)

ΔT = 0.735 J / (0.9916 g × 4.184 J/g °C)

ΔT = 0.735 J / (4.1486 J/g°C × g)

ΔT = 0.177°C (rounded to three significant figures)

Therefore, the temperature of the water will increase by 0.177°C when 0.735 J of heat is added to 0.9916 g of water.

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Answer:

By using a filter.

Explanation:

The chemical and biological indicators of water quality are substances that are used to measure the quality of water.

These indicators help to identify the presence of pollutants and other contaminants in water. Some common chemical indicators of water quality include pH, dissolved oxygen, total dissolved solids, and salinity. pH is a measure of the acidity or alkalinity of water, and can indicate the presence of pollutants such as acid rain.

Dissolved oxygen levels can indicate the level of organic matter in the water, while total dissolved solids can indicate the presence of pollutants such as salt. Salinity levels can indicate the presence of pollutants such as chloride.

Biological indicators of water quality include the presence of bacteria, viruses, and other microorganisms. These indicators can help to identify the presence of pathogens that can cause disease in humans and other animals. Some common biological indicators of water quality include fecal coliform bacteria, E. coli, and Giardia.

The presence of these organisms in water can indicate the presence of pollutants such as sewage or animal waste.

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The mass of chromium that could be deposited by electrolysis of an aqueous solution of Cr2(SO4)3 for 155 min using a constant current of 10.0 A is approximately X grams.

During the electrolysis process, the amount of substance deposited on an electrode can be determined using Faraday's laws of electrolysis. The first law states that the mass of substance deposited is directly proportional to the quantity of electricity passed through the electrolyte. The quantity of electricity can be calculated using the equation Q = I × t, where Q is the quantity of electricity in coulombs, I is the current in amperes, and t is the time in seconds. Since the current is given as 10.0 A and the time is 155 min (or 9300 s), we can calculate Q as Q = 10.0 A × 9300 s.

The next step is to determine the number of moles of chromium ions (Cr³⁺) in Cr2(SO4)3. Cr2(SO4)3 contains 2 moles of chromium ions per formula unit. To find the number of moles of chromium ions, we divide the quantity of electricity (Q) by the Faraday constant (F), which is approximately 96,485 coulombs per mole of electrons. Thus, the number of moles of chromium ions is Q / F.

Finally, to calculate the mass of chromium, we multiply the number of moles of chromium ions by the molar mass of chromium, which is approximately 52.0 g/mol. Therefore, the mass of chromium deposited is equal to the number of moles of chromium ions multiplied by the molar mass of chromium.

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The isomer of the molecule in the depicted image would be A.

Isomers are molecules with the same chemical formula but different arrangements of their component atoms. Isomers also have different properties.

The original molecule in the image has a triple bond with 4 carbon atoms. Thus, its isomers must also have a triple bond with 4 carbon atoms.

The only option with a triple bond and 4 carbon atoms in the image is option A.

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The number of molecules in 1.00 mol of iodine gas is 6.02 × 10²³.

A mole is defined as the amount of a substance which contains 6.022 ×10²³ entities like  atoms, ions, particles, molecules, etc. of the given substance. Mole will measures the number of ions, atoms, or molecules. The number of moles of a substance in a given pure sample will be represented by the following formula:

n = N/NA

Now, a mole is termed as very large collection of particles. In order to have 1.00

moles of iodine crystals, you need to have exactly 6.022 × 10²³ molecules of iodine.

1 mole of I₂ = 6.022 × 10²³ molecules of I₂ (Avogadro's constant)

We have, 1.00 moles of I₂ = 6.02 × 10²³ molecules of I₂.

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Answer:

coulomb force/ electrostatic force

Explanation:

attraction or repulsion of particles

Answer:

coulombe law or coulomb's universe square law

Answer: The volume \((V_{1})\) of 1.00 mol of helium at 273 K is 22.4 L and the final volume \((V_{2})\) is 44.8 L.

Explanation:

Given: \(T_{1}\) = 273 K,   \(n_{1}\) = 1.00 mol

\(T_{2}\) = 273 K,         \(n_{2}\) = 2.00 mol

At the standard pressure, 1 atm the value of \(V_{1}\) and \(V_{2}\) is calculated as follows.

\(V_{1} = \frac{n_{1}RT_{1}}{P}\\= \frac{1.00 mol \times 0.0821 Latm/mol K \times 273 K}{1 atm}\\= 22.4 L\)

Similarly,

\(V_{1} = \frac{n_{2}RT_{2}}{P}\\= \frac{2.00 mol \times 0.0821 Latm/mol K \times 273 K}{1 atm}\\= 44.8 L\)

Thus, we can conclude that the volume \((V_{1})\) of 1.00 mol of helium at 273 K is 22.4 L and the final volume \((V_{2})\) is 44.8 L.

Bigger size elements have low ionization energy and high shielding effect because:

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Explanation:

A prism is a solid geometric figure with two identical parallel bases that are both polygons, and the lateral faces are parallelograms with pairs of opposite sides of equal length.

33.34 J of energy it would take to raise the temperature.

Specific heat = J/g C

(25.35 J/mole C) x (1 mole / 107.8682g) = 0.2350 J/g C

Energy required = (mass in grams)x(Specific heat)x(Temp final - Temp initial)

= (8.20 g)(0.2350 J/g C)(17.3°C) = 33.34 J (units cancel for /g C)

Heat capacity or specific heat is the quantity of heat consistent with unit mass that is required to elevate the temperature with the aid of 1°C. particular warmness is helpful in figuring out the processing temperatures and quantity of warmth vital for processing and may be helpful in differentiating between  polymeric composites.

The specific heat capacity (or in reality, the specific heat), that's the heat potential according to unit mass of a material. Experiments show that the transferred heat relies upon on three factors: (1) The change in temperature, (2) the mass of the system, and (3) the substance and phase of the substance.

The specific heat is the quantity of heat energy per unit mass required to elevate the temperature by way of one degree Celsius. The connection between warmth and temperature change is commonly expressed inside the form shown under where c is the unique warmness .

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The pressure of the gas inside the cylinder is 52.90 atm

Given is the number of moles of gas, the temperature and the volume of the gas and we need to find the pressure of the gas inside the cylinder, for this we can use the ideal gas law equation:

PV = nRT

Where:

P = Pressure of the gas (in units of pressure, such as atm)

V = Volume of the gas (in liters)

n = Number of moles of the gas

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

T = Temperature of the gas (in Kelvin)

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

T = 25.0 °C + 273.15 = 298.15 K

Now we can substitute the values into the ideal gas law equation:

P × 12.5 L = 27.0 moles × 0.0821 L·atm/(mol·K) × 298.15 K

Simplifying the equation:

P × 12.5 L = 661.2587 L·atm

Dividing both sides by 12.5 L:

P = 661.2587 L·atm / 12.5 L

P ≈ 52.90 atm

Therefore, the pressure of the gas inside the cylinder is approximately 52.90 atm.

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We can use the ideal gas law equation to determine the pressure of a gas within a cylinder:

PV = nRT

Where:

P is the pressure of the gas (in units of pressure, such as atm)

V is the volume of the gas (in units of volume, such as liters)

n is the number of moles of the gas

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

T is the temperature of the gas (in units of temperature, such as Kelvin)

we need to convert the temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15

T(K) = 25.0 °C + 273.15

T(K) = 298.15 K

Now we can plug the data into the ideal gas law equation as follows:

P * 12.5 L = 27.0 moles * 0.0821 L·atm/(mol·K) * 298.15 K

Simplifying the equation:

P = (27.0 moles * 0.0821 L·atm/(mol·K) * 298.15 K) / 12.5 L

Calculating the pressure:

P ≈ 5.046 atm

As a result, the gas inside the cylinder is under a pressure of about 5.046 atm.

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Answer:

Californium is a chemical element which is radioactive, its symbol is 'Cf' and atomic number is 98

Answer: the elementale symble is Cf

Explanation: the symble of element is Cf

Answer: True

Explanation: Pressure is not dependent on mass.

Answer:

1/32

Explanation:

Hydrocyanic and hydrofluoric are both weak acids (W), while phenol is a strong acid (S).

Acids are substances in water that can be ionized to release hydrogen ions or hydronium ions. While a base is a substance in water that can be ionized releasing hydroxide ions.

This classification is based on the strength of their conjugate bases. Hydrocyanic acid (HCN) and hydrofluoric acid (HF) have weakly basic conjugate bases (CN⁻and F⁻), meaning they do not readily donate a proton (H⁺) to water molecules. Phenol (C₆H₅OH), on the other hand, has a strongly basic conjugate base (C₆H₅O⁻), meaning it readily donates a proton to water molecules, making it a strong acid.

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Kilocalorie is a unit of measuring the amount of energy of a reaction, but this is not the only unit, we can also have Joules as a unit, and the conversion is:

1 Kcal = 4184 Joules

Therefore if we have 3.25 Kcal, we will have:

3.25 * 4184 = 13600 Joules of energy, therefore letter D

Answer:Because the amount of water vapor in the air increases

Explanation:In humid climates, thunderstorms often cause heavier rain than general wintertime rainfall since moisture content in the air typically is higher in the spring and summer, and since air usually rises at a much more rapid rate within developing thunderstorms than in general winter systems.During very heavy rainfall, where all of the water in the atmosphere is often rained out, the water vapour content governs the amount of rainfall. The higher the humidity the greater the water vapour, and the more rain we're likely to see. HOPE THIS HELPS ^w^

The mass of HI should be present in 0.200l of solution to obtain a solution with pH value's,

(a) pH value is 1.20 the mass is 1.08g

(b) pH value is 1.75 the mass is 0.0066g

(c) pH value is 2.85 the mass is 0.00012g

To solve this problem, we must determine the concentration of H+ ions in the solution using the pH of the solution and the dissociation constant of HI. The concentration of HI and the mass of HI required to make the solution may then be calculated.

The dissociation reaction for HI is:

HI(aq) ↔ H+(aq) + I-(aq)

The dissociation constant, Ka, for this reaction, is:

Ka = [H+][I-]/[HI]

This formula may be simplified by assuming that the starting concentration of HI is equal to the concentration of I- produced, which is equal to the concentration of H+ produced due to the reaction's 1:1 stoichiometry. This results in:

Ka = [H+]^2/[HI]

Solving for [H+], we get:

[H+] = sqrt(Ka*[HI])

Taking the negative log of both sides gives us the pH of the solution:

pH = -log[H+] = -log(sqrt(Ka*[HI]))

pH= -0.5*log(Ka) - 0.5*log([HI])

Rearranging this equation, we get:

[HI] = 10^(-(pH + 0.5*log(Ka)))/V

where V is the volume of the solution.

Now we can calculate the mass of HI required for each pH:

(a) For pH = 1.20:

Ka for HI is 1.3 x 10^-10. Substituting this value into the equation above, we get:

[HI] = 10^(-(1.20 + 0.5*log(1.3 x 10^-10)))/0.200L ≈ 0.0042 M

The mass of HI required is:

mass = concentration x volume x molar mass

     = 0.0042 mol/L x 0.200 L x 127.91 g/mol

     ≈ 1.08 g

Therefore, approximately 1.08 grams of HI is required to prepare a solution with a pH of 1.20.

(b) For pH = 1.75:

[HI] = 10^(-(1.75 + 0.5*log(1.3 x 10^-10)))/0.200L ≈ 0.00026 M

mass = 0.00026 mol/L x 0.200 L x 127.91 g/mol ≈ 0.0066 g

Therefore, approximately 0.0066 grams of HI is required to prepare a solution with a pH of 1.75.

(c) For pH = 2.85:

[HI] = 10^(-(2.85 + 0.5*log(1.3 x 10^-10)))/0.200L ≈ 0.0000047 M

mass = 0.0000047 mol/L x 0.200 L x 127.91 g/mol ≈ 0.00012 g

Therefore, approximately 0.00012 grams of HI is required to prepare a solution with a pH of 2.85.

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The values of the Graduations for A , B and C will be  0.2 ml,  2ml and 1ml respectively.

Graduated Cylinder - A popular piece of scientific equipment used to measure the volume of a liquid is a graduated cylinder, sometimes referred to as a measuring cylinder or mixing cylinder. Its form is slender and cylindrical. Each marked line on the graduated cylinder indicates the volume of liquid that has been measured.

The clear graduated cylinder is widely used for volume measurements and is regarded to be more accurate than a beaker since it contains permanently marked incremental graduations.

To calculate the value of graduations-

1. Subtract the two values to get the value between the labeled graduations.

2. Determine how many spaces there are between the two graduations. Keep in mind that volume equals space.

3. Divide the number of gaps by the value between the graduations.

For cylinder  A.

2 ml- 1ml  = 1ml

Number of spaces = 5

∴1/ 5 = 0.2

Graduation = 0.2ml

For cylinder B -

20-10 = 10

number of spaces = 5

∴10/ 5 =2

Graduation = 2ml

For cylinder C-

10-5 = 5

Number of spaces= 5

∴5/5 = 1

Graduation = 1ml

Hence , Graduations for A , B and C will be  0.2 ml,  2ml and 1ml respectively.

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