The tomato is dropped. What is the velocity, v
, of the tomato when it hits the ground? Assume 86.0 %
of the work done in Part A is transferred to kinetic energy, E
, by the time the tomato hits the ground.
Express your answer with the appropriate units.

The balanced chemical equation is 3 Sr(OH)₂ + Al₂(SO₄)₃\(\rightarrow\) 3 SrSO₄ + 2 Al(OH)₃.

Chemical equation is a symbolic representation of a chemical reaction which is written in the form of symbols and chemical formulas.The reactants are present on the left hand side while the products are present on the right hand side.

The first chemical equation was put forth by Jean Beguin in 1615.By making use of chemical equations the direction of reaction ,state of reactants and products can be stated. In the chemical equations even the temperature to be maintained and catalyst can be mentioned.

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The half reaction with a more positive standard reduction potential will proceed spontaneously in a redox reaction

The half reaction with a more positive standard reduction potential will undergo reduction when compared to the half reaction with a more negative standard reduction potential.

Oxidation-reduction reactions, often known as redox reactions, are a set of chemical reactions that involve electron transfer between reactants. In a redox reaction, one reactant is oxidized, losing electrons, while the other reactant is reduced, gaining electrons.

The oxidation half-reaction is the process of losing electrons and increasing the oxidation number, whereas the reduction half-reaction is the process of gaining electrons and decreasing the oxidation number. The total reaction is referred to as the redox reaction.

Half-reaction:Half-reaction refers to the two parts of an oxidation-reduction reaction that happen separately. A half-reaction must always be either an oxidation reaction or a reduction reaction. It also describes the movement of electrons and hydrogen ions in an equation.

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Answer:

Na⁺ (aq) + OH⁺ (aq) + H⁺ (aq) + Cl⁻ (aq) → Na⁺ (aq) + Cl⁻ (aq) + H₂O (l)

General Formulas and Concepts:

Atomic Structure

Aqueous Solutions

Stoichiometry

Explanation:

Step 1: Define

NaOH reacting w/ HCl

NaOH is soluble

HCl is soluble

[RxN] NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)

Step 2: Total Ionic Equation

Break up soluble compounds into ionic form.

[T.I.E] Na⁺ (aq) + OH⁺ (aq) + H⁺ (aq) + Cl⁻ (aq) → Na⁺ (aq) + Cl⁻ (aq) + H₂O (l)

The molecules of Li₂O that would produce if you used up 6 atoms of Li(s) is 3.

It is an unstable inorganic and radical compound, used in the manufacture of glass and ceramic.

The balanced equation

\(\rm 6 Li(s) + O_2(g) \rightarrow 3 Li_2O(s)\)

If the atoms of lithium moves up to 6 the lithium oxide molecules will be 3, as you can see in the equation.

Thus, the molecules of Li will be 3.

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Benzoic acid and sodium nitrate can be separated through extraction technique.

Extraction involves the transfer of compound from either solid or liquid into a different solvent. This technique uses inorganic chemistry to isolate or remove a target compound.

The separation of sodium nitrate and benzoic acid is based on their polarity differences and any other components of the mixture.

Therefore, extraction technique can be used to separate a mixture of benzoic acid and sodium nitrate.

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Answer:

Starch is long chains of sugar molecules linked together. ... There are two forms of starch: amylose, the branchless form, and amylopectin, the branched form. The main function of starch is as way to store energy for plants. Starch is a source of sugar in an animal's diet.

Answer:

3.3 * 10^-12

Explanation:

The balanced equation of the reaction is;

2H+(aq) + Cu(s) ---------> H2(g) + Cu2+(aq)

Hence two electrons were transferred so n=2

E°cell = E°cathode - E°anode

E°cell = 0 V - 0.34 V

E°cell = - 0.34 V

Then;

E°cell = 0.0592/n log K

Substituting values;

- 0.34 = 0.0592/2 log K

- 0.34/0.0296 = log K

-11.486 = log K

K = Antilog (-11.486)

K = 3.3 * 10^-12

Answer:

14.9 mol

Explanation:

To find the number of moles in a given mass of a sample of sodium chloride (NaCl), we can multiply the number of grams in the sample by the molar mass of sodium chloride, which is 58.44 g/mol.

871 g × (1 mol / 58.44 g)

= 871/58.44 mol

≈ 14.9 mol

Note that we rounded to 3 significant figures in the final answer because that is how many significant figures were given in the mass measurement of the sodium chloride sample.

Answer:

The third option

Explanation:

This reaction is a single replacement reaction because K replaced Mg and a redox reaction because the ion charges of the elements changed from a 0 charge on K to +1 charge, and a +2 charge on Mg to a 0 charge.

Answer:

Reverse faults are caused by a type of stress known as compression where two sections of rock push into one another along a fault. The compression causes one side of rock to move either above or below the other side

The question is incomplete, the complete question is;

In Sample Exercise 10.16, we found that one mole of Cl2 confined to 22.41 L at 0 °C deviated slightly from ideal behavior. Calculate the pressure exerted by 1.00 mol Cl2 confined to a smaller volume, 5.00 L, at 25 °C. (a) First use the ideal-gas equation and (b) then use the van der Waals equation for your calculation. (Values for the van der Waals constants are given in Table 10.3.) (c) Why is the difference between the result for an ideal gas and that calculated using the van der Waals equation greater when the gas is confined to 5.00 L compared to 22.4 L?

Answer:

See explanation

Explanation:

Since we have

For 5.00 L using Ideal gas law;

PV= nRT

P = nRT/V

P = 1 * 0.082 * 298 K/5.00

P= 4.887 atm

Using Van der walls equation;

[P + an^2/v^2] *  [V - nb] = nRT

P = (nRT/ [V - nb]) - (an^2/v^2)

P= 1 * 0.082 * 298/5 - (1 * 0.0562) - (6.49(1^2)/5^2)

P = (24.436/4.9438) - 0.2596

P= 4.683 atm

For 22.4 L

P = nRT/V

P = 1 * 0.082 * 298 K/22.4

P= 1.0909 atm

Using Van der Walls equation;

[P + an^2/v^2] *  [V - nb] = nRT

P = (nRT/ [V - nb]) - (an^2/v^2)

P= 1 * 0.082 * 298/22.4 - (1 * 0.0562) - (6.49(1^2)/22.4^2)

P= (24.436/ 22.3438) - 0.0129

P= 1.081 atm

According to Boyle's law, pressure is inversely proportional to volume. Increase in volume leads to decrease in pressure. Therefore the pressure calculated both by the ideal gas equation and Van der Waals equation at 5.00 L is greater than that calculated using both methods at 22.4 L

Answer:

“You Asked” is a series where Earth Institute experts tackle reader questions on science and sustainability. Over the past few years, we’ve received a lot of questions about carbon dioxide — how it traps heat, how it can have such a big effect if it only makes up a tiny percentage of the atmosphere, and more. With the help of Jason Smerdon, a climate scientist at Columbia University’s Lamont-Doherty Earth Observatory, we answer several of those questions here.

It is the 4th scenario is the dependent event. There are 7 gold tokens and 4 silver tokens in a cup. The first student randomly draws a gold token and keeps it. A second student randomly draws a gold token from the cup.

The fouth scenario is a dependent event because the probability of the second student drawing a gold token is affected by the outcome of the first student's draw.

If the first student draws a gold token, then there are only 6 gold tokens left in the cup, the probability changes. but  if the first student does not draw a gold token, then there are 7 gold tokens left in the cup, the probability will remain the same

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Answer:

23.3mL of 0.150M sodium thiosulfate

Explanation:

The  net reaction of sodium hypochlorite (NaClO) with sodium thiosulfate (Na₂S₂O₃) is:

NaClO + 2 Na₂S₂O₃ + 2H₃O⁺ → NaCl + 3 H₂O  +  Na₂S₄O₆ + 2 Na⁺

2.0mL of the sample of bleach are:

2.0mL ₓ (1.084g / mL) ₓ (6 / 100) ₓ (1 mol / 74.44g) = 1.75x10⁻³ moles of NaClO

As 1 mole of NaClO reacts with 2 moles of thiosulfate:

1.75x10⁻³ moles of NaClO ₓ (2 mol Na₂S₂O₃ / 1 mol NaClO) =

3.50x10⁻³ moles of Na₂S₂O₃

If you have a 0.150M solution of thiosulfate, mL of this solution you need to reach the end point of the titration are:

3.50x10⁻³ moles of Na₂S₂O₃ ₓ (1L / 0.150mol) = 0.0233L =

The bactericidal activity of sanitizers prepared from the water will be reduced if the hardness or pH of the water used to prepare EO water or bleach solutions are increased.

Temperature, pH, relative humidity, and water hardness are other physical and chemical variables that affect disinfection processes. For instance, most disinfectants become more active as the temperature rises, but there are some exceptions.

The harder your water is, the less effective cleaning products will be since it will be difficult to make a soapy lather. Scale, which are crusty deposits made by hard water, can build up in your dishwasher or washing machine. Detergent works effectively when used with soft water.

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26.71 g mass of tin would be required to completely react with 1.20 L of 0.750 M HBr.

What is mass ?

Mass is a measure of the amount of matter in an object or substance. It is often expressed in units of grams (g) or kilograms (kg). Mass is a fundamental property of matter and is different from weight, which is the force exerted on an object by gravity and varies depending on the object's location. Mass can be measured using a balance or scale, and is an important factor in many chemical calculations and experiments, such as determining the amount of reactants needed for a reaction or the concentration of a solution.

First, we need to determine the number of moles of HBr in the solution:

moles of HBr = Molarity x Volume

moles of HBr = 0.750 mol/L x 1.20 L

moles of HBr = 0.900 mol

According to the balanced chemical equation, 1 mole of Sn reacts with 4 moles of HBr to produce 1 mole of SnBr₄. Therefore, the number of moles of Sn required can be calculated as:

moles of Sn = (moles of HBr) / 4

moles of Sn = 0.900 mol / 4

moles of Sn = 0.225 mol

The molar mass of Sn is 118.71 g/mol, so the mass of Sn required can be calculated as:

mass of Sn = moles of Sn x molar mass of Sn

mass of Sn = 0.225 mol x 118.71 g/mol

mass of Sn = 26.71 g

Therefore, 26.71 g of tin would be required to completely react with 1.20 L of 0.750 M HBr.

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Complete question is:  26.71 g mass in grams of tin would be required to completely react with 1.20 L of 0.750 M HBr in the chemical reaction. Sn(s) + 4 HBr(aq) → SnBr₄ (aq) + 2 H₂ (g)

In an acetic acid solution:

1. To make a buffer with pH = 5.00, have a ratio of

\(\frac{[A-]}{[HA]} = 10^{-5.50}\) = 0.316.

The volume of sodium hydroxide needed:

V(NaOH) = (0.316 M - 0.200 M) / 4.50 M = 0.0316 L = 31.6 mL

Therefore, 31.6 mL of 4.50 M sodium hydroxide must be added to 250.0 mL of 0.200 M acetic acid solution to make a buffer with pH = 5.00.

2. The pH of the buffer is calculated as follows:

pH = pKa + log(\(\frac{[A-]}{[HA]}\))

= 4.76 + log(0.2/0.1)

= 4.86

Therefore, the pH of the buffer is 4.86.

3. The mass of salt that must be dissolved in 0.25 dm³ of 1 mol dm³ propanoic acid to give a buffer of pH 4.87:

\(\frac{[A-]}{[HA]} = 10^{-4.87}\) = 0.0114

Therefore, the mass of acetate that must be dissolved:

Mass of acetate = (0.0114 mol dm³)(0.25 dm³) = 0.00285 g

Therefore, 0.00285 g of sodium propanoate must be dissolved in 0.25 dm³ of 1 mol dm³ propanoic acid to give a buffer of pH 4.87.

4. The pH of the buffer is calculated as follows:

pH = pKa + log(\(\frac{[A-]}{[HA]}\))

= 4.74 + log(0.1/0.1)

= 4.74

Therefore, the pH of the buffer is 4.74.

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Answer:

because there are other particles to help it move around in the air

The given instructions outline the steps to follow in a simulation or experiment involving particles A and B. The purpose is to observe and record the number of A and B particles at each minute on the timer. The exact observations and data collection will depend on the specific simulation or experiment being conducted.

Locate the orange reset button: Find the reset button on the bottom right side of the screen and press it to start the reaction over. This ensures that the previous data is cleared and the experiment begins from the initial state.

Drag the ruler: Use the mouse or touch screen to drag the top of the ruler upward until it reaches the 40 mark. This step sets the reference point for measuring the positions of the particles.

Adjust the left platform: Drag the left platform upward until the top of the platform aligns with the 30 mark on the ruler. This step positions the platform for the particles to interact within the desired range.

Toggle the blue play/pause button: Locate the blue play/pause button at the bottom of the screen. Make sure it is in the play position to prevent the reaction from starting before you are ready. This allows you to control the timing of the experiment.

Add 50 A particles: Use the appropriate tool or feature to add 50 A particles to the simulation or experimental setup. This step ensures that the initial condition includes a specific number of A particles.

Start the timer and record observations: Press the play button on the blue box to start the simulation or experiment. Immediately start the timer using the play button on the timer itself. At every minute on the timer, pause the simulation and record the number of A and B particles. This step allows you to track the changes in the particle population over time.

Note: The specific details and actions may vary depending on the simulation or experiment being conducted. It is important to follow the given instructions accurately and record the observations as instructed.

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Our current understanding of the atom is based on Dalton's early 19th-century idea of atomism, which was developed through meteorological investigations. John Dalton is well recognised for developing the atomism hypothesis.

When British chemist John Dalton realised that compounds usually had whole number ratios of atoms, he provided the first modern proof for the existence of atoms. Because of this, it's H2O rather than H20. A novel state of matter has been demonstrated by scientists: an electron circles a nucleus at a considerable distance while being bonded to several other atoms inside the orbit. The Bose-Einstein condensate's electron (blue) orbits the nucleus (red), and the orbit encompasses a large number of other atoms (green). Chemistry's fundamental building component is an atom. It is the lowest fraction of substance into which electrically charged particles cannot be released.

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The least specific heat is possessed by substance A (0.90 J/g-C). Since equal masses of all the substances are heated to the same temperature, material A will therefore cool the fastest.

Also known as specific heat, this is the amount of energy required to increase a substance's temperature by one degree Celsius in one gramme. The units of specific heat are typically calories or joules per gramme per degree Celsius. For instance, water has a specific heat of 1 calorie per gramme per degree Celsius.

According to its 4.186 J/g°C specific heat capacity, water needs 4.186 J of energy (or 1 calorie) to heat a gramme by 1 degree.

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Answer:

Explanation:

Here, we want to get the equation of the reaction between Hydrochloric acid and Zinc metal

Zinc metal displaces the hydrogen from hydrochloric acid to form zinc chloride

We have the equation of reaction as:

The partial charges on each fluorine atom are negative. Option B) The central O atom is the correct answer. Option B

The partial charges in a molecule are determined by the electronegativity values of the atoms involved. Electronegativity is the ability of an atom to attract electrons towards itself in a chemical bond. In the case of \(F_2O\), fluorine (F) is more electronegative than oxygen.

Fluorine is the most electronegative element on the periodic table, meaning it has a high ability to attract electrons. Oxygen is also relatively electronegative but less so than fluorine. When fluorine atoms bond with oxygen, the shared electrons will be pulled more towards the fluorine atoms, creating a polar covalent bond.

In \(F_2O\), each fluorine atom will pull the shared electrons towards itself, resulting in a higher electron density around the fluorine atoms. This creates a region of partial negative charge around the fluorine atoms.

Conversely, the oxygen atom will have a region of lower electron density and, therefore, a partial positive charge. This is because the shared electrons spend more time around the fluorine atoms due to their higher electronegativity.

Option B

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The density of a sample of NH\(_3\)(g) at a pressure of 1.00 atm is 0.869 g/L. 592m/s is the square velocity (in m/s) of the molecules in this sample.

The velocity of a gas molecule at a certain temperature, which corresponds to point P in the graph, is 200 ms. At the given temperature, the rms velocity of a gas molecule is about. Molecular count (A) 163ms (C) 245ms (B) 217 milliseconds; (D) 226 milliseconds. The rms velocity of a particular amount of gas molecules at 27oC and 1.0 105 Nm⁻² pressure is 200 msec-1.

velocity = √3p/density

           =√3/0.869

          =592m/s

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At STP, one mole of any substance contains 22.41 L. 4 moles of HF gives 2 moles of F₂. Then, 0.76 L of HF will give 0.38 L of fluorine gas.

STP condition is the standard temperature and pressure where the temperature is 298 K at 1 atm pressure. At STP one mole of every substance contains 22.41 L volume.

Then, volume of 4 moles of HF = 4 × 22.41 = 89.6 L

volume of 2 moles of F₂ = 44.8 L.

89.6 L of HF gives 44.8 L of the gas product.

Then, 0.76 L of HF will give:

(44.8 × 0.76)/79.6 = 0.38 L

Therefore, the volume of F₂ produced will be 0.38 L.

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Answer:

The colour of the orange solution becomes yellow.  

Explanation:

1. Before adding NaOH

Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.

Both solutions are involved in the same equilibrium:

\(\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O\)

The first beaker contains mostly chromate ions with a few dichromate ions.

The position of equilibrium lies to the left and the solution is yellow.

The second beaker contains mostly dichromate ions with a few chromate ions.

The position of equilibrium lies to the right and the solution is orange.

2. After adding NaOH

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Beaker 1

If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).

The system responds by having the dichromate react with water to replace the H⁺.  

At the same time, the system forms more of the yellow chromate ion.

The position of equilibrium shifts to the left.

However, the solution is already yellow, so you see no change in colour.

Beaker 2

The reaction is the same as in Beaker 1.

This time, however, as the dichromate ion disappears, do does its orange colour.

Also, the yellow chromate is being formed and its yellow colour appears .

The colour changes from orange to yellow.

Answer:

A. Radar would be the best tool for measuring distances to objects in the solar system.

B. A parallax measurement would be best for this nearby star.

C. Cepheids or RR Lyraes would be useful for determining the distance to this cluster.

D. The method using the H-R diagram and getting a spectrum to determine the luminosity class of the star.

Answer:

it is the one below that. NO, because it debt net the octet rule