Which statement BEST describes how a golf club does "work" on a golf ball?
(A) When the club hits the ball the club transfers all of its kinetic energy to the ball.
(B) All of the kinetic energy from the club is transferred to the ball as they both move through the air.
(C)
Some of the kinetic energy from the golf club is transferred to the ball and some transforms into sound
and heat, but the total energy remains the same.
(D) The golf club loses kinetic energy when it hits the ball and the ball gains kinetic energy from the air as it
travels

10366.4J is the amount of energy that the water has in this experiment absorb according to  the calorimeter data.

Energy, which is observable in the execution of labour as well as through the emission of heat and light, is the quantitative quality that is transmitted to a body or into a physical system in physics. Energy is a preserved resource; according to the rule of conservation of energy, energy can only be transformed from one form to another and cannot be generated or destroyed. The joule (J) is the unit of quantification for electricity in the internationally recognised System of Units (SI). A moving object's kinetic energy and an object's stored potential energy are examples of common types of energy.

q = m×c×ΔT

q = 100×4.18×(46-21.2)

   = 10366.4J

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Answer:

1.8 J/g·°C (2 sig.figs. per ΔT)

Explanation:

Given:

mass (m) = 78.9 grams

specific heat (c) = ?

Temp. Change (ΔT) = 109°C - 18°C = 91°C

Heat flow (q) = 13,240 Joules

q = m·c·ΔT => c = q/m·ΔT

∴c = 13,240J / 78.9g·91°C = 1.844 J/g·°C ≅ 1.8 J/g·°C (2 sig.figs. per ΔT)

Answer:

First, let's calculate the number of moles of H₂SO₄:

Number of moles of H₂SO₄ = mass of H₂SO₄ / molar mass of H₂SO₄

Number of moles of H₂SO₄ = 583 g / 98.079 g/mol

Number of moles of H₂SO₄ = 5.944 mol

Next, let's calculate the number of moles of water:

Number of moles of water = mass of water / molar mass of water

Number of moles of water = 1.50 kg / 18.0153 g/mol

Number of moles of water = 83.277 mol

The mole fraction of H₂SO₄ can now be calculated as follows:

Mole fraction of H₂SO₄ = moles of H₂SO₄ / total moles

Mole fraction of H₂SO₄ = 5.944 mol / (5.944 mol + 83.277 mol)

Mole fraction of H₂SO₄ = 0.0666

Therefore, the mole fraction of H₂SO₄ in the solution is 0.0666 (rounded to four significant figures).

The word "symptom" refers to a specific manifestation or indication of a condition, disease, or disorder that is experienced or observed by an individual.

Symptoms are subjective or objective changes in the body's normal functioning that may be recognized as abnormal, uncomfortable, or problematic. Symptoms can manifest in various ways depending on the nature of the underlying condition. They can be physical, such as pain, rash, cough, fever, or fatigue, indicating an illness or injury affecting the body. Symptoms can also be psychological, such as anxiety, depression, or confusion, reflecting disturbances in mental health.

Symptoms serve as important clues for medical professionals to identify and diagnose diseases or disorders. They provide valuable information about the nature, severity, and progression of an illness, helping healthcare providers formulate appropriate treatment plans. Additionally, symptoms may also be important for individuals to self-assess their own health status and seek appropriate medical attention.

It is essential to note that symptoms alone may not provide a definitive diagnosis, as they can overlap across different conditions. Further evaluation, including medical tests and examinations, is often necessary to confirm a diagnosis and determine the appropriate course of action.

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Answer:b

Explanation:

Answer:

heat the solution

Explanation:

i think

Answer:

The way to make a supersaturated solution is to add heat, but just a little heat won't do the job. You have to heat the water close to the boiling point. When the water gets this hot, the water molecules have more freedom to move around, and there is more space for solute molecules between them.

Answer: D. The energy needed to change a liquid into a gas

The amount of heat energy produced when 1 g of hydrogen and 2 g of nitrogen are reacted, is -6.61 KJ

First, we shall obtain the limiting reactant. Details below:

3H₂ + N₂ -> 2NH₃

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂

Therefore,

2 g of N₂ will react with = (2 × 6) / 28 = 0.43 g of H₂

We can see that only 0.43 g of H₂ is needed in the reaction.

Thus, the limiting reactant is N₂

Finally, we the amount of heat energy produced. Details below:

3H₂ + N₂ -> 2NH₃ ΔH = -92.6 KJ

From the balanced equation above,

When 28 grams of N₂ reacted, -92.6 KJ of heat energy were produced.

Therefore,

When 2 grams of N₂ will react to produce = (2 × -92.6) / 28 = -6.61 KJ

Thus the heat energy produced from the reaction is -6.61 KJ

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Answer:

47.25 moles

Explanation:

First we find the amount of O2 in mass in air

That will be = (21/100) x 3.6

This will give us=0.756 kg

Now we want to find the moles, so we convert this mass into grams since we are going to be working with grams= 0.756 x 1000 = 756 grams

Now, to find the number of moles:

moles= mass / molar mass

in this case mass is 756 grams and molar mass of oxygen is 16

So, the mass divided by molar mass would be =756/16= 47.25 moles of O2

The best way to balancing a combustion equation of an organic compound, like the one in the question, is to start balancing the carbon, then the hydrogen and finish with the oxygen. For that you have to split the hydrocarbon into individual elements. And once you have finished you shall always try to reduce the coefficient if possible. Therefore, the answers are A. and C.

Explanation:

Sydney has been the work that is a great place 56 to be able and a day and day for a week or two types and we will have a very nice time of course but 6yy6667 has 7575665yy44y

The final value for the enthalpy change of the formation of CO2 and 2H2O from their elements (C, H2, and O2) would be -1410 kJ per mole of CO2 and 2 moles of H2O formed.

The enthalpy change (ΔH) for the reaction CO2 + 2H2O → H2CO3 can be calculated by multiplying the stoichiometric coefficients of the balanced equation by the enthalpy values of the corresponding compounds involved in the reaction.

In the given reaction, the enthalpy change is -1410 kJ. However, it's important to note that this enthalpy change corresponds to a specific reaction and may not directly apply to the formation of CO2 and 2H2O from another reaction or process.

If we assume that the reaction is the formation of one mole of CO2 and two moles of H2O, we can say that the enthalpy change for this specific formation reaction is -1410 kJ.

Therefore, the final value for the enthalpy change of the formation of CO2 and 2H2O from their elements (C, H2, and O2) would be -1410 kJ per mole of CO2 and 2 moles of H2O formed.

It's worth mentioning that the enthalpy change can vary depending on the specific conditions (temperature, pressure, etc.) and the reactants involved in the reaction. Therefore, it's crucial to specify the conditions and reaction context when referring to enthalpy values.

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Explanation:

this equation is balanced

if you look at it carefully

k=1

f=1

o=2

we do not have any opposing element

Answer:

this equation is balanced

if you look at it carefully

k=1

f=1

o=2

we do not have any opposing element

Multiplying by 1000 will allow you to convert 35 liters to milliliters

1 Liters = 1000 milliliters

Thus, multiply the value in liter by 1000 will convert it to milliters

With the above scale, we can convert 35 liters to milliliters. Details below:

We can convert 35 liters to milliliters as illustrated below:

1 Liters = 1000 milliliters

Therefore,

35 liters = (35 liters × 1000 milliliters) / 1 liters

35 liters = 35000 milliliters

Thus, 35 liters is equivalent to 35000 milliliters

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Answer:

1

Explanation:

To determine the value of K (equilibrium constant) at 25 °C, we can use the Nernst equation, which relates the cell potential (E) to the equilibrium constant (K) and the standard cell potential (EºCell). The Nernst equation is given by:

E = EºCell - (RT / nF) * ln(K)

Where:

E = cell potential

EºCell = standard cell potential

R = gas constant (8.314 J/(mol·K))

T = temperature in Kelvin (25 °C = 298 K)

n = number of electrons transferred in the balanced redox equation

F = Faraday's constant (96,485 C/mol)

ln = natural logarithm

In this case, the given standard cell potential (EºCell) is -0.37 V.

The balanced redox equation for the cell reaction is:

Pt + Cr²+ -> Pt + Cr³+

Since there is no change in the oxidation state of Pt, no electrons are transferred in the reaction (n = 0).

Substituting the known values into the Nernst equation, we have:

E = -0.37 V - (8.314 J/(mol·K) * 298 K / (0 * 96,485 C/mol)) * ln(K)

E = -0.37 V

Since n = 0, the term (RT / nF) * ln(K) becomes 0, and we are left with:

-0.37 V = -0.37 V - 0

This implies that the value of K is 1, since any number raised to the power of 0 is equal to 1.

Therefore, the value of K at 25 °C for the given cell is 1.

Answer:

Question

Explanation:

Identification of problem is the first scientific method. This is overtaken by asking questions

29 is not the best answer depends on the context and the rules for significant figures.

What is best answer?

The best answer depends on the context and the rules for significant figures. If we assume that we need to round to three significant figures:

Therefore, 29 is not the best answer.

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The concentration of hydrogen ions [H⁺] ions is 10⁻⁵ and hydroxide ions [OH⁻] ions is 10⁻⁹ in a solution of pH of 5. A) a solution that has a pH of 5 has more hydrogen ions than hydroxide ions.

C)when dealing with pH of 5, you are dealing with acid.

pH = - log [H⁺]

pH is given = 5

therefore,

5 = - log [H⁺]

[H⁺] = 10⁻⁵

now,  [H⁺]   [OH⁻] = 10⁻¹⁴

           10⁻⁵  [OH⁻] = 10⁻¹⁴

             [OH⁻] = 10⁻⁹

A solution that has pH below 7 is acidic solution with higher hydrogen ion concentration.

Thus, The concentration of hydrogen ions [H⁺] ions is 10⁻⁵ and hydroxide ions [OH⁻] ions is 10⁻⁹ in a solution of pH of 5. A) a solution that has a pH of 5 has more hydrogen ions than hydroxide ions.

C)when dealing with pH of 5, you are dealing with acid.

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Convert cm³ to ft³, then multiply by the density to get the mass in lb, then convert to g.

\(8.00\,\mathrm{cm}^3 \cdot \left(\dfrac{1\,\rm in}{2.54\,\rm cm} \cdot \dfrac{1\,\rm ft}{12\,\rm in}\right)^3 \cdot \dfrac{455\,\rm lb}{1\,\mathrm{ft}^3} \cdot \left(\dfrac{0.4536\,\rm kg}{1\,\rm lb}\cdot\dfrac{1000\,\rm g}{1\,\rm kg}\right) \approx \boxed{58.3 \mathrm g}\)

Answer:

B. 1,1,2

Explanation:

If you plug in the coefficients into the reaction

1\(Cl_{2} O_{5}\) + 1\(H_{2}O\)→ 2\(HClO_{3}\)

You have 2 Cl, 6 O, and 2H on the left side and 2Cl, 6 O, and 2 H on the right side.

The coefficients of reactants in balanced reaction are the stoichiometric ratios of these reactants. The balanced equation of the given reaction is written as: \(Cl_{2}O_{5} + H_{2}O \rightarrow 2 HClO_{3}\).  Hence, the coefficients are 1,1,2.

In a balanced reaction, each of the groups or elements in a reaction will be equal in number on both side. The balanced reaction thus a shows all reactants and products in proper stoichiometric proportions.

To balance a reaction, we can multiply by suitable integers on either side. In the given reaction there are two chlorines in the reactant side and only one in the product side so by multiplying by 2 on the product side make the number of chlorines equal.

Adding 2 to the product side also make the all the elements in equal number on both side. Two hydrogen on both side and 6 oxygens in both sides.

Hence, the balanced reaction is written as follows:

\(Cl_{2}O_{5} + H_{2}O \rightarrow 2 HClO_{3}\)

Therefore, the coefficients are 1,1, 2 and option B is correct.

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The balanced redox reaction is

\(3F(e)_3+ + 2N O^{-2} + 4H_2O\)→ \(3F(e)_2+ + 2N O^{-3} + 4H+\)

Let's start balancing the redox reaction. Here is the given redox reaction:

\(3F(e)_3+ + 2N O^{-2} + 4H_2O\) → \(F(e)_2+ + H+ + N O^{-3}\)

Step 1: Separate the redox reaction into two half-reactions.Half-reaction 1: \(F(e)_3+\)→ \(F(e)_2+\) Half-reaction 2: \(N O^{-2}+ H_2O\) → \(N O^{-3}+ H+\)

Step 2: Balance the elements other than H and O in each half-reaction. Balanced Half-reaction 1: \(F(e)_3+\) →\(F(e)_2+\)Balanced Half-reaction 2: \(N O^{-2} + H_2O\) → \(N O^{-3} + H+\)

Step 3: Balance the oxygen atoms by adding H2O to the side that needs oxygen . Balanced Half-reaction 1: \(F(e)_3+\) → \(F(e)_2+\) Balanced Half-reaction 2: \(N O^{-2}+ H_2O\) →\(N O^{-3} + H+ + H_2O\)

Step 4: Balance the hydrogen atoms by adding H+ to the side that needs hydrogen . Balanced Half-reaction 1: \(F(e)_3+\) → \(F(e)_2+\) + e− + H+Balanced Half-reaction 2: \(N O^{-2} + H_2O + H+ - > N O{-3}+ H+ + H_2O + e^-\)

Step 5: Determine the number of electrons required to balance the half-reaction . Balanced Half-reaction 1: \(F(e)_3+ - > F(e)_2+ + e^{-}+ H+\) [3 electrons are required]Balanced Half-reaction 2: \(N O^{2} + H_2O + H+ - > N O^{-3} + H+ + H_2O + e^-\)[2 electrons are required]

Step 6: Multiply the half-reactions by integers to make the number of electrons equal for both half-reactions.Balanced Half-reaction 1:\(3(F(e)_3+ - > F(e)_2+ + e^{-} + H+\) ).Balanced Half-reaction 2: \(2(N O^{-2} + H_2O + H+ - > N O^{-3}+ H+ + H_2O + e^{-})\)

Step 7: Add the balanced half-reactions together\(3F(e)_3+ + 2N O^{-2} + 4H+ - > 3F(e)_2+ + 2N O^{-3} + 2H_2O\) .The balanced redox reaction is:\(3F(e)_3+ + 2N O{-2} + 4H_2O - > 3F(e)_2+ + 2N O^{-3} + 4H+\)

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Answer: 24.305052 (8 sig figs)  [Mg]

Explanation:  The AMU of each isotope is multiplied by it's percent abundance to yield a weighted average for each isotope.  The sum of these weighted averages will be the wighted average of the element's isotopes:

Isotope    Mass (AMU)     %            Weighted AMU

24Mg 23.985042 78.99% 18.94578468

25Mg 24.985837 10.00%   2.4985837

26Mg 25.982593 11.01%   2.860683489

                                   24.305052 (8 sig figs)

You take the mass of carbon dioxide, 56.8g, divide by its molar mass, 44.01g/mol, to produce the moles of carbon dioxide. This is multiplied by the molar ratio of butane/CO2, (2/8) = 1/4, which gives the moles of butane required to produce the carbon dioxide.

The mass of a substance is the product of the moles and the molar mass. 65.5 g of carbon dioxide will be able to produce 21.62 grams of butane.

The mass of a reactant or a product is estimated by the number of moles and the molar mass of the substance.

The combustion reaction of methane is shown as,

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Given,

Molar mass of butane = 58.12 g/mol

Mass of carbon dioxide = 65.5 g

Molar mass of carbon dioxide = 44.01 g/mole

Moles of carbon dioxide are calculated as,

Moles = mass ÷ molar mass

= 65.5 ÷ 44.01

= 1.488 moles

From the above reaction, two moles of butane produce eight moles of carbon dioxide. So, 1.488 moles of carbon dioxide will be produced from,

2 × 1.488 ÷ 8 = 0.372 moles

The moles of butane required is 0.372 moles

The mass of butane from moles is calculated as,

Mass = moles × molar mass

= 0.372 × 58.12 g/mole

= 21.62 gms

Therefore, 21.62 gm of butane is required to produce 65.5 g of carbon dioxide.

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Answer:

correct will be from back to the start:

grass-grasshopper-toad-snake-hawk

Explanation

ood chain begin with plant-life and end with animal-life.

Answer:

D. 6.3 x 10-5 mol/L NaOH

Explanation:

Alizarin yellow is an indicator that is yellow when pH < 10.1. In the same way, thymol blue is blue when pH > 9

That means the pH of the solution is between 9 - 10.1

Any acid as HCl could have a pH of these.

The solution of 3.2x10⁻⁴M NaOH has a pH of:

pOH = -log[OH-] = 3.49

pH = 14-pOH = 10.51. The pH of the solution is not 10.5

Now, the solution of 6.3x10⁻⁵M NaOH has a pH:

pOH = -log[OH-] = 4.2

pH = 14-pOH = 9.8

The pH of the solution could be 9.8. Right option is:

Answer

720.9 mmHg

Explanation

Gas collected over water must always account for the saturated vapor pressure.

Note that the saturated vapor pressure of water at 23 °C = 21.1 mm Hg.

Therefore atmospheric pressure is

The atmospheric pressure given = 742 mmHg, Psvp = 21.1 mmHg, thus, we