A golf ball and bowling ball are moving and both have the same kinetic energy.Which one is moving faster? If they move at the same speed, which one has more kinetic energy?

Answer:the answer is the Velocity of molecules in the body

Explanation:hope this helped have a great day

Answer:

The lengths of the C-C bonds increases with the decrease in the resistance of said bond, for example, a triple bond has a shorter length than in the case of a single bond. Butane has the single bonds and the CC bond is hybridized with sp3 hybridization, however in the butylcyclohexane structure the CC bond is also sp3 and the angle is 120°, however the angle shown is equal to 109.5°, so there is a certain angular tension and it is very unstable with respect to butane

Explanation:

Covalent compounds can be identified by their low melting and boiling points, non-conductivity of electricity, and solubility in nonpolar solvents.

Covalent compounds, also referred to as molecular compounds, are usually identified through several physical properties such as their low melting and boiling points, non-conductivity of electricity, and their solubility in nonpolar solvents.

Low melting and boiling points: Covalent compounds don't have strong intermolecular forces, resulting in low melting and boiling points.

Non-conductivity of electricity: Since most covalent compounds don't dissociate into ions in solutions, they typically do not conduct electricity.

Solubility in nonpolar solvents: Due to the 'like soluble like' rule, covalent compounds tend to be soluble in nonpolar solvents rather than polar solvents such as water.

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The correct answer is water.

During hydrolysis, _____water______ must be added before the bonds can be broken.

For example, sucrose undergoes hydrolysis to break into glucose and fructose. Here sucrose is a disaccharide and glucose and fructose are monosaccharides.  

Sucrose + H₂O → glucose + fructose

Esters are formed by the combination of carboxylic acid and alcohol.

But the hydrolysis of ester causes the release of carboxylic acid and alcohol.

RCOOR'(ester) + H₂O → RCOOH(carboxylic acid) + R'OH(alcohol)

129 molecules of ATP is produced if all resulting molecules of acetyl-coa are converted into β-hydroxybutyrate.

This is a metabolic process involving multiple steps by which fatty acid molecules are broken down to produce energy.

Palmitic acid is a 16 carbon fatty acid and undergoes 7 beta-oxidation reactions which is shown below

7 NADH + 7 FADH2 + 8 acetyl-CoA

Acetyl-CoA being converted to acetoacetate with β-hydroxybutyrate being the reduced form.

The average ATP production is 131 minus 2 ATP which was used up for the initial activation of every fatty acid thereby bringing the amount to 129ATP.

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Answer:

D.

The molecular structure has changed.

Explanation:

I got it right on Plato.

ANSWER: There is only one path to get a career in chemistry.

EXPLANATION: There are many paths of career in Chemistry and not only one. It would not be justified to say that there is only one path to get a career in chemistry. Careers in chemistry would include Biochemistry, Forensic Scientist, Research Scientist, Chemical Engineer, Chemical Plant Operator and even a Science teacher or Chemistry professor in a school or university.

Answer : The perchloric acid and potassium hydroxide base is used to prepare the salt of potassium perchlorate, [tex](KClO_4)[/tex]

Explanation :

when the perchloric acid react with the potassium hydroxide as a base to form a salt of potassium perchlorate, [tex](KClO_4)[/tex]

The balanced chemical reaction will be,

[tex]HClO_4+KOH\rightarrow KClO_4+H_2O[/tex]

By the stoichiometry, we can say that 1 mole of perchloric acid react with the 1 mole of potassium hydroxide base to give 1 mole of potassium perchlorate and 1 mole of water as a product.

Hence, the perchloric acid and potassium hydroxide base is used to prepare the salt of potassium perchlorate, [tex](KClO_4)[/tex]

Answer:

True

Explanation:

Final answer:

To find the mass of H3PO4 in 521 ml of a 9.30 M solution, convert the volume to liters, multiply by the molarity to get moles, and then convert moles to grams using the molar mass of H3PO4. Approximately 474.6434 grams of H3PO4 are in 521 ml of a 9.30 M solution.

Explanation:

The student has asked how many grams of H3PO4 are in 521 ml of a 9.30 M solution. To find this, we use the formula:

Calculate the number of moles of H3PO4 by multiplying the molarity (M) of the solution by the volume of the solution in liters.

Convert the moles of H3PO4 to grams using the molar mass of H3PO4 (which is approximately 98 g/mol).

First, we convert 521 ml to liters (521 ml * 0.001 L/ml = 0.521 L). Next, we calculate the moles of H3PO4 (9.30 M * 0.521 L = 4.8433 mol). Finally, we multiply the number of moles by the molar mass of H3PO4 to get the mass in grams (4.8433 mol * 98 g/mol = 474.6434 g).

Therefore, there are approximately 474.6434 grams of H3PO4 in 521 ml of a 9.30 M solution.

The molar mass of carbon dioxide is 44.0 g/mol. a mass of 150.0 grams of carbon dioxide is equivalent to 3.41 mol of carbon dioxide.

In the International System of Units, the mole is the unit of substance quantity (SI). How so many elementary units of a certain substance are present in an item or sample is determined by the quantity of that material. There are precisely 6.022×10²³ elementary entities in a mole.

For instance, although having differing volumes and weights, 10 moles containing water along with ten moles of mercury both contain the same quantity of material, and the mercury includes exactly 1 atom for every molecule of water.

M(CO₂)=44.0 g/mol

m(CO₂)=150.0 g

n(CO₂)=m(CO₂)/M(CO₂)

n(CO₂)=150.0/44.0=3.41 mol

Therefore, 3.41 mol of carbon dioxide is there.

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Explanation:

According to the the ideal gas law, PV= nRT.

This means that pressure is directly proportional to temperature.

So, when HCl and zinc are combined in a flask then it will lead to the formation of zinc chloride and hydrogen gas.

The reaction equation will be as follows.

             [tex]HCl + Zn \rightarrow ZnCl_{2} + H_{2}[/tex]

Since, the flask is sealed hence, hydrogen gas will not be able to move out of the flask.

So, when it will behave ideally then due to directly proportional relation between pressure and temperature there will occur a decrease in temperature with decrease in pressure.

To calculate the pH of the given solution, use an ICE table to determine the concentrations of HC2H3O2 and C2H3O2-. Substitute these concentrations into the Ka expression to calculate the Ka value. Finally, use the definition of pH to determine the pH of the solution.

To calculate the pH of a solution that is 0.195 M in HC2H3O2 and 0.125 M in KC2H3O2, we can use an ICE table. First, we set up the ICE table to determine the concentrations of HC2H3O2 and C2H3O2- at equilibrium. Once we have these concentrations, we substitute them into the Ka expression to calculate the Ka value. Finally, we use the definition of pH to determine the pH of the solution.

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The concentration of H+ ions in the solution of 2.7 molar HClO₄ is 2.7 molar. Thus the pH of the acid is -0.43.

pH of a solution is the measure of its H+ ions concentration. It determines how much acidic or basic the solution is. Mathematically it is the negative logarithm of H+ ion concentration.

A pH of 7 indicates that the solution is neutral and a pH below 7 is acidic and pH of above 7 is for basic solution. Thus strong acids such as chloric acid HClO₄ ,  have a pH very less. But it depends on the concentration of solute in the solution.

Given that the molarity of the acid is 2.7 molar. It furnish equal number of moles H+ ions and anions. Thus the concentration of H+ ion is 2.7 M.Now the pH is calculated as follows:

pH = -log [H+]

     = -log (2.7)

     = -0.43

Hence, the pH of the solution is -0.43.

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Mass of (NH4)2SO4  is 0.6025g

Using the equation:

Number of moles = Mass /Molar Mass

Given,

0.00456 as Number of moles of (NH4)2SO4

and Molar mass as

N =  14.007 g/mol=2(14.007)

H = 1.008 g/mol= 8(1.008)

S = 32.065 g/mol= 32.065

O = 16 g/mol=           4(16)

(NH4)2SO4 = 132.143 g/mol

Mass of  (NH4)2SO4 = 132.143 g/mol x 0.00456=0.6025g

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To find the Ka value for the weak acid, we can use the given pH and concentration of the acid solution. The Ka value is calculated using the equation Ka = ([H3O+][A-])/[HA].

To find the value of Ka for the weak acid, we can use the given pH and concentration of the acid solution. We know that the pH is a measure of the concentration of H3O+ ions in a solution, so we can use the pH to calculate the [H3O+] concentration. From the given pH of 3.25, we can determine that the [H3O+] concentration is 10^(-pH). So, [H3O+] = 10^(-3.25) M.

Now, the equilibrium equation for the dissociation of the weak acid is HA(aq) + H2O(l) -> H3O+(aq) + A-(aq). Since we know the [H3O+] concentration, we can assume that the [HA] concentration is equal to the [H3O+] concentration. So, [HA] = [H3O+] = 10^(-3.25) M.

The Ka value is calculated using the equation Ka = ([H3O+][A-])/[HA]. Substituting the given values, Ka = (10^(-3.25)^2)/10^(-3.25). Simplifying this expression gives us the value of Ka for the acid.

Calcium, with an atomic number of 20, forms ions by losing 2 electrons, leaving it with 18 electrons and 20 protons. This gives it a net charge of +2.

The electrical charge of a calcium ion can be determined by looking at its number of electrons compared to its atomic number. Calcium's atomic number is 20, meaning it has 20 protons and normally 20 electrons. However, the question states that it forms ions with 18 electrons, so it has lost 2 electrons to become a calcium ion.

When an atom loses electrons, it becomes positively charged. Since calcium has lost 2 electrons, its electrical charge is +2.

Therefore, the electrical charge of a calcium ion is +2.

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The equilibrium constant for isomerization reaction is [tex]\boxed{9.615}[/tex]

Further Explanation:

The standard Gibbs free energy change in a reaction [tex]\left( {{{\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }}} \right)[/tex] is the difference of sum of the standard free energies of formation of product molecules and sum of standard free energies of formation of reactant molecules at the standard conditions. The formula used to calculate the value of standard Gibbs free energy  change for a reaction [tex]\left( {{{\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }}} \right)[/tex] is as follows:

[tex]\Delta\text{G}_{\text{rxn}}^{\circ}=\sum\text{n}\Delta\text{G}_{\text{f}(\text{products})}^{\circ}-\sum\text{m}\Delta\text{G}_{\text{f}(\text{reactants})}^{\circ}[/tex]

Here, n is the stoichiometric coefficients of products, and m are the stoichiometric coefficients of reactants in a balanced chemical equation.

The formula to determine the relationship between change in standard Gibbs free energy [tex]\left( \Delta{\text{G}^{\circ}} \right)[/tex] and equilibrium constant [tex]\left({\text{K}}\right)[/tex] is given as follows:

[tex]{\Delta }}{{\text{G}}^{{^\circ }}} = - {\text{RTlnK}}[/tex]       ......(1)

Here,

[tex]\Delta{\text{G}^{\circ}[/tex] is the standard Gibbs free energy change.

[tex]{\text{R}[/tex] is the gas constant.

[tex]{\text{T}}[/tex] is the temperature in Kelvin.

[tex]{\text{K}}[/tex] is the equilibrium constant.

The isomerization of glucose-1-phosphate to fructose-6-phosphate occurs in 2 steps:  

The reaction of step 1 is as follows:

[tex]{\text{glucose - 1 - phosphate}} \to {\text{glucose - 6 - phosphate}}[/tex]

                                       ......(2)

[tex]\Delta{\text{G}^{\circ}_{1}[/tex] for equation (2) is  [tex]- 7.28\;{\text{kJ/mol}}[/tex]

The reaction of step 2 is as follows:

[tex]{\text{fructose - 6 - phosphate}} \to {\text{glucose - 6 - phosphate}}[/tex]

                                                   ......(3)

[tex]\Delta{\text{G}^{\circ}_{2}[/tex] for equation (3) is  [tex]- 1.67\;{\text{kJ/mol}}[/tex]

Reverse the reaction of step 2.

[tex]{\text{glucose - 6 - phosphate}} \to {\text{frutcose - 6 - phosphate}}[/tex]

                                                ......(4)

[tex]\Delta{\text{G}^{\circ}_{3}[/tex] for equation (4) is [tex]+ 1.67\;{\text{kJ/mol}}[/tex]

Add equation (1) and (3) to get the final equation.

[tex]{\text{glucose - 1 - phosphate}} \to {\text{frutcose - 6 - phosphate}}[/tex]

To calculate [tex]\Delta {\text{G}}_{{\text{rxn}}}^{^\circ }}[/tex], add [tex]\Delta{\text{G}^{\circ}_{1}[/tex] and [tex]\Delta{\text{G}^{\circ}_{3}[/tex] as follows:  

[tex]\Delta{\text{G}^{\circ}_{\text{rxn}}=\Delta{\text{G}^{\circ}_{1}+\Delta{\text{G}^{\circ}_{3}[/tex]                       ......(5)

Substitute [tex]- 7.28\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}_{1}[/tex] and [tex]+ 1.67\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}_{3}[/tex] in equation (5).

[tex]\begin{aligned}\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }} &=  - 7.28\;{\text{kJ/mol + }} + 1.67\;{\text{kJ/mol}}\\{\text{}}&= - 5.61\;{\text{kJ/mol}}\\\end{aligned}[/tex]

For equilibrium constant (K), rearrange equation (1)

[tex]{\text{K}}={\text{e}}\frac{-\Delta{\text{G}}^{\circ}}{\text{RT}}[/tex]    ......(6)

Substitute [tex]- 5.61\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}[/tex],[tex]8.314\;{\text{J/mol}} \cdot {\text{K}}[/tex] for R and [tex]298\;{\text{K}}[/tex] for T in equation (6)

[tex]\begin{aligned} {\text{K}}&= {{\text{e}}^{\frac{{ - \left( { - 5.61\;{\text{kJ/mol}}} \right)}}{{\left( {8.314\;{\text{J/mol}} \cdot {\text{K}}} \right)\left( {\frac{{{\text{1J}}}}{{1000{\text{kJ}}}}} \right)\left( {298\;{\text{K}}} \right)}}}}\\&= {{\text{e}}^{2.2634}}\\&= 9.615\\\end{aligned}[/tex]

The equilibrium constant for the reaction is 9.615.

Learn more:

1. The change in standard gibbs free is for a reaction: https://brainly.com/question/10838453

2. Determination of the equilibrium constant for pure water: https://brainly.com/question/3467841

Answer details:

Grade: Senior Secondary School

Subject: Chemistry

Chapter: Chemical Equilibrium

Keywords: Standard Gibbs free energy, equilibrium, constant, glucose-1-phosphate and fructose-6-phosphate.

Final answer:

The energy released from converting 1.0 g of mass into energy is 9 × 10¹³ joules (J), using the equation E = mc², where c is the speed of light.

Explanation:

According to Einstein's famous equation E = mc², where E represents energy, m is mass, and c is the speed of light in a vacuum, the energy released from completely converting 1.0 g of mass into energy is tremendously large. Since the speed of light, c, is approximately 3 × 10⁸ meters per second, and the mass m is 1.0 g (which is 1/1000 of a kilogram), the calculation is E = (1.0 g / 1000) × (3 × 10⁸ m/s)². This results in an energy release of 9 × 10¹³ joules (J), which is equivalent to about twice the energy released by the atomic bomb dropped on Hiroshima.