Convert 55 kilometers per hour into meters per second. Use the conversion factor 1 km = 1000 m.55 ______/ _____ ´ 1000 _____/1 ______ ´ 1 ______/60 ______
´ 1 _____/60 _____ = 15 _____

Answer:

CaCO3 + 2HCL -> CO2 +H2O + CaCl2

Explanation:

When balancing an equation, it is important to first ensure that the chemical formulas in the equation are correct. In the above equation, the first modification required is for CaCl to be corrected to CaCl2. This is because the calcium ion has a charge of 2+ which has to be balanced by 2 chloride ions, each of charge 1-

Next, the overall atoms of the equations should be balanced. The number of atoms of each element on the side of reactants should be equal to their number on the side of the products. In this case, the number of HCL molecules should be 2.

Before balancing equation:

Reactants: Ca: 1, C: 1, O: 3, H: 1, Cl: 1

Products: Ca: 1, C: 1, O: 3, H: 2, Cl: 1

After balancing equation:

Reactants: Ca: 1, C: 1, O: 3, H: 2, Cl: 2

Products: Ca: 1, C: 1, O: 3, H: 2, Cl: 2

Answer:

Percentage of a sample remains after 60.0 min is 13.03%.

Explanation:

• It is known that the decay of isotopes of C-11 obeys first order kinetics.

• Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

• Half-life time (t1/2) in first order reaction = 0.693/k, where k is the rate constant.

∴ k = 0.693/(t1/2) = 0.693/(20.4 min) = 0.03397 min⁻¹.

• The integrated law for first order reaction is:

kt = ln[A₀]/[A],

where, k is the rate constant (k = 0.03397 min⁻¹).

t is the time of the reaction (t = 60.0 min).

[A₀] is the initial concentration of C-11 ([A₀] = 100.0 %).

[A] is the remaining concentration of C-11 ([A] = ???%).

∵ kt = ln[A₀]/[A]

∴ (0.03397 min⁻¹)(60.0 min) = ln(100%)/[A]

∴ 2.038 = ln(100%)/[A]

• Taking e for both sides:

∴ 7.677 = (100%)/[A]

∴ [A] = (100%)/(7.677) = 13.03%.

So, percentage of a sample remains after 60.0 min is 13.03%.

Final answer:

The half-life of Carbon-11 is 20.4 minutes, meaning the quantity reduces by half every 20.4 minutes. So, after approximately 60.0 minutes (roughly three half-lives), approximately 12.5% of the original sample would remain.

Explanation:

Carbon-11 is a radioisotope that's widely used in medical imaging due to its radioactive properties. It has a half-life of 20.4 minutes, which means that the quantity of Carbon-11 reduces to half in every 20.4 minutes.

If we denote one half-life as t, after three half-lives (60.0 min or 3t), the ratio of the remaining sample of Carbon-11 would be 1/2 * 1/2 * 1/2 = 1/8. In other words, approximately 12.5% of the original sample would be remaining. This percentage changes a bit due to the fact that 20.4 min is slightly less than 22.2 min (which would be the exact 3 half lives).

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Answer:

The answer to your question is 0.14 M

Explanation:

Data

Molarity = ?

number of moles = 0.25

volume = 1.8 L

Formula

Substitution

Simplification and result

      Molarity = 0.14 M

The electrolysis describe " electrical energy is used to produce a chemical change."

What is electrolysis?

The process of dissolving ionic compounds into various constituent components by running a continuous electric current through into the complex in a fluid form is known as electrolysis.

What iselectrical energy?

The energy produced by the flow of electrons from one location to another is referred to as electrical energy. Current or electricity is the transportation of charged particles across a medium.

The electrolysis describe " electrical energy is used to produce a chemical change."

To know more about electrolysis and electrical energy.

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