calculate the amount of daughter isotope that has been produced after 1, 2, 3, and 4 half-lives. include 4 decimal places in your answers daughter present: 0.0000 unit (i.e., no daughter isotope is present) after 1 half-life: 0.5000 units

Answers

Answer 1

The amount of daughter isotope that has been produced after 1, 2, 3, and 4 half-lives is 0.9375.

remaining after 1 half life = 1.0000/2 = 0.5000

remaining after 2 half lives = 0.5000/2 = 0.2500

remaining after 3 half lives = 0.2500/2 = 0.1250

remaining after 4 half lives = 0.1250/2 = 0.0625

For parent isotope:

In each half life the amount of remaining parent isotope will be half of the remaining of previous half life.

Therefore,

Parent present = 1.0000

present after 1 half life = 0.0000+ (1.0000 - 0.5000)=0.5000

present after 2 half lives = 0.5000+ (0.5000 -0.2500) = 0.7500

present after 3 half lives = 0.7500+ (0.2500 -0.1250) = 0.8750

remaining after 4 half lives = 0.8750+ (0.1250 - 0.0625)=0.9375

For daughter isotope:

The amount of daughter isotopes will be found by adding the value by which the number of parent isotope is decreased by in the relevant half life to the available daughter isotopes.

Daughter isotopes initially present: 0.0000

                        Daughter isotope         Parent isotope       Daughter to

                                 present                 remaining               parent ratio                                                                                

Starting state:           0.0000                1.0000                 0.0000/1.0000 =0

After 1 half life:          0.5000               0.5000                  0.5000/0.5000=1

After 2 half lives:     0.7500                0.2500                   0.7500/0.2500=3

After 3 half lives:     0.8750                0.1250                     0.8750/ 0.1250=7    

After 4 half lives:    0.9375               0.0625                     0.9375/0.0625=15

the amount of daughter isotope that has been produced after 1, 2, 3, and 4 half-lives is 0.9375.

and Ratio of the amount of daughter to the amount of parent isotope is 15.

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Answers

Answer:

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Explanation:

From the question given above, the following data were obtained:

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Answer:

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Answer:

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[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \underline{ \large \orange{\tt{ ꧁ A \: N\: S \: W \: E \: R ꧂}}}}[/tex]

✧ [tex] \large{ \red{\tt{E\:X \: O \: T \: H \: E \: R \: M \: I \: C \: \: R \: E \: A \: C \: T\: I \: O \: N}} : }[/tex]

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[tex] \underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}}[/tex] ✎

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Love you

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Answer:

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Answers

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Answers

Answer:

320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.

Explanation:

The balanced reaction is:

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By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

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By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

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Then you can apply the following rule of three: if 1 mole of sulfuric acid is produced by the reaction of 80 grams of sulfur trioxide, 4 moles of sulfuric acid is produced from how much mass of sulfur trioxide?

[tex]mass of sulfur trioxide= \frac{4 moles of sulfuric acid* 80 grams of sulfur trioxide}{1 mole of sulfuric acid }[/tex]

mass of sulfur trioxide= 320 grams

320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.

The mass of the required compound is 320 grams

Chemical reactions

Given the chemical equation expressed as:

[tex]SO_3 + H_2O \rightarrow H_2SO_4[/tex]

According to stochiometry, 1 mole of SO3  produces 1 mole of H2SO4, hence the moles of SO3 will also be 4.00moles

Determine the mass of SO3

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Mass = 320 grams

The mass of the required compound is 320 grams

Learn more on stoichiometry here:  https://brainly.com/question/16060223

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Explanation:

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Answers

Answer:

The molarity of the solution is 0.335 [tex]\frac{moles}{liter}[/tex]

Explanation:

Molar concentration is a measure of the concentration of a solute in a solution, be it some molecular, ionic, or atomic species.

Molarity is the number of moles of solute that are dissolved in a certain volume and is calculated by:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].

Being the molar mass of BaBr2 equal to 297.14 g/mole, that is to say that 1 mole contains 297.14 grams, the mass of 198 grams are contained in:

[tex]198 grams*\frac{1 mole}{297.14 grams} = 0.67 moles[/tex]

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[tex]Molarity=\frac{0.67 moles}{2 L}[/tex]

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Answer:

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