Thallium-201 has a half-life 73 hours. If 4.0 mg of thallium-201 disintegrates over a period of 6.0 days and 2 hours, how many milligrams of thallium-201 will remain?

Answers

Answer 1

Answer:

[tex]\huge\boxed{\sf 1.0 \ mg}[/tex]

Explanation:

Half-life = 73 hours

Disintegration period = 6 days 2 hours = (6*24) hours + 2 hours

= 144 hours + 2 hours

= 146 hours

No. of Half lives = Disintegration period / Half-life

No. of Half lives = 146 hours / 73 hours

No. of Half lives = 2

Hence, there are 2 half lives passed in 6 days 2 hours.

Amount of thallium-201 left after first half-life:

= 4.0 mg / 2

= 2.0 mg

Amount of thallium-201 left after second half-life:

= 2.0 mg / 2

= 1.0 mg

Hence, the amount of thallium-201 left after 6.0 days and 2 hours is 1.0 mg.

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

~AH1807

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

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3 of 12
Assessment: Force and Motion CBA
Four students recorded the time it took them to walk to school from their homes as shown in the table
below.
Student
1
2
3
Motion of Four Students
Distance (meters) Time (minutes)
1,100
20
1,026
19
848
16
795
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4
Which student had the greatest average speed?
Student 1
Student 2
Student 3
Student 4
1
2
3
4
5
6
7
8
9
10
Next

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Select the correct answer.
John is riding a ski lift to the top of Wildcat Mountain. He removes his gloves and rapidly rubs his hands together to warm them up. What
happens when John rubs his hands?
O A. The skin on his hands rapidly conducts heat, similar to metal.
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O D. Thermal energy moves from his fingertips to his palms.
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Therefore, 1 mole of hydrogen gas (H2) will produce;

= 1 × 2/3

= 2/3

= 0.6666

= 0.67moles of ammonia (NH3)

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Answer: [tex]C_7H_{16}+11O_2\rightarrow 7CO_2+8H_2O[/tex]

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2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔG°f (kJ/mol) -110.9 87.6 51.3 -237.1

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Hello there!

In this case, according to the definition of the Gibbs free energy of reaction in terms of the Gibbs free energy of formation of the involved reactants and products, it is possible for us to compute it as shown below:

[tex]\Delta G_{rxn}=3\Delta _f G_{NO_2}+\Delta _f G_{H_2O}-2\Delta _f G_{HNO_3}-\Delta _f G_{NO}[/tex]

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[tex]\Delta G_{rxn}=3(51.3)+(-237.1)-2(-110.9)-(87.6)\\\\\Delta G_{rxn}=51.0kJ/mol[/tex]

Best regards!

The standard Gibbs free energy change (ΔG°rxn) for the reaction is 76.7 kJ/mol.

The balanced equation for the reaction is:

2 HNO₃(aq) + NO(g) → 3 NO₂(g) + H₂O(l)

The standard free energy change for a reaction ( ΔG0rxn) can be found out using the equation,

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To learn more about the Gibbs free energy, follow the link:
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A student weighs an empty flask and stopper and finds the mass to be 53.256 g. She then adds about 5 mL of an unknown liquid and heats the flask in a boiling water bath at 98.8°C. After all the liquid is vaporized, she removes the flask from the bath, stoppers it, and lets it cool. After it is cool, she momen- tarily removes the stopper, then replaces it and weighs the flask and condensed vapor, obtaining a mass of 53.780 g. The volume of the flask is known to be 231.1 mL. The absolute atmospheric pressure in the laboratory that day is 728 mm Hg. a. What was the pressure of the vapor in the flask in atm?

Answers

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