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skad [1K]
3 years ago
15

Thallium-201 has a half-life 73 hours. If 4.0 mg of thallium-201 disintegrates over a period of 5.0 °, how many milligrams of th

allium-201 will remain?
Chemistry
1 answer:
Anika [276]3 years ago
5 0

<u>Given:</u>

Half life of Thallium-201 (T1/2)= 73 hrs

Initial amount of thallium (A₀) = 4.0 mg

Disintegration Time period (t) = 5.0 hrs

<u>To determine:</u>

Amount of thallium remaining, A

<u>Explanation:</u>

The radioactive half-life (T1/2) can be related to the amounts of the radio active element through the relation:

A = A_{0} e^{-0.693t/T1/2}

A = 4.0 * e^{-0.693*5/73} = 3.815 mg

Ans: Amount of thallium-201 remaining is 3.815 mg

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Answer:

Enthalpy of vaporization = 30.8 kj/mol

Explanation:

Given data:

Mass of benzene = 95.0 g

Heat evolved = 37.5 KJ

Enthalpy of vaporization = ?

Solution:

Molar mass of benzene = 78 g/mol

Number of moles = mass/ molar mass

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Number of moles = 1.218 mol

Enthalpy of vaporization =  37.5 KJ/1.218 mol

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In the disproportionation reaction CI2 + H2Omc021-1.jpgHCIO + HCI, what describes the oxidation states of the substance Cl?
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<span>Reaction: CI2 + H2O    ---->  HCIO + HCI

Oxidations states:

The oxitation state of Cl2 = 0, because the oxidation state of an atom alone or a molucule with one kind of atom is always 0.

The oxidation state of Cl in HClO is +1 because the oxidation state of H is + 1, the oxidation state of O is - 2, and the molecule is neutral, so  +1 + 1 - 2 = 0

The oxidation state of Cl in HCl is - 1, because the oxidation state of H is +1 and the molecule is neutral, so - 1 + 1 = 0.

Also, you shall remember that when an atom increases its oxidation state is is oxidized and when an atoms reduces its oxidations state it is reduced.

With that you conclude that the right option is the last statement: </span>Cl has an oxidation number of 0 in Cl2. It is then reduced to CI- with an oxidation number of –1 in HCl and is oxidized to Cl+ with an oxidation number +1 in HClO.
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How many grams are in 4.07x10^15 molecules of calcium hydroxide
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There are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the number of moles in the substance by its molecular mass.

However, given that the number of molecules in calcium hydroxide is 4.07 x 10¹⁵molecules, we need to calculate the number of moles in Ca(OH)2 as follows:

no. of moles of Ca(OH)2 = 4.07 x 10¹⁵ ÷ 6.02 × 10²³

no. of moles = 0.676 × 10-⁸

no. of moles = 6.76 × 10-⁹ moles.

Molar mass of Ca(OH)2 = 74.093 g/mol

Mass of Ca(OH)2 = 74.093 × 6.76 × 10-⁹ moles

Mass = 5.01 × 10-⁷grams.

Therefore, there are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

Learn more about how to calculate mass at: brainly.com/question/8101390?referrer=searchResults

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