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gladu [14]
2 years ago
11

Identify the nuclide produced when plutonium-239 decays by alpha emission: 239 94pu→42he + ? express your answer as an isotope u

sing prescripts. 235 92u submithintsmy answersgive upreview part incorrect; try again; 4 attempts remaining part b identify the nuclide produced when thorium-234 decays by beta emission: 234 90th→ 0−1e + ? express your answer as an isotope using prescripts. submithintsmy answersgive upreview part part c identify the nuclide produced when fluorine-18 decays by positron emission: 18 9f→01e + ? express your answer as an isotope using prescripts. submithintsmy answersgive upreview part part d identify the nuclide produced when thallium-201 decays by electron capture: 201 81tl+ 0−1e→00γ + ? express your answer as an isotope using prescripts.
Chemistry
1 answer:
Galina-37 [17]2 years ago
6 0

1. Answer is ₉₂²³⁵U<span>

<span>Since this is an alpha emission the atomic number of the daughter nucleus decreases by 2 while mass number decreases by 4 compared to parent atom. Since parent atom has 94 as atomic number the daughter atom should have </span>94 - 2 = 92 as atomic number<span> <span>and </span></span>239 - 4 = 235 as mass number. <span>

</span></span>₉₄²³⁹Pu → ₂⁴He + ₉₂²³⁵U<span>

2. </span><span>Answer is </span>₉₁²³⁴Pa<span>

</span><span>Since this is a beta emission, a neutron is converted into a proton while emitting an electron. Hence atomic number increases by 1 compared to mass number but mass number remains as same. Hence, the </span>atomic number of the daughter atom<span> <span>should be </span></span>90 + 1 = 91<span> <span>which belongs to </span></span>Pa<span>. But the </span>mass number is same as 234.<span>

</span>₉₀²³⁴Th → ₋₁⁰e + ₉₁²³⁴Pa<span>
</span><span>
3.<span> <span>Answer is </span></span></span>₈¹⁸O<span>

</span><span>Since this is a positron emission, a proton is converted into a neutron while emitting an positron. Hence atomic number decreases by 1 compared to mass number but mass number remains as same. Hence, the </span>atomic number of the daughter atom<span> should be </span>9- 1 = 8 <span>which belong to </span>O<span>. But the </span>mass number is same as 18.<span>

</span>₉¹⁸F → ₊₁⁰e + ₈¹⁸O<span>

4. </span>Answer is ₈₀²⁰¹Hg<span>

</span><span>This is an </span><span>electron capture decay. </span>A<span> proton is converted into a neutron by emitting a gamma ray. In this process </span>mass number remains as same<span> <span>as parent atom which is </span></span>201<span>, but the </span>atomic number is decreased by 1<span> <span>than parent atom. Hence atomic number of daughter nucleus is 81 -1 = </span></span>80 <span>which belongs to </span>Hg.<span>
</span><span>
</span>₈₁²⁰¹Tl +  ₋₁⁰e →  ₀⁰γ + ₈₀<span>²⁰¹Hg</span>

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Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh
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Answer:

1.82 g   is the maximum mass of CuS.

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

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1 mL = 10⁻³ L

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Moles=0.500 \times {38.0\times 10^{-3}}\ moles

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<u>For (NH_4)_2S : </u>

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The conversion of mL to L is shown below:

1 mL = 10⁻³ L

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According to the given reaction:

CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}

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0.019 mole of CuCl_2 reacts with 0.019 mole of (NH_4)_2S

Moles of (NH_4)_2S = 0.019 mole

Available moles of (NH_4)_2S = 0.0252 mole

<u>Limiting reagent is the one which is present in small amount. Thus, CuCl_2 is limiting reagent.</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of CuCl_2 gives 1 mole of CuS

0.019 mole of CuCl_2 gives 0.019 mole of CuS

Moles of CuS formed = 0.019 moles

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Mass of CuS = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

<u>1.82 g   is the maximum mass of CuS.</u>

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Predict all bond angles in the following compound. Be sure to answer all parts. HCa≡CbCcH2OH
morpeh [17]

Answer:

See explanation below

Explanation:

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Hybridization   # bonds                            Angle

sp³                         4                                   109.5º

sp²                         3 +   1 pi bond              120º

sp                           2 + 2 pi bonds             180º

Carbon atom (a) is bonded to two atoms: Carbon (b) and one Hydrogen. It has a triple bond to Carbon (b). Therefore its hybridization is sp with two pi bonds, and for sp hybridization we know the angle is 180 º.

The same hybridization sp happens to carbon (b) bonded to Carbon (c) and C(a) using one sp bond to Carbon (a) and 2 pi bonds; it is also bonded using the other sp  to Carbon (c). The angle is therefore 180 between Carbons b and c.

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