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gladu [14]
3 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]3 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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3 years ago
(c) O2 gas is transferred from a 3 L vessel containing oxygen at 4 atm to an evacuated 20 L vessel at a constant temperature of
Fudgin [204]

Answer:

After the transfer the pressure inside the 20 L vessel is 0.6 atm.

Explanation:

Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

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Therefore, for this problem the step by step explanation is:

P_{1} xV_{1} = P_{2} xV_{2}

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The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

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