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pav-90 [236]
3 years ago
8

이

Chemistry
1 answer:
Nataliya [291]3 years ago
4 0

The daughter isotope  : Radon-222 (Rn-222).

<h3>Further explanation</h3>

Given

Radium (Ra-226) undergoes an  alpha decay

Required

The daughter nuclide

Solution

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,  

  • alpha α particles ₂He⁴
  • beta β ₋₁e⁰ particles
  • gamma particles ₀γ⁰
  • positron particles ₁e⁰
  • neutron ₀n¹

The decay reaction uses the principle: the sum of the atomic number and mass number before and after decay are the same

Radium (Ra-226) : ₈₈²²⁶Ra

Alpha particles : ₂⁴He

So Radon-226 emits alpha α particles ₂He⁴  , so the atomic number decreases by 2, mass number decreases by 4

The reaction :

₈₈²²⁶Ra ⇒ ₂⁴He + ₈₆²²²Rn

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3 years ago
A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

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E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

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n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

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5 0
3 years ago
What volume of 0.120 M HNO3(aq) is needed to completely neutralize 150.0 milliliters of 0.100 M NaOH(aq)?
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(.120M HNO3)(volume HNO3)
volume HNO3 = .125L HNO3, or 125mL
HNO3
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