The daughter isotope (a decay product)of O-15 = N-15(Nitrogen 15)
<h3>Further explanation
</h3>
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 γ
- positron particles ₁e⁰
O-15 emits positron particles ₁e⁰, so the atomic number decreases by 1, the mass number is the same
Reaction
The mass number of the daughter isotope = 15, atomic number = 7
If we look at the periodic system, the element with atomic number 7 is Nitrogen (N)
2H2O+O2--->2H2O2
8.5 gm H2O2=0.25 mole
hence H2O is also 0.25 mole i.e.4.5 gm
O2is 0.125 mole i.e.4 gm
Pv=nRT
where,p=199, R(constant)=8.314, V=4.67 T=30C=293K
n=pv/RT=0.38 moles
Answer:
Explanation:
<u>1) Data:</u>
a) Hypochlorous acid = HClO
b) [HClO} = 0.015
c) pH = 4.64
d) pKa = ?
<u>2) Strategy:</u>
With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.
<u>3) Solution:</u>
a) pH
b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)
c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]
d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M
e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M
f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46
Given what we know about ratios, we can confirm that in comparison to the iron pan, the aluminum pan will receive twice as much heat.
<h3>Why would the aluminum pan receive double the heat?</h3>
- This has to do with the ratio being presented.
- A ratio of 2:1 means that for every unit of heat to the iron, the aluminum receives 2.
- In other words, the aluminum pan receives twice as much heat when compared to the iron pan.
Therefore, we can confirm that because the ratio of heat to the aluminum in comparison to the iron pan is 2:1, this means that for every 1 unit of heat to the iron pan, the aluminum pan will receive 2. This results in double the total heat received by the aluminum pan.
To learn more about ratios visit:
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