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Luda [366]
3 years ago
11

The sodium isotope with 13 neutrons. Express your answer as an isotope

Chemistry
1 answer:
wariber [46]3 years ago
6 0

Answer:

          \large\boxed{\large\boxed{^{24}_{11}Na}}

Explanation:

The question is asking to write the <em>isotopic symbol </em>of the form ^A_ZX for the <em>sodium isotope with 13 neutrons</em>.

In general, the isotopic symbol in the given form shows:

  • The element's chemical symbol: X

  • A: mass number of the isotope, written as a superscript to the left of the element's simbol, and

        A = number of protons + number of neutrons

  • Z: atomic number of the isotope, written as a subscript to the left of the elements's symbol,

        Z = number of protons

The atomic symbol of sodium is Na.

The atomic number, or number of protons, is the same for every isotope of the element, and you can find it in any periodic table. Tha atomic number of sodium is 11. Thus:

  • Z = 11

The mass number is the number of protons plus neutrons, hence:

  • A = 11 + 13 = 24

Now you can write the isotope symbol for the sodium isotope with 13 neutrons:

         ^{24}_{11}Na

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Balanced chemical equation for the reaction is:

2SO_{2} (g) + O_{2} (g)+ 2H_{2}O (l) ⇒H_{2} S_{} O_{4}

Moles of H_{2} S_{} O_{4} formed is 5.75 moles.

Moles of oxygen used is 5.75 moles in the reaction.

Explanation:

Data given:

moles of SO_{2} = 11.5 moles

moles of H_{2} S_{} O_{4} = ?

Moles of O_{2} needed =?

balanced equation with states of matter =?

Balanced chemical reaction under STP condition is given as:

2SO_{2}(g) + O_{2} (g) + 2H_{2}O (l) ⇒H_{2} S_{} O_{4}

From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:

so, from 11.5 moles of SO_{2}, x moles of H_{2} S_{} O_{4} is formed

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2x = 11.5

x = 5.75 moles of sulphuric acid formed.

From the balanced reaction 1 mole of oxygen reacted to form 1  mole of sulphuric acid.

when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.

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<span>44.0 kJ is the answer</span>
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For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.

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