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Ulleksa [173]
3 years ago
13

Prove that an element has an order in an abelian group

Chemistry
1 answer:
Allushta [10]3 years ago
4 0

<span>2down votefavoriteI will quote a question from my textbook, to prevent misinterpretation:Let <span>GG</span> be a finite abelian group and let <span>mm</span> be the least common multiple of the orders of its elements. Prove that <span>GG</span> contains an element of order <span>mm</span>.I figured that, if <span><span>|G|=n</span><span>|G|=n</span></span>, then I should interpret the part with the least common multiple as <span><span>lcm(|<span>x1</span>|,…,|<span>xn</span>|)=m</span><span>lcm(|<span>x1</span>|,…,|<span>xn</span>|)=m</span></span>, where <span><span><span>xi</span>∈G</span><span><span>xi</span>∈G</span></span> for <span><span>0≤i≤n</span><span>0≤i≤n</span></span>, thus, for all such <span><span>xi</span><span>xi</span></span>, <span><span>∃<span>ai</span>∈N</span><span>∃<span>ai</span>∈N</span></span> such that <span><span>m=|<span>xi</span>|<span>ai</span></span><span>m=|<span>xi</span>|<span>ai</span></span></span>. I guess I should use the fact that <span><span>|<span>xi</span>|</span><span>|<span>xi</span>|</span></span> divides <span><span>|G|</span><span>|G|</span></span>, so <span><span>∃k∈N</span><span>∃k∈N</span></span> such that <span><span>|G|=k|<span>xi</span>|</span><span>|G|=k|<span>xi</span>|</span></span> for all <span><span><span>xi</span>∈G</span><span><span>xi</span>∈G</span></span>. I'm not really sure how to go from here, in particular how I should use the fact that <span>GG</span> is abelian.</span>
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Net ionic of ammonium sulfide added to iron (ll) chloride
meriva

Answer:

Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)  

Step-by-step explanation:

Molecular Equation:

(NH₄)₂S(aq) + FeCl₂(aq) ⟶ 2NH₄Cl(aq) + FeS(s)

Ionic equation :

2NH₄⁺(aq) + S²⁻(aq) + Fe²⁺(aq) + 2Cl⁻(aq) ⟶ 2NH₄⁺(aq) + 2Cl⁻(aq) + FeS(s)

Net ionic equation :

Cancel all ions that appear on both sides of the reaction arrow (underlined).

<u>2NH₄⁺(aq)</u> + S²⁻(aq) + Fe²⁺(aq) + <u>2Cl⁻(aq)</u> ⟶ <u>2NH₄⁺(aq) </u>+ 2<u>Cl⁻(aq) </u>+ FeS(s)

Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)

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3 years ago
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Which solution has the same boiling point as 0.25 mol CaCl2 dissolved in 1000 g water?
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</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
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Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.

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