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andre [41]
3 years ago
8

Two samples of potassium iodide are decomposed into their constituent elements. The first sample produced 13.0 g of potassium an

d 42.3 g of iodine. If the second sample produce 24.4 kg of potassium, how many kg of iodine were produced?
Chemistry
1 answer:
Cerrena [4.2K]3 years ago
5 0

Answer:

79.4 kg

Explanation:

The formula for potassium iodide is:-

KI

This means that they are present in the same mole ration, i.e. 1 : 1

Also for 2 samples, the ratio of the masses must be equal .

So,

\frac{Mass\ of\ potassium\ in\ sample\ 1}{Mass\ of\ Iodine\ in\ sample\ 1}=\frac{Mass\ of\ potassium\ in\ sample\ 2}{Mass\ of\ Iodine\ in\ sample\ 2}

Thus,

Mass of potassium in sample 1 = 13.0 g

Mass of Iodine in sample 1 = 42.3 g

Mass of potassium in sample 2 = 24.4 kg

Applying in the above formula, we get that:-

\frac{13.0\ g}{42.3\ g}=\frac{24.4\ kg}{Mass\ of\ Iodine\ in\ sample\ 2}

Mass of iodine in sample 2 = 79.4 kg

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expeople1 [14]

The expression for  S°(Cl2O)(g) is given by ,

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The general standard entropy of the reaction is given by ,

ΔS°rxn = sum of the S°(Product ) - sum of the S°(reactants )

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where , m and n are the stoichiometry coefficients of each product and reactants .

Thus , the standard entropy of the given reaction is given by ,

ΔS°rxn = 2×S° (HClO)(g)  - [ S°(H2O)(g) + S°(Cl2O)(g) ]

ΔS°rxn = 2×S° (HClO)(g) - S°(H2O)(g) - S°(Cl2O)(g)

Thus the S°(Cl2O) is given by ,

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Hence the expression for the standard entropy of Cl2O (g) is given by ,

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6 0
1 year ago
Which are the hybrid orbitals of the carbon atoms in the following molecules? (a) H3C―CH3 sp sp2 sp3 (b) H3C―CH═CH2 sp sp2 sp3 (
kherson [118]

Answer:

(a)  sp³    sp³

   H₃<u>C</u> - <u>C</u>H₃

(b)     sp³           sp²

     H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂

                sp²

(c)     sp³        sp    

    H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH

              sp         sp³

(d)     sp³    sp²    

       H₃<u>C</u> - <u>C</u>H=O

Explanation:

Alkanes or the carbons with all the single bonds are sp³ hybridized.

Alkenes or the carbons with double bond(s) are sp² hybridized.

Alkynes or the carbons with triple bond are sp hybridized.

Considering:

(a) H₃C-CH₃ , Both the carbons are bonded by single bond so both the carbons are sp³ hybridized.

Hence,

 sp³    sp³

H₃<u>C</u> - <u>C</u>H₃

(b) H₃C-CH=CH₂ , The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp² hybridized because they are bonded by double bond.

Hence,

   sp³           sp²

H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂

         sp²

(c) H₃C-C≡C-CH₂OH , The carbons of the methyl group and alcoholic group are sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp hybridized because they are bonded by triple bond.

Hence,

   sp³        sp    

H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH

         sp         sp³

(d)CH₃CH=O, The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The other carbon is sp² hybridized because it is  bonded by double bond to oxygen.

Hence,

   sp³    sp²    

 H₃<u>C</u> - <u>C</u>H=O

     

8 0
3 years ago
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