Question

Question 1
Coding Challenge
Reverse 11
Problem Statement: N people are standing in a queue. The line administrator orders instructions of reversing first k people. The instruction is given M times! What will be the alignment of people after all the operations?
Input Format The first line contains two space separated integers N and M The second line contains N space separated integers denoting Next M lines contains the value of K for each reverse operation​

Answer 1

Answer:

35.5 g of chlorine occupies 11,200 cm³ at S.T.P.

35.5 g of chlorine occupies 11,200 cm³ at S.T.P. We have to calculate the atomicity of chlorine.

35.5 g of chlorine occupies 11,200 cm³ at S.T.P. We have to calculate the atomicity of chlorine. As we know 1 mole of any particular substance is equal to 1 g molecular mass of it.

i.e., 1 mole of substance = 1 g molecular mass of it

1 gram of chlorine occupies how much ?

1 gram of chlorine occupies how much ? ==> 11200cm³

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200 ==> 22400 × 355 / 11200 × 10

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200 ==> 22400 × 355 / 11200 × 10 ==> 224 × 355 / 112 × 10

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200 ==> 22400 × 355 / 11200 × 10 ==> 224 × 355 / 112 × 10 ==> 224 × 71 / 112 × 2

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200 ==> 22400 × 355 / 11200 × 10 ==> 224 × 355 / 112 × 10 ==> 224 × 71 / 112 × 2 ==> 112 × 71 / 112

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200 ==> 22400 × 355 / 11200 × 10 ==> 224 × 355 / 112 × 10 ==> 224 × 71 / 112 × 2 ==> 112 × 71 / 112 ==> 71

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Thus, gram molecular mass would be 71g .

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula:

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic mass

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values:

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values: ==> 71 / 35.5

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values: ==> 71 / 35.5 Solving now,

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values: ==> 71 / 35.5 Solving now,==> 71 × 10 / 355

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values: ==> 71 / 35.5 Solving now,==> 71 × 10 / 355==> 71 × 2 / 71

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values: ==> 71 / 35.5 Solving now,==> 71 × 10 / 355==> 71 × 2 / 71 ==> 2

=> atomicity of chlorine is 2