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Vaselesa [24]
3 years ago
7

You are performing an experiment in a lab to attempt a new method of producing pure elements from

Chemistry
1 answer:
9966 [12]3 years ago
8 0

Answer:

  • <u>Molybdenum</u>

Explanation:

Since, the atomic mass of the elements is a characteristic property of the elements, you can use the data given, number of moles and mass in grams of the product, to calculate the atomic mass of the product, and then compare with the atomic masses of the elements (information foun in any periodic table).

<u>1) Atomic mass of the product</u>:

  • Atomic mass = mass in grams / number of moles

  • Atomic mass = 604.4 g / 6.3 mol = 95.94 g/mol ≈ 94.9 g/mol (rounded to three significant figures)

<u>2) Periodic table:</u>

  • Molybdenum, Mo, the element with atomic number 42, has atomic mass equal to 95.94 g/mol.

<u>3) Conclusion</u>:

So, you can conclude safely that the element you have produced is Molybdenum.

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Answer:

A:Double replacement

Explanation:

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4 0
3 years ago
8. A sample of potassium chlorate (KCIO,) was heated in a test tube and decomposed 2KC?(s) 302 (g) + 2KCIO, (s) The oxygen was c
dangina [55]

Answer:

Partial pressure of O_{2} in the gas was 733 torr and mass of KClO_{3} in the sample was 2.12 g.

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of O_{2})

Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.

So, partial pressure of O_{2}= (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr

b) Lets assume that O_{2} behaves ideally. Hence-

                                            PV=nRT

where P is pressure of O_{2}, V is volume of O_{2} , n is number of moles of O_{2} , R is gas constant and T is temperature in kelvin

here P = 733 torr = (733\times 0.001316)atm = 0.9646 atm

        V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K

   So, n=\frac{PV}{RT}

                   = \frac{(0.9646 atm)\times (0.65 L)}{(0.082 L.atm/(mol.K))\times (295 K)}

                   = 0.0259 moles

As 3 moles of O_{2} are produced from 2 moles of KClO_{3} therefore 0.0259 moles of O_{2} are produced from (\frac{2\times 0.0259}{3}) moles or 0.0173 moles of KClO_{3}.

Molar mass of KClO_{3}= 122.55 g

So mass of KClO_{3} in sample = (0.0173\times 122.55)g

                                                                    = 2.12 g

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