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Elina [12.6K]
3 years ago
7

How many grams of Cl2 are in 1.20 x 1024 Cl atoms?

Chemistry
1 answer:
Vaselesa [24]3 years ago
5 0
<h3>Answer:</h3>

70.906 g

<h3>Explanation:</h3>

We are given;

  • Atoms of Chlorine = 1.2 × 10^24 atoms

We are required to calculate the mass of Chlorine

  • We know that 1 mole of an element contains atoms equivalent to the Avogadro's number, 6.022 × 10^23.
  • That is , 1 mole of an element = 6.022 × 10^23 atoms
  • Therefore; 1 mole of Chlorine = 6.022 × 10^23 atoms

But since Chlorine gas is a molecule;

  • 1 mole of Chlorine gas = 2 × 6.022 × 10^23 atoms

But, molar mass of Chlorine gas = 70.906 g/mol

Then;

70.906 g Of chlorine gas = 2 × 6.022 × 10^23 atoms

                                          = 1.20 × 10^24 atoms

Thus;

For 1.2 × 10^24 atoms ;

= ( 70.906 g/mol × 1.2 × 10^24 atoms ) ÷ (1.20 × 10^24 atoms)

<h3>=  70.906 g </h3>

Therefore, 1.20 × 10^24 atoms of chlorine contains a mass of 70.906 g

=  

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If 14.5 g of MnO4- (permanganate) react with manganese (II) hydroxide how many grams of manganese (IV) oxide will be produced? T
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A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c
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Answer:

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Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat  capacity and the temperature change. Working with this equation, and assuming no heat is lost to  the surroundings, we write :

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Note we expressed the temperature change in K, because the heat capacity is written in K.

<u> Now that we have the heat of combustion, we need to calculate the molar heat.  </u>

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write  the <u>conversion factor as 135.58 kJ/ 4.40 g</u>.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

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Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58  kJ/mol.

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