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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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The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of B
loris [4]

I believe here is the right question, so will just ignore the rest of the junk information from the previous message

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT

Answer:

C_{15}H_{24}0

Explanation:

A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).

If all the carbon in BHT is present in CO_2 and also, all the hydrogen in BHT is  present in H_2O, Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:

moles of  CO_2 = 8.990 g*(\frac{1mole}{44.01g})

                       =  0.2043 moles

∴ moles of C =  0.2043 moles

moles of H_2O = 2.944 g *(\frac{1mole}{18.01g} )

                       = 0.1635 moles

∴ moles of H = 2 × 0.1635 moles

                      = 0.327 moles

Since number of moles= \frac{mass}{molarmass}

number of moles of H =  0.327 moles

molar mass of H = 1.008 g/mol

∴  mass of H in the sample = 0.327 moles × 1.008 g/mol

                                             = 0.329616g

                                             

mass of C in the sample can be calculated as = 0.2043 moles × (\frac{12.01g}{1 mole} )

= 2.453643 g

mass of C+H in the sample = 2.453643g + 0.329616g

= 2.783259 g

mass of O can be calculated as = 3.001 g - 2.783259 g

= 0.217741 g

∴ moles of O = 0.217741g × (\frac{1mole}{16.0g})

= 0.0136 moles

Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.

C                                          H                                    O

0.2043                                0.327                             0.0136

Dividing by the least number (0.0136) , we have :

\frac{0.2043}{0.0136}                                     \frac{0.327}{0.0136}                               \frac{0.0136}{0.0136}

15.02                                      24.04                             1

≅

15                                             24                                  1

Therefore, the empirical formula would be : C_{15}H_{24}0

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