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jolli1 [7]
2 years ago
10

At 580 k, 6. 42% of the molecules are in the chair form. calculate the value of the equilibrium constant for the reaction as wri

tten above.
Chemistry
1 answer:
Ann [662]2 years ago
5 0
Equilibrium chemical reaction between chair and boat forms is:
Kc = \frac{[Boat form]}{[Chair form]}
Suppose the total number of molecules is x 
Number of molecules of chair form is 6.42/100 x
Number of molecules of boat form is 93.58/100 x
Kc = \frac{(0.9358x)}{(0.0642x)} = 14.6
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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
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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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