Answer:
4>3>1>2.
That is, the C-C bond length in ethane > benzene > ethene > ethyne.
Explanation:
The C-C bond in ethane is single, the C-C bond in ethene is double and the C-C bond in ethyne is triple. As the number of bonds between a C-C increases, the length of the bond decreases with an increase in strength. This explains why the C-C bond length in ethane > ethene > ethyne. For benzene, all the C-C bonds in the aromatic compound has been found to have an identical length of 1.40 Å, compared to ethane (1.54Å), ethene (1.34Å) and ethyne (1.20Å). Hence the trend in bond lengths: ethane > benzene > ethene > ethyne.
Answer:
A. P₂ / P₁ = 2
B. P₂ / P₁ = 1.1
Explanation:
A. Determination of the ratio P₂/P₁
Volume = constant
Initial temperature (T₁) = 46 K
Final temperature (T₂) = 92 K
Final pressure /Initial pressure (P₂/P₁) =?
P₁/T₁ = P₂/T₂
P₁/46 = P₂/92
Cross multiply
46 × P₂ = P₁ × 92
Divide both side by P₁
46 × P₂ / P₁ = 92
Divide both side by 46
P₂ / P₁ = 92 / 46
P₂ / P₁ = 2
B. Determination of the ratio P₂/P₁
Volume = constant
Initial temperature (T₁) = 35.4 °C = 35.4 + 273 = 308.4 K
Final temperature (T₂) = 69.0 °C = 69 + 273 = 342 K
Final pressure /Initial pressure (P₂/P₁) =?
P₁/T₁ = P₂/T₂
P₁/308.4 = P₂/342
Cross multiply
308.4 × P₂ = P₁ × 342
Divide both side by P₁
308.4 × P₂ / P₁ = 342
Divide both side by 308.4
P₂ / P₁ = 342 / 308.4
P₂ / P₁ = 1.1
The reactants, products, coefficients, subscripts. ( I forgot the rest lol)
Answer:
- Empirical:
- Molecular:
Explanation:
Hello,
In this case, based on the information regarding the combustion, the moles of carbon turn out:
Moreover, the moles of hydrogen:
Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:
Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:
In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:
Which is hexane.
Best regards.
