Decreasing order of C-C bond length is 1) ethene 2)ethyne 3) benzene 4) ethane

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.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol