List the substances nacl, cl2, ch3cl, and ch3cooh in order of increasing strength of intermolecular attractions. list the substances , , , and in order of increasing strength of intermolecular attractions. nacl

The order will be:

Cl₂ < CH₃Cl < CH₃COOH < NaCl

Explanation:

More the polarity in a compound more the intermolecular interactions between the molecules.

Out of the given molecules, NaCl is the ionic compound and will have strong ion -ion interactions thus will have highest order of strength of intermolecular interactions.

Then CH₃COOH may also dissociate to give acetate ion and hydrogen ion thus it will also show ionic interactions but as it is a weak electrolyte it will have less interactions then NaCl

CH₃Cl is slightly more polar than Cl₂ (which is non polar so will have only london dispersion forces which are weakest intermolecular interactions), so CH₃Cl will have dipole dipole interactions

The order will be:

Cl₂ < CH₃Cl < CH₃COOH < NaCl

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

it would be a gas formation