The arrangement in which the X substituents are at maximum separation is called the anti conformation. What is the dihedral angle between the X substituents for the anti conformation

The dihedral angle between the X substituents for the anti conformation is 180°

Explanation:

Note: I think this question is not complete. Still, it can be answered. But for your concept and understanding, I am completing this question and then I will try to solve this question for you.

Other part of this question is about Gauche conformations and angle between the x substituents.

So, I will be drawing and answering both.

Answer to the original question: 180°

So, the dihedral angle between the X substituents for the anti conformation is 180°

Answer to the Gauche Conformation: 60°

Here, in this question we are discussing about the possible different staggered conformations of CH2XCH2X

You can see the angles between the X substitutes in anti conformations and Gauche Conformations for yourself in the attachment below.

Furthermore, this CH2XCH2X has 3 possible different staggered conformations. Out of which, one is anti conformation and other two are Gauche conformations.

Please refer to the attachment to see the conformations. Check the angles between X-X. In anti conformation, angle between X-X is 180° and energy is lower, because of the less interaction between H and X. However, in Gauche conformation, angle between X-X is 60° and energy is higher, because of the more interactions between H and X than anti conformation.

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