The molecule cyanazine, once used as an herbicide, has now been phased out by the EPA because of its significant human toxicity. Add lone pairs, as needed, to the structure of cyanazine. Indicate the hybridization of the specified atoms. Be sure to consider any lone pairs you added to the structure. The structure of cyanazine has 5 atoms highlighted. Atom 1 is an N atom bonded to 2 C atoms and 1 H atom through single bonds. Atom 2 is an N atom bonded to one C atom through a single bond and one C atom through a double bond. Atom 3 is a C atom bonded to one C l atom and one N atom through a single bond and one N atom through a double bond. Atom 4 is a C atom that is bonded to three C atoms and one N atom through single bonds. Atom 5 is a C atom that is bonded to one C atom through a single bond and one N atom through a triple bond. What is the hybridization of atom 1

Cyanazine is used as an herbicide to control the growth of weeds and annual grasses. The herbicide has been phased out due to increased toxicity to the environment and humans.

The hybridization of the molecule is sp³.

The given information in the question suggests that Nitrogen (atom 1) to two Carbon atoms and single bonded to 1 hydrogen atom.

The nitrogen is bonded with three single bonds, two with carbon and one with a hydrogen bond.

Based on the given data, the hybridization of the molecule will be sp³ and will have tetrahedral geometry. The lone pair present on the nitrogen atom will occupy one of the hybrid sp³ orbitals.

Therefore, the sp3 hybridization of the cyanazine molecule will exist.

To know more about the hybridization of cyanazine refer to the following link:

link

Sp3

Explanation:

According to the question, atom 1 is an N atom bonded to 2 C atoms and 1 H atom through single bonds.

Whenever nitrogen is bonded to three atoms via single bonds (in this case two carbon atoms and one hydrogen atom) the nitrogen atom is sp3 hybridized and in tetrahedral geometry. The lone pair on nitrogen atom occupies one of the hybrid sp3 orbitals.

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