Using the bond energies provided below, calculate δh° for the reaction ch4(g) + 4cl2(g) → ccl4(g) + 4hcl(g) bond energies: c–h = 413 kj/mol, cl–cl = 243 kj/mol, c–cl = 339 kj/mol, h–cl = 427 kj/mol

a. 1422 kj

b. 110 kj

c. 440 kj

d. – 440 kj

e. –110 kj

option d= -440 kj

Explanation:

chemical equation:

CH4 + 4Cl2 → CCl4 + 4HCl

Given data:

Bond broken energies:

C-H = 413 Kj/mol

Cl-Cl =243 Kj/mol

Bond formation energies:

C-Cl = 339 Kj/mol

H-Cl = 427 Kj/mol

Formula:

δh° = ∑n (bonds broken) - ∑m (bonds formation)

Solution:

Total bonds broken energy:

C-H = 413 Kj/mol        4×413 Kj/mol  = 1652 Kj/mol

Cl-Cl = 243 Kj/mol      4×243 Kj/mol =  272Kj/mol

                                                   total = 2624 Kj/mol

Total bond formation energy:

C-Cl = 339 Kj/mol       4×339 Kj/mol  = 1356 Kj/mol

H-Cl = 427 Kj/mol        4×427 Kj/mol = 1708 Kj/mol

                                                  total   = 3064 Kj/mol

Now we will put the values in formula to calculate the δh° for reaction.

δh° = ∑n (bonds broken) - ∑m (bonds formation)

δh° =                      (2624) - (3064)

δh° =                      -440 kj

under a microscope, cancer cells may look very different from normal cells. they often have different sizes and some may be larger than normal while others are smaller. cancer cells are often abnormally shaped and the control centre of the cell (the nucleus) may have an abnormal appearance.

i think? ?