Consider the reaction chcl3(g) + cl2(g)  ⟶  ccl4(g) + hcl(g).use the following rate data to determine the order of the reaction with respect to cl2      [chcl3] (m)                                [cl2] (m)                        initial rate (m/s)      0.010                                                   0.010                                                    0.0035      0.020                                                    0.010                                                    0.0070      0.020                                                    0.020                                                    0.0099121.40.5

The reaction is half-order in [Cl₂].  

For the reaction A + B ⟶ products

The rate law is r = k [A]^m[B]^n

Consider two Experiments 1 and 2.

r₂/r₁ = {k[A]₂^m[B]₂^n}/{k[A]₁^m[B]₁^n}  

r₂/r₁ = {[A]₂/[A]₁}^m × {[B]₂/[B]₁}^n

Let A = CHCl₃ and B = Cl₂

You must find two experiments in which [B] changes but [A] does not.

Consider Experiments 3 and 2.

r₃/r₂ = (0.020/0.020)^m × (0.020/0.010)^n = 0.0099/0.0070

1.0^m × 2.0^n = 1.41

2.0^m = 1.41

mlog2 = log 1.41

m = log1.41/log2 = 0.50 ≈ ½

The reaction is half-order in [Cl₂].

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.