Given the following balanced equation, determine the rate of reaction with respect to [cl2]. if the rate of cl2 loss is 4.44 × 10-2 m/s, what is the rate of formation of nocl? 2 no(g) + cl2(g) → 2 nocl(g)1.11 × 10-1 m/s2.22 × 10-2 m/s8.88 × 10-2 m/s1.61 × 10-2 m/s4.44 × 10-2 m/s

8.88 x 10⁻² M/s

Explanation:

The rate  of reaction for:

                    NO(g) + Cl₂ (g) ⇒  2NOCl(g)

is rate = -ΔNO/Δt = -ΔCl2/Δt = 1/2 ΔNOCl/Δt

so  ΔNOCl/Δt = 2 ΔCl2/Δt  = 2  x 4.44 × 10⁻²  M/s = 8.88 x 10⁻² M/s

In general given a reaction

                            aA + bB ⇒ cC + dD

rate = -1/a ΔA/Δt = -1/b ΔB/Δt = 1/c ΔC/Δt = 1/d ΔD/Δt

i think it is true since you can test the answer and can use observations to complete it.

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

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~ trueboss72