Answer:
(a) Pair 1: H₂S and HS⁻
Pair 2: NH₃ and NH₄⁺
(b) Pair 1: HSO₄⁻ and SO₄⁻
Pair 2: NH₃ and NH₄⁺
(c) Pair 1: HBr and Br⁻
Pair 2: CH₃O⁻ and CH₃OH
(d) Pair 1: HNO₃ and NO₃⁻
Pair 2: H₃O⁺
Explanation:
When an acid loses its proton (H⁺), a conjugate base is produced.
When a base accepts a proton (H⁺), it forms a conjugate acid.
(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.
NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺
(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.
The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.
(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.
CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.
(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.
H₂O gains a proton to form the conjugate acid H₃O⁺.
Answer:
the sun Solar energy
Explanation:
Energy from the sun. Solar energy is the original source of most energy on Earth. There are many ways we use energy from the Sun.
Answer:
They both have 7 electrons in their outer shell
The result of Moseley's revisions were that the elements were arranged in atomic number order rather than atomic mass order.
Answer:
Kc = 4.774 * 10¹³
Explanation:
the desired reaction is
2 NO₂(g) ⇋ N₂(g) + 2 O₂(g)
Kc =[N₂]*[O₂]² /[NO₂]²
Since
1/2 N₂(g) + 1/2 O₂(g) ⇋ NO(g)
Kc₁= [NO]/(√[N₂]√[O₂]) → Kc₁²= [NO]²/([N₂][O₂])
and
2 NO₂(g) ⇋ 2 NO(g) + O₂(g)
Kc₂= [NO]²*[O₂]/[NO₂]² → 1/Kc₂= [NO₂]²/([NO]²[O₂])
then
Kc₁²* (1/Kc₂) = [NO]²/([N₂]*[O₂]) *[NO₂]²/([NO]²[O₂]) = [NO₂]²/([N₂]*[O₂]²) = 1/Kc
Kc₁² /Kc₂ = 1/Kc
Kc= Kc₂/Kc₁² =1.1*10⁻⁵/(4.8*10⁻¹⁰)² = 4.774 * 10¹³
