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:
86,400 seconds in a day.
Explanation:
There are 24 hrs in a day. So there are 2460 mins in a day. ( 1hr=60 mins) So there are 246060 seconds in a day. (1 min = 60 seconds) Therefore 86,400 seconds in a day.
With computer you have a wire you connect the nuclear and the computer together that is how it works in the in the computer there is a stopwatch of 30 minutes
Answer:
T₂ = 721 k
Explanation:
Given data:
Initial volume = 285 mL
Initial pressure = 1.88 atm
Initial temperature = 355 K
Final temperature = ?
Final volume = 435 mL
Final pressure = 2.50 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁ / P₁V₁
T₂ = 2.50 atm × 435 mL × 355 K / 1.88 atm × 285 mL
T₂ = 386062.5 atm. mL. K /535.8 atm. mL
T₂ = 721 k