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 mass of one mole of a substance is equal to that substance's molecular weight. ... water is 18.015 atomic mass units (amu), so one mole of water weight 18.015 grams. ... Avogadro's number is a proportion that relates molar mass on an atomic ... one molecule of water (H2O), one mole of oxygen (6.022×1023 of O atoms)
Answer:
18.066 × 10²³ particles
Explanation:
Given data:
Number of moles = 3 moles
Number of particles = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
For 3 moles of substance:
One mole = 6.022 × 10²³ particles
3 mol × 6.022 × 10²³ particles/ 1 mol = 18.066 × 10²³ particles
Answer:
Given, 0.29 g of hydrocarbon produces 448ml of CO2 at STP. then, C2H5 is the emperical formula of hydrocarbon . n = 2 , hence, molecular formula will be C4H10
<span>Answer: D. They all have the same number of electrons in the electron cloud</span>
