About <u>1 in 500,000,000</u> molecules in a glass of water are dissociated.
<h3>What is water?</h3>
Water can be defined as a liquid molecule or chemical compound that typically comprises two (2) main chemical elements in a ratio of two (2) atoms of hydrogen to one (1) atom of oxygen. This ultimately implies that, these two (2) chemical elements include the following:
Water is an amphoteric chemical compound (substance), which ultimately implies that it can accept a proton that is also acting as a base, while also donating a proton that acts as an acid as shown in the balance chemical reaction below:
2H₂O(l) ⇄ H₃O⁺(aq) + OH⁻(aq)
Based on scientific research and experiments, approximately one (1) water molecule in half a billion (500,000,000) dissociate into an OH⁻ ion, especially by losing a proton to another water molecule.
Read more on water here: brainly.com/question/15164981
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The solid is ice because it has london disperson forces due to electron revolving around nucleus in atom of hydrogen and oxygen
It also has dipole-dipole attraction due to differnce in electronegativity among hydrogen and oxygen atom
It also has hydrogen bonding between oxygen and hydrogen
The C-H bonds (in methane) break and so do the O-O bonds (in oxygen)
<span>The reaction is :</span>
CH4 +2O2 -> CO2 + 2H2O
Answer:
mass of one carbon atom = g
Explanation:
1 mole of carbon contains atoms of carbon
therefore mass of atoms of carbon is 12.011 grams. therefore mass of one carbon atom will be g
mass of one carbon atom = g
Answer:
1. V₁ = 2.0 mL
2. V₁ = 2.5 mL
Explanation:
<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>
To calculate the volume required (V₁) in each dilution we will use the dilution rule.
C₁ . V₁ = C₂ . V₂
where,
C are the concentrations
V are the volumes
1 refers to the initial state
2 refers to the final state
<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>
C₁ . V₁ = C₂ . V₂
(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL
V₁ = 2.0 mL
<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>
C₁ . V₁ = C₂ . V₂
(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL
V₁ = 2.5 mL