Answer:
In an acid-base equilibrium, acid becomes a conjugate base and base becomes a conjugate acid.
Explanation:
Let's remember the Bronsted-Lowry theory to answer this specific question. According to the theory, acid is a proton donor, while a base is a proton acceptor.
Consider an acid in a form HA (aq) and base in a form of B (aq). Since acid is a proton donor, it will donate its hydrogen ion to the base, B. The resultant products would be
(aq) and
(aq).
Remember that an acid-base reaction is an equilibrium reaction. This means we may also look at this proton transfer reaction from the product side towards the reactants. Summarizing what has been said, we may write the equilibrium as:
⇄ 
Now acid, HA, donates a proton to become a conjugate base. The conjugate base, if we look from the reverse equation side, is actually a base, since it can accept a proton to become HA. Similarly, B accepts a proton to become a conjugate acid. Looking from the reverse reaction, it can now donate a proton, so in reality we can consider it a base.
To summarize, your logic is correct.
ones that are answered through observing :)
Answer: The suffixes of the names of polyatomic ions have a pattern associated with them. ... For example, the sulfite ion has three oxygen atoms whereas the sulfate ion has four oxygen atoms.
Temperature, salinity, and density are the group of factors are most important in determining the composition of ocean water.
a.)temperature, salinity, and density
<u>Explanation:</u>
The three fundamental factors that help in determining the composition of ocean water are temperature, salinity, and density. Temperature, saltiness, salinity, and density influence the thickness of seawater.
Enormous water masses of various densities are significant in the layering of the sea water (increasingly thick water sinks). As temperature builds water turns out to be less thick. As saltiness builds water gets denser. The temperature helps in deciding the pace of vanishing of the ocean.
<h3>
Answer:</h3>
0.50 mol SiO₂
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
30 g SiO₂ (sand)
<u>Step 2: Identify Conversions</u>
Molar Mass of Si - 28.09 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of SiO₂ - 28.09 + 2(16.00) = 60.09 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig figs and round. We are given 2 sig figs.</em>
0.499251 mol SiO₂ ≈ 0.50 mol SiO₂