Explanation:
The reaction equation will be as follows.

Calculate the amount of
dissolved as follows.

It is given that
= 0.032 M/atm and
=
atm.
Hence,
will be calculated as follows.
=
= 
= 
or, = 
It is given that 
As, ![K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%7D%7B%5BCO_%7B2%7D%5D%7D)
= 
= 
Since, we know that pH = ![-log [H^{+}]](https://tex.z-dn.net/?f=-log%20%5BH%5E%7B%2B%7D%5D)
So, pH = 
= 5.7
Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.
<h3>
Answer:</h3>
1.85 M
<h3>
Explanation:</h3>
<u>We are given;</u>
- Number of moles as 0.50 mol
- Volume of the solution is 270 ml
But, 1000 mL = 1 L
- Thus, volume of the solution is 0.27 L
We are required to calculate the molarity of the solution;
- Molarity refers to the concentration of a solution in moles per liter.
- It is calculated by dividing number of moles with the volume.
Molarity = Moles ÷ Volume
In this case;
Molarity = 0.50 moles ÷ 0.27 L
= 1.85 Mol/L or 1.85 M
Therefore, molarity of the solution is 1.85 M
<h3>
Answer:</h3>
5.55 mol C₂H₅OH
<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 Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
[Given] 500. g C₆H₁₂O₆ (Glucose)
[Solve] moles C₂H₅OH (Ethanol)
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH
[PT] Molar mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:

- [DA} Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH
Answer:
A.'C
Explanation:
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