Answer:
first choice
Explanation:
energy is written on the left side is its absorbed
2 elements came together so bonds are formed.
Answer:
Write a hypothesis statement about the relationship between the number of mushrooms and the concentration of heavy metals in the soil ...
<h3>
Answer:</h3>
0.0463 mol KCl
<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>
3.45 g KCl
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of KCl - 39.10 + 35.45 = 74.55 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.046278 mol KCl ≈ 0.0463 mol KCl
Answer:
121.37 moles.
Explanation:
Always one mole of any molecule contains Avogadro number of molecules.
So one mole of methane contains 6.023*10^23 molecules.
Now we need number of moles of methane in 7.31*10^25 molecules.
This is just cross multiplication stuff.
6.023*10^23 molecules —————- 1 mole
7.31*10^25 molecules ——————- ?
= (7.31*10^25) / (6.023*10^23)
= 121.37 moles.
Answer:

Explanation:
1. Mass of acetylsalicylic acid (ASA)

2. Moles of ASA
HC₉H₇O₄ =180.16 g/mol

3. Concentration of ASA

4. Set up an ICE table

5. Solve for x
![K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}]\text{A}^{-}]} {\text{[HA]}} = 3.33 \times 10^{-4}\\\\\dfrac{x^{2}}{0.01757 - x} = 3.33 \times 10^{-4}\\\\\textbf{Check that }\mathbf{x \ll 0.01757}\\\\\dfrac{ 0.01757 }{3.33 \times 10^{-4}} = 53 < 400\\\\\text{The ratio is less than 400. We must solve a quadratic equation.}\\\\x^{2} = 3.33 \times 10^{-4}(0.01757 - x) \\\\x^{2} = 5.851 \times 10^{-6} - 3.33 \times 10^{-4}x\\\\x^{2} + 3.33 \times 10^{-4}x - 5.851 \times 10^{-6} = 0](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Ba%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BH%7D_%7B3%7D%5Ctext%7BO%7D%5E%7B%2B%7D%5D%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%20%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%203.33%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.01757%20-%20x%7D%20%3D%203.33%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5C%5Ctextbf%7BCheck%20that%20%7D%5Cmathbf%7Bx%20%5Cll%200.01757%7D%5C%5C%5C%5C%5Cdfrac%7B%200.01757%20%7D%7B3.33%20%5Ctimes%2010%5E%7B-4%7D%7D%20%3D%2053%20%3C%20400%5C%5C%5C%5C%5Ctext%7BThe%20ratio%20is%20less%20than%20400.%20We%20must%20solve%20a%20quadratic%20equation.%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%203.33%20%5Ctimes%2010%5E%7B-4%7D%280.01757%20-%20x%29%20%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%205.851%20%5Ctimes%2010%5E%7B-6%7D%20-%203.33%20%5Ctimes%2010%5E%7B-4%7Dx%5C%5C%5C%5Cx%5E%7B2%7D%20%2B%203.33%20%5Ctimes%2010%5E%7B-4%7Dx%20-%205.851%20%5Ctimes%2010%5E%7B-6%7D%20%3D%200)
6. Solve the quadratic equation.


7. Calculate the pH
![\rm [H_{3}O^{+}]= x \, mol \cdot L^{-1} = 0.002258 \, mol \cdot L^{-1}\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.002258} = \mathbf{2.65}\\\text{The pH of the solution is } \boxed{\textbf{2.65}}](https://tex.z-dn.net/?f=%5Crm%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D%20x%20%5C%2C%20mol%20%5Ccdot%20L%5E%7B-1%7D%20%3D%200.002258%20%5C%2C%20mol%20%5Ccdot%20L%5E%7B-1%7D%5C%5C%5Ctext%7BpH%7D%20%3D%20-%5Clog%7B%5Crm%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%20%3D%20-%5Clog%7B0.002258%7D%20%3D%20%5Cmathbf%7B2.65%7D%5C%5C%5Ctext%7BThe%20pH%20of%20the%20solution%20is%20%7D%20%5Cboxed%7B%5Ctextbf%7B2.65%7D%7D)