A) The answer to whether an average-size, single-family home be built with one day’s output from the saw mill is; No, it can't be built from one days output
B) The number of days will it take for the saw mill to produce enough lumber to build a neighborhood of 100 average-size homes is; 1100 days
A) From research, the average size capacity saw mill usually produces around 1500 board feet of boards per day.
Now, converting board ft to cubic meter gives;
1 board feet = 1/424 m³
Thus; 1500 board feet = 1500 * 1/424 ≈ 3.54 m³
Now, we are told that it takes 38 m³ of boards and plywood to build an average size single family home.
Since one days output = 3.54 m³
It is less than the volume required for the average sized house and so the daily output will not be sufficient.
B) Number of days to produce enough boards and plywood for the average sized family home = 38/3.54 ≈ 11 days
Thus, for 100 homes;
Number of days = 100 * 11 = 1100 days
Read more about production capacity at; brainly.com/question/25899399
Answer:
46 g
Explanation:
First we <u>convert 1.2 x 10²⁴ atoms of sodium into moles</u>, using <em>Avogadro's number</em>:
- 1.2x10²⁴ atoms ÷ 6.023x10²³ atoms/mol = 2.0 mol
Then we <u>convert 2.0 moles of sodium into grams</u>, using <em>sodium's molar mass</em>:
- 2.0 mol Na * 23 g/mol = 46 g
Thus, there are 46 grams in 1.2x10²⁴ atoms of sodium.
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)
