Answer:
Mole fraction of methanol will be closest to 4.
Explanation:
Given, Mass of methanol = 128 g
Molar mass of methanol = 32.04 g/mol
The formula for the calculation of moles is shown below:
Thus,
Given, Mass of water = 108 g
Molar mass of water = 18.0153 g/mol
The formula for the calculation of moles is shown below:
Thus,
So, according to definition of mole fraction:
<u>Mole fraction of methanol will be closest to 4.</u>
Answer:
b) 0.47
Explanation:
MwC5H12 = 72.15g/mol
⇒mol C5H12 = (10.0)*(mol/72.15)=0.1386molC5H12
MwC6H14=86.18g/mol
⇒molC6H14=(20.0)*(mol/86.18)=0,232
MwC6H6=78.11g/mol
⇒molC6H6=(10.0)*(mol/78.11)=0.128molC6H6
<h3>XC6H14=(0.232)/(0.1386+0.232+0,128)=0.465≅0.47</h3>
Answer:
The tissue cells
Explanation:
I think you mean this
It all starts from Carbondioxide. This Carbondioxide is dissolved in the blood and taken by red blood cell and converted into carbonic acid. It then dissociates to form a bicarbonate ion 
and a hydrogen ion 
This <--> means that the whole process is reversible. It is a buffer system to maintain the pH in the blood and duodenum. And also to support proper metabolic function.
Answer:
Close to the calculated endpoint of a titration - <u>Partially open</u>
At the beginning of a titration - <u>Completely open</u>
Filling the buret with titrant - <u>Completely closed</u>
Conditioning the buret with the titrant - <u>Completely closed</u>
Explanation:
'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.
As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.
Answer:
1.8 × 10⁻¹⁶ mol
Explanation:
(a) Calculate the solubility of the Sr₃(PO₄)₂
Let s = the solubility of Sr₃(PO₄)₂.
The equation for the equilibrium is
Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹
1.2 + 3s 2s
![K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} = 1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BSr%24%5E%7B2%2B%7D%24%5D%24%5E%7B3%7D%24%5BPO%24_%7B4%7D%5E%7B3-%7D%24%5D%24%5E%7B2%7D%24%7D%20%3D%20%281.2%20%2B%203s%29%5E%7B3%7D%5Ctimes%20%282s%29%5E%7B2%7D%20%3D%20%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5C%5Ctext%7BAssume%20%7D%203s%20%5Cll%201.2%5C%5C1.2%5E%7B3%7D%20%5Ctimes%204s%5E%7B2%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5C6.91s%5E%7B2%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5Cs%5E%7B2%7D%20%3D%20%5Cdfrac%7B1.0%20%5Ctimes%2010%5E%7B-31%7D%7D%7B6.91%7D%20%3D%201.45%20%5Ctimes%2010%5E%7B-32%7D%5C%5C%5C%5Cs%20%3D%20%5Csqrt%7B%201.45%20%5Ctimes%2010%5E%7B-32%7D%7D%20%3D%201.20%20%5Ctimes%2010%5E%7B-16%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C)
(b) Concentration of PO₄³⁻
[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹
(c) Moles of PO₄³⁻
Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol
