Answer:
A supersaturated solution made out of sugar and oil is crystallized on a surface suitable for crystal nucleation such as a stick.
I hope this helps.
Answer:
Increasing pressure shifts rxn right (lower molar volume side of equation).
Explanation:
In general, if a stress is applied to a gas phase reaction, the reaction will shift away from the applied stress and establish a new equilibrium for the given reaction. The applied stress factors include ...
changes in masses of reactant or product,
changes in applied temperature values, and/or
changes in applied pressure values for a reaction confined in a reaction vessel.
For this problem, if pressure is increased on the reaction N₂(g) + 3H₂(g) ⇄ 2NH₃(g), the reaction will shift away from the applied stress, that is, in the direction of the product side of the reaction in order to relieve the applied stress as there are fewer number of moles of gas on the product side of the equation.
Answer:
14.3 mL
Explanation:
Assume the student used 0.113 g ascorbic acid and 0.0900 mol·L⁻¹ NaOH.
1. Balanced chemical equation.
The formula of ascorbic acid is H₂C₆H₆O₆ (MM = 176.12 g/mol).
However, for the balanced equation, let's write it as H₂A.
![\rm H_{2}A + 2NaOH \longrightarrow Na_{2}A + 2H_{2}O](https://tex.z-dn.net/?f=%5Crm%20H_%7B2%7DA%20%2B%202NaOH%20%5Clongrightarrow%20Na_%7B2%7DA%20%2B%202H_%7B2%7DO)
2. Moles of ascorbic acid
![\text{Moles of H$_{2}$A} =\text{0.113 g H$_{2}$A} \times \dfrac{\text{1 mmol H$_{2}$A}}{\text{0.176 12 mg H$_{2}$}A} = \text{0.6416 mmol H$_{2}$A}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20H%24_%7B2%7D%24A%7D%20%3D%5Ctext%7B0.113%20g%20H%24_%7B2%7D%24A%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mmol%20H%24_%7B2%7D%24A%7D%7D%7B%5Ctext%7B0.176%2012%20mg%20H%24_%7B2%7D%24%7DA%7D%20%3D%20%20%5Ctext%7B0.6416%20mmol%20H%24_%7B2%7D%24A%7D)
3. Moles of NaOH
The molar ratio is 2 mmol NaOH:1 mmol H₂A.
![\text{Moles of NaOH}= \text{0.6416 mmol H$_{2}$A} \times \dfrac{\text{2 mmol NaOH}}{\text{1 mmol H$_{2}$A}} =\text{1.283 mmol NaOH}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20NaOH%7D%3D%20%20%5Ctext%7B0.6416%20mmol%20H%24_%7B2%7D%24A%7D%20%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20mmol%20NaOH%7D%7D%7B%5Ctext%7B1%20mmol%20H%24_%7B2%7D%24A%7D%7D%20%3D%5Ctext%7B1.283%20mmol%20NaOH%7D)
4. Volume of NaOH
![V = \text{1.283 mmol NaOH}\times \dfrac{\text{1 mL NaOH}}{\text{0.0900 mmol NaOH}} = \textbf{14.3 mL NaOH}\\\\\text{The student will need $\large \boxed{\textbf{14.3 mL NaOH}}$}](https://tex.z-dn.net/?f=V%20%3D%20%5Ctext%7B1.283%20mmol%20NaOH%7D%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mL%20NaOH%7D%7D%7B%5Ctext%7B0.0900%20mmol%20NaOH%7D%7D%20%3D%20%20%20%5Ctextbf%7B14.3%20mL%20NaOH%7D%5C%5C%5C%5C%5Ctext%7BThe%20student%20will%20need%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B14.3%20mL%20NaOH%7D%7D%24%7D)
Answer:
The required flow rate of air in mol/h is 20144 and the percent by mass of oxygen in the product gas is 23,1%.
Explanation:
The air of 9 mole% methane have an average molecular weight of:
![0,09*16,04g/mol + 0,91*29g/mol = 27,8g/mol](https://tex.z-dn.net/?f=0%2C09%2A16%2C04g%2Fmol%20%2B%200%2C91%2A29g%2Fmol%20%3D%2027%2C8g%2Fmol)
And a flow of 700'000g/h÷27,8g/mol = 25180 mol/h
In the reactor where methane solution and air are mixed:
In = Out
Air balance:
91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)
<em>Where X is the flow rate of air in mol/h.</em>
<em>Solving, </em><em>X = 20144 mol air/h</em>
The air in the product gas is:
95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = 289 kg O₂
43058 mol air×29g/mol 1249 kg air
Percent mass of oxygen in the product gas is: =0,231 kg O₂/ kg air* 100 = 23,1%
I hope it helps!
Answer:
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