Calculate the molarity when 28.91 grams of nacl is dissolved in 352 ml of water
1 answer:
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<span>the pH of a 0.050 M triethylamine, is 11.70
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For triehtylamine,
, the reaction will be
and we know, pH = -log[H+] and pOH = -log[OH-]
And![kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}](https://tex.z-dn.net/?f=kb%20%3D%20%20%5Cfrac%7B%28%20%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%5D%2A%20%20OH%5E%7B-%7D%20%29%7D%7B%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%5D%7D%20)
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
</span>
For triehtylamine,
and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4And
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
Answer:
Explanation:
The air 9% mole% methane have an average molecular weight of:
9%×16,04g/mol + 91%×29g/mol = 27,8g/mol
And a flow of 700000g/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)
Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>
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The air in the product gas is
95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>
43058 mol air×29g/mol <em>1249 kg air</em>
Percent of oxygen is: =<em>0,231 kg O₂/ kg air</em>
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I hope it helps!