Explanation:
here's the answer to your question
Answer:
Ka = 1.52 E-5
Explanation:
- CH3-(CH2)2-COOH ↔ CH3(CH2)2COO- + H3O+
⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]
mass balance:
⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M
charge balance:
⇒ [H3O+] = [CH3(CH2)2COO-]
⇒ Ka = [H3O+]²/(1 - [H3O+])
∴ pH = 2.41 = - Log [H3O+]
⇒ [H3O+] = 3.89 E-3 M
⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )
⇒ Ka = 1.519 E-5
Answer:
At one atmosphere and twenty-five degrees Celsius, could you turn it into a liquid by cooling it down? Um, and the key here is that the triple point eyes that minus fifty six point six degrees Celsius and it's at five point eleven ATMs. So at one atmospheric pressure, there's no way that you're ever going to reach the liquid days. So the first part of this question is the answer The answer to the first part of a question is no. How could you instead make the liquid at twenty-five degrees Celsius? Well, the critical point is at thirty-one point one degrees Celsius. So you know, if you're twenty-five, if you increase the pressure instead, you will briefly by it, be able to form a liquid. And if you continue Teo, you know, increase the pressure eventually form a salad, so increasing the pressure is the second part. If you increase the pressure of co two thirty-seven degrees Celsius, will you ever liquefy? No. Because then, if you're above thirty-one point one degrees Celsius in temperature. You'LL never be able to actually form the liquid. Instead, you'LL only is able Teo obtain supercritical co too, which is really cool thing. You know, they used supercritical sio tu tio decaffeinated coffee without, you know, adding a solvent that you'LL be able to taste, which is really cool. But no, you can't liquefy so two above thirty-one degrees Celsius or below five-point eleven atmospheric pressures anyway, that's how I answer this question. Hope this helped :)
