<span>When pKas of polyprotic intermediates have a difference of 2 or more you just average them using the equation: pH = (pKa2 + pKa3) / 2 </span>
<span>pKa2 = -log(Ka2) ; pKa3 = -log(Ka3) </span>
<span>so, for this problem, REGARDLESS OF THE CONCENTRATION GIVEN, the answer is: </span>
<span>pH = (7.2076+12.3767) / 2 </span>
<span>pH = 9.79</span>
A closed circuit will light the bulb because electricity flows through it.
But on the other hand in an open circuit there is no flow of electricity current hence no light is produced
Answer:
<h2>
#1</h2><h3>
Which example below would not be found on primitive Earth?</h3><h3 /><h3>
<em><u>Oxygen Gas</u></em></h3>
<h2>
#2</h2>
Which event would have occurred by the end of the Hadean era?
<h3>
<em><u>Earth’s crust had cooled and solidified</u></em></h3>
<h2>
#3</h2><h3>
Classify Earth’s current atmosphere</h3>
<h3>
<em><u>Secondary Atmosphere</u></em></h3>
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
first determine the equation. KHP is a monoprotic acid, so it reacts with NaOH in a 1:1 ratio.
<span>a) HKC8H4O4 + NaOH → NaKC8H4O4 + H2O </span>
<span>This allows you to calculate the molarity of NaOH, from the equation V1M1n2 = V2M2n1 </span>
<span>Molarity times volume will give you the number of moles of acid. You can work this out by dividing the mass used by the molar mass. For this question, you will have 1.0115/204.2285 moles KHP, so this value can be substituted for V1M1 in the equation. You need 38.46 mL of NaOH to titrate all the acid, so this will be V2. Now you find M2. </span>
<span>b) 1.0115/204.2285 × 1 = 38.46 × M2 × 1, so M2 = 0.0001288M </span>
<span>Now go on to part 3. Here, you use a similar calculation. The acid and base react in a 1:1 ratio, so n1 = n2 = 1. M2 = 0.1905 and V2 = 25.62. V1M1 will be calculated. </span>
<span>V1M1 × 1 = 25.62×0.0001288×1 = 0.003299856 </span>
<span>Now mass divided by molar mass = V1M1, so molar mass = mass/V1M1. </span>
<span>1.2587/0.003299856 = 381.44 g/mol</span>
