<span>Step 1: Ask a question or identify a problem. ...Step 2: Background research. ...Step 3: Form a hypothesis. ...Step 4: Experiment and observe. ...<span>Step 5: Draw a conclusion.</span></span>
Answer:
Use a ratio of 0.44 mol lactate to 1 mol of lactic acid
Explanation:
John could prepare a lactate buffer.
He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C3.5%20%3D%203.86%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%203.5%20-%203.86%20%3D%20-0.36%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%2010%5E%7B-0.36%7D%20%3D%20%5Cmathbf%7B0.44%7D)
He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.
For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.
Answer:
The element is CARBON
The number 6 refers to the ATOMIC NUMBER
the numbers 12, 13, and 14 refer to the ATOMIC MASS
how many protons and neutrons are in the first isotope?
<u>6</u><u>. </u><u> </u><u> </u><u> </u><u> </u><u>6</u>
how many protons and neutrons are in the second isotope?
<u>6</u><u>. </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>7</u>
<u>how many protons and neutrons are in the </u><u>t</u><u>h</u><u>i</u><u>r</u><u>d</u><u> </u><u>isotope?</u>
<u>6</u><u>. </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>8</u>
<u>y</u><u>o</u><u>u</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>w</u><u>e</u><u>l</u><u>c</u><u>o</u><u>m</u><u>e</u><u> </u><u>:</u><u>)</u>
Q1. TI (210/81Thallium)
Q2.
The answers are opposite from each other
The answer is chemical weathering