M=mol/L, 0.323M=mol/0.01325. Rework to solve for mol and bam! (I.e. times the two numbers)
Answer:
Energy
Explanation:
A sugar group would be used for carbohydrates or nucleic acids. An adenosine group would be used in ATP formation if I recall correct. And disaccharides are just two monosaccharides linked together, so that would also be for carbohydrates. Therefore, energy is the answer.
To find moles in this sample, you would divide grams by molar mass of ethyl alcohol
(18.0g)/(46.07g/mol) = 0.391mol C2H6O


- <u>We </u><u>have </u><u>250g </u><u>of </u><u>liquid </u><u>water </u><u>and </u><u>it </u><u>needs </u><u>to </u><u>be </u><u>cool </u><u>at </u><u>temperature </u><u>from </u><u>1</u><u>0</u><u>0</u><u>°</u><u> </u><u>C </u><u>to </u><u>0</u><u>°</u><u> </u><u>C</u>
- <u>Specific </u><u>heat </u><u>of </u><u>water </u><u>is </u><u>4</u><u>.</u><u>1</u><u>8</u><u>0</u><u>J</u><u>/</u><u>g</u><u>°</u><u>C</u>

- <u>We </u><u>have </u><u>to </u><u>find </u><u>the</u><u> </u><u>total</u><u> </u><u>number </u><u>of </u><u>joules </u><u>released</u><u>. </u>

<u>We </u><u>know </u><u>that</u><u>, </u>
Amount of heat energy = mass * specific heat * change in temperature
<u>That </u><u>is, </u>

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u><u>:</u><u>-</u>




Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

<u>We </u><u>have </u><u>to </u><u>tell </u><u>whether </u><u>the </u><u>above </u><u>process </u><u>is </u><u>endothermic </u><u>or </u><u>exothermic </u><u>:</u><u>-</u>
Here, In the above process ΔT is negative and as a result of it Q is also negative that means above process is Exothermic
- <u>Exothermic </u><u>process </u><u>:</u><u>-</u><u> </u><u>It </u><u>is </u><u>the </u><u>process </u><u>in </u><u>which </u><u>heat </u><u>is </u><u>evolved </u><u>. </u>
- <u>Endothermic </u><u>process </u><u>:</u><u>-</u><u> </u><u>It </u><u>is </u><u>the </u><u>process </u><u>in </u><u>which </u><u>heat </u><u>is </u><u>absorbed </u><u>.</u>
Use PV = mRT/M and solve for R. R = PVM/RT. Since you have the same gas under two sets of conditions then you can write
<span>P1V1M1/m1T1 = P2V2M2/m2T2 </span>
<span>Since P, M and T are constant, the equation becomes </span>
<span>V1/m1 = V2/m2 </span>
<span>Now plug in your values and solve for V2</span>