

- <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>
<span>Valence electrons are important because these are the only electrons involved with an atom's chemical reactivity.</span>
Answer:
The ionization equation is
⇄
(1)
Explanation:
The ionization equation is
⇄
(1)
As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.
The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,
![Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BHPO_%7B4%7D%5E%7B-2%7D%5D%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BH_%7B2%7DPO_%7B4%7D%5E%7B-%7D%5D%20%5BH_%7B2%7DO%5D%7D%20%3D%206.2x10%5E%7B-8%7D)
The pKa is

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.
In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.
If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.
Its C, Outermost Electrons.