A. Large atoms have valence electrons farther from the nucleus and lose them more readily, so they are more reactive than small atoms.
For example, the valence electron of a small atom like Li is tightly held. <em>Lithium gently fizzes</em> on the surface as it reacts with the water to produce hydrogen.
In contrast, the valence electron of a large atom like Cs is so loosely held that <em>cesium exlodes </em>on contact with water.
Answer:
CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base or vice versa.
For the acetic acid buffer, CH₃CO₂H is the weak acid and its conjugate base is the ion without H⁺, that is CH₃CO₂⁻. The equilibrium equation in water knowing this is:
<h3>CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺</h3>
<em>In the equilibrium, the acid is dissociated in the conjugate base and the hydronium ion.</em>
Answer:
D. To ensure the cooling process is not affected by surrounding temperature
Explanation:
The conical flask acts as a <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u><u>j</u><u>a</u><u>c</u><u>k</u><u>e</u><u>t</u><u>.</u>
Answer:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
Yup go this website for more information http://dwb.unl.edu/calculators/activities/BalEqn.html
