For bleaching a stain, there will be an identifiable odor and it will easily absorb colors and/or stains. Burning a match also has odor involved, but the smoke that is released is from a hot fire that is chemically produced by the formula to make the tip of the match.
Answer:
Magnesium
Explanation:
Magnesium is in group 2 of the periodic table. The group in which an element is found in tells us the number of valence eletrons it has.
HOPE THIS HELPED
Answer is: <span>decomposition.
Balanced chemical reaction: H</span>₂CO₃ → CO₂ + H₂O.
H₂CO₃ is carbonic acid.
CO₂ is carbon (IV) oxide or carbon dioxide.<span>
Chemical decomposition is the separation of
a single chemical compound (in this example </span>carbonic acid<span>) into
its two or more simpler compounds (in this example water and
carbon dioxide).</span>
Answer:
94.325 g
Explanation:
We'll begin by converting 350 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
350 mL = 350 mL × 1 L /1000 mL
350 mL = 0.35 L
Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:
Volume = 0.35 L
Molarity of KC₂H₃O₂ = 2.75 M
Mole of KC₂H₃O₂ =?
Molarity = mole /Volume
2.75 = Mole of KC₂H₃O₂ / 0.35
Cross multiply
Mole of KC₂H₃O₂ = 2.75 × 0.35
Mole of KC₂H₃O₂ = 0.9625 mole
Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:
Mole of KC₂H₃O₂ = 0.9625 mole
Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)
= 39 + 24 + 3 + 32
= 98 g/mol
Mass of KC₂H₃O₂ =?
Mass = mole × molar mass
Mass of KC₂H₃O₂ = 0.9625 × 98
Mass of KC₂H₃O₂ = 94.325 g
Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g
The given above pretty much states already that with the presence of the calcium carbonate which acts as the buffer will allow the solution to withstand changes in acidity. The greater the amount, the higher chances that it will be able to withstand the said changes. Therefore, if Lake X had greater ppm of CaCO3 then, it will be able to withstand greater amount of acid rain.
