- <em>Expansion </em><em>of </em><em>particles</em><em> </em><em>of</em><em> </em><em>substances.</em><em> </em>
- <em>Increase</em><em> </em><em>in </em><em>temperature</em><em>.</em>
- <em>Change</em><em> </em><em>in </em><em>state</em><em>.</em>
- <em>Change</em><em> </em><em>in </em><em>physical</em><em> </em><em>property</em>
- <em>It </em><em>may </em><em>bring</em><em> </em><em>out </em><em>chemical</em><em> </em><em>changes</em><em>.</em>
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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.
Answer:
oceans is the answer that I got
Answer:
8.44 atm
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 2.25 L
Initial temperature (T₁) = 350 K
Initial pressure (P₁) = 1.75 atm
Final volume (V₂) = 1 L
Final temperature (T₂) = 750 K
Final pressure (P₂) =?
The final pressure of the gas can be obtained as illustrated below:
P₁V₁/T₁ = P₂V₂/T₂
1.75 × 2.25 / 350 = P₂ × 1 / 750
3.9375 / 350 = P₂ / 750
Cross multiply
350 × P₂ = 3.9375 × 750
350 × P₂ = 2953.125
Divide both side by 350
P₂ = 2953.125 / 350
P₂ = 8.44 atm
Thus, the final pressure of the gas is 8.44 atm.
Answer:
142.82 g
Explanation:
The following data were obtained from the question:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Density of gol= 19.3 g/cm³
Mass of gold =.?
Next, we shall determine the volume of the gold. This can be obtained as follow:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Volume of gold =.?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 19.4 – 12
Volume of gold = 7.4 mL
Finally, we shall determine the mass of the gold as follow:
Note: 1 mL is equivalent to 1 cm³
Volume of gold = 7.4 mL
Density of gol= 19.3 g/cm³ = 19.3 g/mL
Mass of gold =?
Density = mass /volume
19.3 = mass of gold /7.4
Cross multiply
Mass of gold = 19.3 × 7.4
Mass of gold = 142.82 g
Therefore, the mass of the gold pebble is 142.82 g
