The answer is the lake.
A lake is a large expanse of terrestrial water, consisting of streams that are linked to the lake. Although some dimensions of some salt-water lakes are considered island seas; they are therefore distinct from the lagoons, and are larger and deeper than the ponds, while remaining a water body by definition.
Answer:
34.15% is the mass percentage of calcium in the limestone.
Explanation:
Mass of precipitate that is calcium oxalate = 140.2 mg = 0.1402 g
1 mg = 0.001 g
Moles of calcium oxalate = 
1 mole of calcium oxalate have 1 mole of calcium atom.
Then 0.001095 moles of calcium oxalate will have 0.001095 moles of calcium atom.
Mass of 0.001095 moles of calcium :
0.001095 mol × 40 g/mol = 0.04381 g
Mass of sample of limestone = 128.3 mg = 0.1283 g
Percentage of calcium in limestone:

34.15% is the mass percentage of calcium in the limestone.
<span>Atmospheric Pressure</span>
Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place