The correct term for each statement is as follows;
- Caves being formed by acid rain dissolving underground limestone - weathering
- Deltas forming at the mouths of rivers - deposition
- Glaciers scraping rocks across the earth’s surface - erosion
- Rocks being made smooth by tumbling across a stream bed - weathering
- Ponds filling up with sediment and becoming marshes - deposition
<h3>What is weathering?</h3>
Weathering in geology is the mechanical or chemical breaking down of rocks in situ by weather or other causes. options 7 and 14 are the case in this scenario as there is a breakdown of limestone and rocks.
Erosion is the result of having been worn away or eroded, as by a glacier on rock or the sea on a cliff face. Agents of erosion are water, ice or wind. Options 12 is an example of erosion because glaciers are wearing off rocks.
Deposition is the laying down of sediment carried by wind, flowing water, the sea or ice. Sediment can be transported as pebbles, sand and mud, or as salts dissolved in water. Options 8 and 15 are examples of deposition.
Learn more about weathering and deposition at: brainly.com/question/367069
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The answer is B because isotopes have the same number of protons and neutrons.
Answer:When heat is added to a substance, the molecules and atoms vibrate faster. As atoms vibrate faster, the space between atoms increases. The motion and spacing of the particles determines the state of matter of the substance. ... They contract when they lose their heat.
Explanation:Google
Answer:
Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.
Explanation:
Molarity of HCl solution = 0.1174 M
Volume of HCl solution = 83.15 mL = 0.08315 L
Moles of HCl = n
According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then 0.009762 mol of HCl will recat with:
Moles of Sodium carbonate = 0.004881 mol
Volume of the sodium carbonate containing solution taken = 1L
Concentration of sodium carbonate in the solution before the addition of HCl:
![[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D%5Cfrac%7B0.004881%20mol%7D%7B1%20L%7D%3D0.004881%20mol%2FL)
