Robert A. Millikan and Harvey Fletcher performed the oil drop experiment.
Answer:
Option (D)
Explanation:
Weathering is usually defined as the disintegration of rocks at the surface of the earth. This break down of rocks is mainly caused by the geological processes that occur on the earth's surface. This process results in the formation of sediments that are transported and deposited in a new environment.
This weathering process primarily takes place in three different ways such as-
- Physical weathering- Here, the rocks are broken down by the physical agents such as wind, water, ice.
- Chemical weathering- Here the rocks are broken down when interacts with the chemical containing water.
- Biological weathering- here, the rocks are broken down due to the activities done by organisms such as plants and animals.
In the given condition, Quincy can see a real example of rock weathering in the high mountainous region, as in the mountainous region the rocks are frequently weathered and eroded by the agents such as wind, water, and ice.
Thus, the correct answer is option (D).
M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C
m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C
T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C
<span>Divide the number of grams present in the sample by copper's gram atomic weight to find the number of gram atomic weights present. Then multiply that result by Avogadro's Number: 6.022137 x 10^23 atoms/gram atomic weight.1,200 g/(63.54 g/gram atomic weight) ? 18.885741 gram-atomic weights. Hope this helps. </span>
Answer:
T2 = 94.6 C
Explanation:
Use Clausius-Clayperyon equation.
ln P1/P2 = ∆Hvap/R (1/T2 - 1/T1) where R = 8.314 J/mol-K and T is in degrees K
P1 = 760 mmHg
P2 = 630 mmHg
T1 = 373 K
T2 = ?
∆Hvap = 40.7 kJ/mole
R = 0.008314 kJ/mole-K (NOTE: change R to units of kJ)
Plug in and solve for T2
ln 760 mmHg/630 mmHg = 40.7 kJ/mole (1/T2 - 1/373K)
T2 = 367.74 K = 94.6 C
