Is there a picture of the isotope or?- becaue I can’t help if I don’t have a visual.
<h3><u>Answer;</u></h3>
exceeds evaporation over land
Precipitation<u> exceeds evaporation over land </u>
<h3><u>Explanation;</u></h3>
- <em><u>In order to maintain earths water balance, evaporation exceeds precipitation over oceans but precipitation exceeds evaporation over land.</u></em>
- Water evaporates into the atmosphere from the ocean and to a much lesser extent from the continents. Winds transport this moisture-laden air, often great distances, until conditions cause the moisture to condense into clouds and to precipitate and fall.
- Most precipitation originates by evaporation from the oceans. Over time, water evaporated from the oceans is replenished by inflow of freshwater from rivers and streams.
Answer:
1.20atm
Explanation:
Given parameters:
Partial pressure of gas 1 = 0.35atm
Partial pressure of gas 2 = 0.20atm
Partial pressure of gas 3 = 0.65atm
Unknown:
Total pressure of the gas mixture = ?
Solution:
To solve this problem, we need to recall and understand the Dalton's law of partial pressure.
Dalton's law of partial pressure states that "the total pressure of a mixture of gases is equal to the sum of the partial pressure of the constituent gases".
Total pressure =Pressure of gas(1 + 2 + 3)
The partial pressure is the pressure a gas would exert if it alone occupied the volume of the gas mixture.
Now we substitute;
Total pressure = (0.35 + 0.20 + 0.65)atm = 1.20atm
Answer:
Fe₂O₃
Explanation:
To solve this question we must find the moles of Iron in 1.68g. With the difference of the masses we can find the moles of oxygen. The formula will be obtained with the ratio of both amount of moles:
<em>Moles Fe:</em>
1.68g * (1mol / 56g) =0.03moles
<em>Moles O:</em>
2.40g-1.68g = 0.72g * (1mol/16g) = 0.045moles
The ratio O/Fe is:
0.045moles / 0.03moles = 1.5 moles. this ratio is obtained if the formula is:
<h3>Fe₂O₃</h3>
Answer:
2.5 atm
Explanation:
P2 = 1.2 atm x 2.65 L x 345 K / 1.5 L x 298 K
Esto debería ser igual...
2.45 atm (personajes importantes)
