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sashaice [31]
2 years ago
7

SOMEONE HELP WITH THAT THING PLSSS( I also didn’t know what to put as the topic)

Chemistry
1 answer:
rewona [7]2 years ago
6 0

Answer:

hope this helps

Explanation:

Water was on earth with the dinosaurs and almost at the very start.

Water starts anywhere from clouds to ponds

I choose ponds, water evaporates from the sun and moves its way through the air being collected into clouds in the atmosphere, clouds are evaporated water molecules, then the water combines to form rain wich then end up on a hill and travel back to a water source or goes into the ground and makes groundwater and if there is enough rain overtime will form a pond.

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Is the narrator male or female in the story leaving?​
AlladinOne [14]
The narrator is a female

Hope this helps
8 0
3 years ago
The Dew Point... a. is the same as temperature when the relative humidity is 100%. b. can only be given in Kelvin degrees. c. is
telo118 [61]

Answer:

Option (A)

Explanation:

The dew point refers to the temperature at which the amount of water vapor present in the air is so high that the relative humidity becomes 100%, and with the increasing rate of cooling, the condensation process takes place and dew is formed.

So for dew point to occur, the air temperature must reach a condition where the air is fully saturated or the relative humidity is 100%.

Thus, the correct answer is option (A).

6 0
3 years ago
List the path of the respiratory system
Lena [83]
nasal cavities (or oral cavity) > pharynx > trachea > primary bronchi (right & left) > secondary bronchi > tertiary bronchi > bronchioles > alveoli (site of gas exchange)
3 0
3 years ago
A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas that react to form water vapor. Assume the volume of the con
maxonik [38]

Mole fraction of Oxygen=0.381

Mole fraction of Oxygen= (range of moles of oxygen) ÷(general moles)

also, mole fraction of oxygen = (partial stress of oxygen) ÷ (total strain)

consequently , mole fraction of Oxygen= (2.31 atm)÷(2.31 atm + 3.75 atm)

= 0.381

The mole fraction may be calculated by means of dividing the variety of moles of 1 element of a solution by the entire quantity of moles of all the additives of a solution. It is cited that the sum of the mole fraction of all of the components inside the solution should be identical to 1.

Mole fraction is a unit of awareness. in the solution, the relative amount of solute and solvents are measured by way of the mole fraction and it's far represented through “X.” The mole fraction is the variety of moles of a selected aspect inside the answer divided by way of the entire range of moles in the given answer.

Mole fraction is the ratio between the moles of a constituent and the sum of moles of all ingredients in a mixture. Mass fraction is the ratio between the mass of a constituent and the full mass of a mixture.

The question is incomplete. Please read below to find the missing content.

Assuming that only the listed gases are present, what would the mole fraction of oxygen gas be for each of the following situations? A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas react to form water vapor. Assume the volume of the container and the temperature inside the container does not change.

Learn more about the mole fraction here brainly.com/question/14783710

#SPJ1

4 0
2 years ago
Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s
Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

                        N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where N is amount of radioactive nuclide after t time, N_{0} is initial amount of radioactive nuclide and t_{\frac{1}{2}} is half life of radioactive nuclide

Here N_{0}=4.60 mg, t=4.1\times 10^{9}yr and t_{\frac{1}{2}}=4.5\times 10^{9}yr

So,N=(4.60mg)\times (\frac{1}{2})^{\frac{4.1\times 10^{9}}{4.5\times 10^{9}}}

so, N = 2.446 mg

mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg

3 0
3 years ago
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