Answer:
Atoms
Explanation:
Energy, potential energy, is stored in the covalent bonds holding atoms together in the form of molecules. This is often called chemical energy.
Answer:
It is mostly water (up to 95% by volume), and contains dissolved proteins (6–8%) (e.g. serum albumins, globulins, and fibrinogen), glucose, clotting factors, electrolytes (Na+, Ca2+, Mg2+, HCO3−, Cl−, etc.), hormones, carbon dioxide (plasma being the main medium for excretory product transportation) and oxygen.
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can i pls have a brainilist!
Answer:
The pressure of a given amount of gas is directly proportional to iys absolute temperature provided that that the volume does not change
Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
- P is the gas pressure.
- V is the volume that occupies.
- T is its temperature.
- R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
- n is the number of moles of the gas.
Then, in this case:
- P= 1 atm
- V= 44.1 L
- n= ?
- R= 0.082

- T= 0°C =273 K
Replacing in the expression for the ideal gas law:
1 atm× 44.1 L= n× 0.082
× 273 K
Solving:

n=1.97 moles
Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:
= 63.04 g ≈ <u><em>63 g</em></u>
Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
Learn more about the ideal gas law:
To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:
45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present
Hope this answers the question. Have a nice day.