The volume of a 14.00g of nitrogen at 5.64atm and 315K is 4.59L.
<h3>How to calculate volume?</h3>
The volume of an ideal gas can be calculated using the following ideal gas equation formula;
PV = nRT
Where;
- P = pressure (atm)
- V = volume (L)
- n = number of moles
- R = gas law constant
- T = temperature
An ideal gas is a hypothetical gas, whose molecules exhibit no interaction, and undergo elastic collision with each other and with the walls of the container.
The number of moles in 14g of nitrogen can be calculated as follows:
moles = 14g ÷ 14g/mol = 1mol
5.64 × V = 1 × 0.0821 × 315
5.64V = 25.86
V = 25.86 ÷ 5.64
V = 4.59L
Therefore, 4.59L is the volume of the gas
Learn more about volume at: brainly.com/question/12357202
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Answer:
Li₂CO₃ ⇒ Li₂O + CO₂ is a decomposition reaction.
Explanation:
In a decomposition reaction, a single substance decomposes, producing two or more different substances. In other words, in this type of reaction two or more substances are formed from a compound. The atoms that form a compound separate to give the products according to the formula:
AB → A + B
It can occur spontaneously, or be aided by a catalyst, heat, or electrolysis.
In this case:
Li₂CO₃ ⇒ Li₂O + CO₂ It is a type of chemical reaction in which a single compound decomposes into two or more compounds. So it is a decomposition reaction.
3.
∆E = ∆m x c ² ∆m = E / c ² ∆m = 3,83•10^-12 / 3•10^8 ² ∆m = 4,256•10^-29 kg
Taking this class as well
Answer: The distance is slightly less than 3.5 m
Explanation: assuming wall and target are the same thing, and the bullet has constant velocity, the bullet will travel 7 m in half a second, so half that distance is 3.5 m.
In reality, the bullet is decelerating (at an unknown rate) so the distance is slightly less than 3.5 m.
There is also a vertical velocity component, which means it hits the target/wall at an angle. The trajectory is such that it hits the wall above the shooter because the ricochet hits at ~the level at which it left the firearm.
If the wall was absent, the bullet would have described a parabola which brough it back to the initial level after 7m. This could be calculated, but it means that the actual distance between the shooter and the wall is slightly less than 3.5 m
In addition, the collision with the wall is not 100% elastic, so the velocity aftercthe ricochetvis further reduced.
A calculation would be complex because these confounding factors are not completely independent of each other, but all reduce the average velocity and therefore the distance.
Therefore it is only possible to say that the distance was somewhat less than 3.5 m
