Answer:
0.683 moles of the gas are required
Explanation:
Avogadro's law relates the moles of a gas with its volume. The volume of a gas is directely proportional to its moles when temperature and pressure of the gas remains constant. The law is:
V₁n₂ = V₂n₁
<em>Where V is volume and n are moles of 1, initial state and 2, final state of the gas.</em>
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Computing the values of the problem:
1.50Ln₂ = 5L*0.205mol
n₂ = 0.683 moles of the gas are required
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<span>Avogadro's law applies at STP where P is 1 atm and T is 273K. From Avogadro's law; 1 mole of gas takes up 22.4 L of volume at STP. I. e 1 moles = 22.4 L. Hence 1.2 moles of water vapor will take up. 1.2 * 22.4 = 26.88L. Or using ideal ga s eqn PV = NRT. We have P = 1 atm. N = 1.2 moles. R = 0.0821 L and T =273 K. So V = NRT/P.Then we have 1.2 * 0.0821 * 273 = 26.88L.</span>
Answer:
C.sterols
Explanation:
Sterols or steroid alcohols are type of lipids. In plants they are present as phytosterols and in animals as zoosterols. They maintain the fluidity of cell membrane, act as signalling molecules and also form the skin oils in animals.
Cholesterol is an important zoosterol. It is a fatty waxy substance. It is present in cell membrane. It is also a precursor for vitamin D. It is precursor for steroid hormones like cortisol and aldosterone. When it is non esterified, it gets converted to bile.
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
