Givens
=====
V
= 4.00 L
T
= 273oK We're assuming the temperature does not change, just the
pressure.
n
= 0.864 moles
R
= 8.314 joules / mole * oK
P
= ?????
Formula
======
PV
= n*R*T
P
= n*R*T/V
P
= 0.864 * 8.314 * 273 / 4
P
= 490 kpa
You
have to add 1.6 – 0.864 = 0.736 moles of gas.
We
have to assume that the temperature and pressure remain the same when
we add the 0.736 moles of gas. We are now looking for the volume.
PV
= n*R*T
<span>
V
= 0.736 * 8.314 * 273 / 490</span>
V
= 3.41 L Remember this is at about 4 atmospheres so we have to
convert to Standard Pressure.
Total
Volume = 3.41 + 4.00 = 4.41
V1
* P1 = V2 * P2
P1
= 490 kPa
P2
= 101 kPa
V1
= 7.41 L
V2
= ????
<span>
<span>
7.41*
490 = V2 * 101
V2
= 7.41 * 490 / 101
V2
= 35.94 L
</span>
</span>
<span>You
had 4 L now you need 31.94 more.</span>
Answer:
≈ 20.35 N [newton's of tension]
Explanation:
( (2.9 × 9.8) ÷ cos(35.6°) ) × sin (35.6°) =
( (28.42) ÷ (≈0.813) ) × (≈0.582) =
(≈34.96) × (≈0.582) = 20.3449446.... ≈ 20.35
Here you go mate. Hope it helps u. Pls follow me in reddit lol username: RobloxNoob2006
Answer:
The atmospheric pressure and boiling point are directly proportional
Increasing atmospheric pressure increases the boiling point also
Explanation:
The atmosphere contain molecules that are in constant motion. They exert a downward force on a liquid’s surface. The higher the air pressure, the harder it is for the liquid to evaporate. Therefore, the boiling point of a solvent or liquid is affected by the atmospheric pressure and boiling point is raised.
A liquid in a high pressure environment boils at a higher temperature.
When placed in a lower pressure environment it boils at a lower temperature.
