Answer:
0.170 M
Explanation:
As this is a <em>series of dilutions</em>, we can continuosly<em> use the C₁V₁=C₂V₂ formula </em>to solve this problem:
For the first step:
- 59.0 mL * 1.80 M = 258 mL * C₂
Then for when 129 mL of that 0.412 M are diluted by adding 183 mL of water:
- V₂ = 129 mL + 183 mL = 312 mL
Using <em>C₁V₁=C₂V₂:</em>
- 129 mL * 0.412 M = 312 mL * C₂
Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=
V
nRT
. Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}
mol⋅K
atm⋅L
:
P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=
V
nRT
=
15.0
L
(0.33
mol
)(0.0821
mol⋅K
atm⋅
L
)(248.15
K
)
=0.4482atm
In conclusion, the pressure of this gas is P=0.4482 atm.
Reference:
Chang, R. (2010). Chemistry. McGraw-Hill, New York.
Answer:
The second friend is mostly correct.
Explanation:
When pressure decreases, gas solubility decreases. The bubbles are gas molecules that were dissolved in the water at atmospheric pressure. The bubbles started coming out as a result of the decrease pressure. The second bubbles would be water vapor because as the pressure continue to decrease, the amount of vapor pressure ( vapor leaving the surface of the liquid ) equals the external pressure and water begin to boil. So the second bubbles must be water vapor.
