Answer:
676mmHg
Explanation:
Using the formula;
P1/T1 = P2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 725.0mmHg
P2 = ?
T1 = 20°C = 20 + 273 = 293K
T2 = 0°C = 0 + 273 = 273K
Using P1/T1 = P2/T2
725/293 = P2/273
Cross multiply
725 × 273 = 293 × P2
197925 = 293P2
P2 = 197925 ÷ 293
P2 = 676mmHg.
The resulting pressure is 676mmHg
Answer is: <span>the volume of water after the solid is added</span> is 4.5 ml.
d(gold) = 8.0 g/cm³; density of gold.
m(gold) = 4 g; mass of gold.
V(gold) = m(gold) ÷ d(gold); volume of gold.
V(gold) = 4 g ÷ 8 g/cm³.
V(gold) = 0.5 cm³ = 0.5 ml.
V(water) = 4.00 ml = 4.00 cm³.
V(flask) = V(gold) + V(water).
V(flask) = 0.5 cm³ + 4 cm³.V = 4.5 cm³.
Answer:
3.711 L
Explanation:
The formula you need to use is the following:

3.4L / 298 K = V2 / 273 K
V2 = 3.711 L
Answer:
12 atm
Explanation:
First, let us convert Celcius into Kelvin: 28.0 °C = 301.15 K and 129.0 °C = 402.15 K
For this question we must employ the Combined Gas Law:
, where
is the initial pressure and
is the new pressure.
We know that intitially, P=9 atm, V=30 L, and T=301.15K. From our problem, only temperature and pressure changes, while the number of moles, volume and the gas constant, R, stay the same, so they are irrelevant.
Thus, the filled out Combined Gas Law would be:
=
, where the volume, moles of gas, and R are cancelled out.
We can manipulate this equation to derive the new pressure. We find that
9atm≈0.74885
.
This means that
≈9/0.74885≈12 atm