Answer:
0.8078 Kg
Explanation:
Pressure of water = 0.15 MPa = 1.5 bar
At critical point of water ,temperature = 647 K=374°C
From the ideal gas equation
P×V= m×R×T
Let us assume volume = 1 m^3
1.5 x 105 x 1 = m x 287 x 647
m= 0.8078 kg
the fraction of mass of liquid at 25°C.
Answer:
Salt domes result when <u><em>the pressure of overlying rock forces the salt to rise. (Option 2)</em></u>
Explanation:
In geology it is called the gently wavy and rounded relief dome.
Salt has some special properties like rock:
- Salt has a lower specific gravity in relation to a common mineral.
- Salts deform plastically and are very mobile.
- Salts have a high water solubility.
These properties allow, if the pressure is very high, that the salt layers move upwards (due to their lower density). That is, the internal forces produce the elevation of the strata by means of the pressure they exert towards a higher point, generating that the salt looks for its way towards the surface [that is, the salt ascends through the sedimentary layers of the earth's crust, crossing them and deforming them] and causing the bulging structure. The oldest strata are located in the central area of the dome, while the most modern are distributed in the farthest radius. The structure is called salt or diapiro dome, the phenomenon by which it is formed is called diapirism.
Finally, you can say that <u><em>Salt domes result when the pressure of overlying rock forces the salt to rise.</em></u>
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
Answer:
yes, if they were in rest they would have potential energy
Explanation:
D = m / V
D = 8.0 g / 25 cm³
D = 0.32 g/cm³
Answer A