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:
At one atmosphere and twenty-five degrees Celsius, could you turn it into a liquid by cooling it down? Um, and the key here is that the triple point eyes that minus fifty six point six degrees Celsius and it's at five point eleven ATMs. So at one atmospheric pressure, there's no way that you're ever going to reach the liquid days. So the first part of this question is the answer The answer to the first part of a question is no. How could you instead make the liquid at twenty-five degrees Celsius? Well, the critical point is at thirty-one point one degrees Celsius. So you know, if you're twenty-five, if you increase the pressure instead, you will briefly by it, be able to form a liquid. And if you continue Teo, you know, increase the pressure eventually form a salad, so increasing the pressure is the second part. If you increase the pressure of co two thirty-seven degrees Celsius, will you ever liquefy? No. Because then, if you're above thirty-one point one degrees Celsius in temperature. You'LL never be able to actually form the liquid. Instead, you'LL only is able Teo obtain supercritical co too, which is really cool thing. You know, they used supercritical sio tu tio decaffeinated coffee without, you know, adding a solvent that you'LL be able to taste, which is really cool. But no, you can't liquefy so two above thirty-one degrees Celsius or below five-point eleven atmospheric pressures anyway, that's how I answer this question. Hope this helped :)
Answer:
The u (amu is the old unit name) is 1/12 of the weight of an 12C atom. The way the u is chosen ensures that all core and atom masses are multiples of 1(±0.1) u.
Explanation:
Further explanation if needed...
Carbon 12 was chosen because the chemical atomic weights based on C12 are almost identical to the chemical atomic weights based on the natural mix of oxygen. Simply because the atomic mass is defined as 1/12 of the mass of 12C. Others isotopes of carbon (13C mostly, with an abundance of 1.1% approximately) account for an average atomic mass slightly above 12.
First of all, the problem says that you have to convert the acid salt to its acidic form. If you take it to the neutral pH, that won't be acidic at all. As simple as that, you don't take it to neutral pH because it would lose its definition of being acidic afterall.
