Answer:
the mass of CaO present at equilibrium is, 0.01652g
Explanation:
= 3.8×10⁻²
Now we have to calculate the moles of CO₂
Using ideal gas equation,
PV =nRT
P = pressure of gas = 3.8×10⁻²
T = temperature of gas = 1000 K
V = volume of gas = 0.638 L
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mole.k

Now we have to calculate the mass of CaO
mass = 2.95 * 10 ⁻⁴ × 56
= 0.01652g
Therefore,
the mass of CaO present at equilibrium is, 0.01652g
The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is : - 0.69°C
<h3>Determine the freezing point of the solution </h3>
First step : Calculate the molality of NaCl
molality = ( 2.5 grams / 58.44 g/mol ) / ( 230 * 10⁻³ kg/ml )
= 0.186 mol/kg
Next step : Calculate freezing point depression temperature
T = 2 * 0.186 * kf
where : kf = 1.86°c.kg/mole
Hence; T = 2 * 0.186 * 1.86 = 0.69°C
Freezing point of the solution
Freezing temperature of solvent - freezing point depression temperature
0°C - 0.69°C = - 0.69°C
Hence the Freezing temperature of the solution is - 0.69°C
Learn more about The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is : - 0.69°C
Answer: False
Explanation: False, what if its in space?
Answer:
ΔG° = 1022. 8 kJ
Explanation:
ΔH° = –199 kJ/mol
ΔS° = –4.1 J/K·mol
T = 25°C = 25 + 273 = 298K (Converting to kelvin temperature)
ΔG° = ?
The relationship between these varriables are;
ΔG° = ΔH° - TΔS°
ΔG° = –199 - 298 (–4.1)
ΔG° = -199 + 1221.8
ΔG° = 1022. 8 kJ