To balance equations you have to have same number of atoms on both sides of the equation just multiply with a suitable digit
Answer:
672 g
Explanation:
We can calculate the mass of water that can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories using the following expression.
where,
c: specific heat of the water
m: mass
ΔT: change in the temperature
The mass of water that can be warmed under these conditions is 672 grams.
•boiling and freezing point
•surface tension
•vapor pressure
• solid state
•liquid state
•gaseous state
Answer:
T2 = 94.6 C
Explanation:
Use Clausius-Clayperyon equation.
ln P1/P2 = ∆Hvap/R (1/T2 - 1/T1) where R = 8.314 J/mol-K and T is in degrees K
P1 = 760 mmHg
P2 = 630 mmHg
T1 = 373 K
T2 = ?
∆Hvap = 40.7 kJ/mole
R = 0.008314 kJ/mole-K (NOTE: change R to units of kJ)
Plug in and solve for T2
ln 760 mmHg/630 mmHg = 40.7 kJ/mole (1/T2 - 1/373K)
T2 = 367.74 K = 94.6 C
The volume is 19.76987448 by taking the known variables mass=18.9g and density=0.956g/ml
To get volume you divide the mass by the density which gives you about
19.77 ml in volume