Answer:
5746.0 mL.
Explanation:
We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 6193.0 mL, T₁ = 62.3°C + 273 = 335.3 K.
V₂ = ??? mL, T₂ = 38.1°C + 273 = 311.1 K.
<em>∴ V₂ = V₁T₂/T₁ </em>= (6193.0 mL)(311.1 K)/(335.3 K) = <em>5746.0 mL.</em>
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.
where,
- i: van 't Hoff factor (1 for non-electrolytes)
- Kf: cryoscopic constant
- m: molality
The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
Learn more: brainly.com/question/2292439
Answer:
B. a strongly basic solution
Explanation:
Kb is base dissociation constant, which indicates how completely a base dissociates into its component ions in water. The greater the Kb value, the greater the alkalinity of the solution and vice versa.
Therefore, a solution with a Kb value much greater than 1, indicates a strongly basic solution, while a solution with a Kb value less than 1, indicates a weakly basic solution.
Explanation:
Given
The enthalpy of formation of RbF (s) is –557.7kJ/mol
The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol
The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)
= -583.8 - (-557.7) kJ/mol
= -26.1 kJ/mol
The enthalpy is negative which means that the temperature will rise when RbF is dissolved.
Answer:
The new temperature of the water bath 32.0°C.
Explanation:
Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)
Initial temperature of the water = 
Final temperature of the water = 
Specific heat capacity of water under these conditions = c = 4.18 J/gK
Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J
( 1kJ=1000 J)
The new temperature of the water bath 32.0°C.
