Answer:
A) if the system is isothermal then all the heat added to the system will be used to do work (since none is used to raise the temperature of the gas). The heat added will be equal to the work done = 340 J
B) change in internal energy of the system of the process is isothermal will be zero, since there is no rise in temperature.
C) an adiabatic process is one involving no heat loss or gain through the system, Therefore heat gain will be zero
D) if the process is adiabatic then there is no heat loss or gain through the system and hence there is no change in temperature. Change in internal energy will be zero
E) if the process is isobaric then, there is no work done and the total heat to the system is equal zero
F) if there is no work done, and no heat added, then the internal energy will be equal zero.
An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.
Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]
moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles
moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles
pKa of phenol = 9.98
We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation
pH = pKa + log [salt] / [acid]
volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula
pH = 9.98 + log [12.93 / 42.55] = 9.46
Answer:
Condensation reaction/ direct synthesis reaction
Explanation:
Combines simple molecules to form complex molecules producing water
Answer:
1. First
2. Third
3. Fourth
4.remain the same as
Explanation:
Given the reaction equation;
Rate= k[A] [B]^3
We can see that the order of reaction is first order with respect to reactant A and third order with respect to reactant B. This gives an overall fourth order reaction.
If the concentration of A is doubled and that of B is halved. The rate of reaction remains the same.
