Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K
Explanation :
Formula used :
where,
= change in entropy
= change in enthalpy of fusion = 3.17 kJ/mol
As we know that:
= freezing point temperature = 
Now put all the given values in the above formula, we get:
Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K
Answer:
Total mass of the reactant = 2+2.5 =4.5 mg
Total mass of product = 4.15 mg
therefore, mass of unreacted oxygen = 4.50-4.15 = 0.35 g
Answer:
6.48 L
Explanation:
From the question,
Applying
PV/T = P'V'/T'......................... Equation 1
P = initial pressure of the helium balloon, V = Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.
make V' the subject of the equation
V' = PVT'/P'T......................... Equation 2
Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm
Substitute these values into equation 2
V' = (4.5×1×253)/(0.6×293)
V' = 1138.5/175.8
V' = 6.48 L
Answer:There are two atoms in the molecule.
Explanation: All gases are diatomic i.e they are covalently bonded to another atom of the same element in order to attain stability. O atom is usually unstable because of it's incompletely filled outermost shell.
