Answer:
The enthalpy of vaporization of water at 273 K and 1 bar = 44.9 KJ/mol
Explanation:
Enthalpy of vaporization of water at 273 K, ΔHvap(T₂) is given as;
ΔHvap(T₂) = ΔHvap(T₁) + ΔCp * (T₂ - T₁)
where ΔCp = molar heat capacity of gas - molar heat capacity of liquid
Therefore, ΔCp = (33.6 - 75.3) = -41.70 J/(mol K) = 0.0417 kJ/(molK)
substituting ΔCp = 0.0417 kJ/(mol K) in the initial formula
;
ΔHvap(T) = ΔHvap(T1) + ΔCp * (T₂ - T₁)
ΔHvap(T₂)= 40.7 kJ/mol + {-0.0417 kJ/(mol K) * (272 - 373 K)}
ΔHvap(T₂) = 44.9 kJ/mol
Therefore, enthalpy of vaporization of water at 273 K and 1 bar = 44.9kJ/mol
Answer:
-205.7kj
Explanation:
Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of the reaction.
The expression for enthalpy for the following reaction will be,
where,
n = number of moles
Now put all the given values in the above expression, we get:
Therefore, the enthalpy of the following reaction is, -205.7kj
Answer:
It is present in the second group of the periodic table
Explanation:
German scientist Döbereiner was one responsible for grouping elements into triads based on most notably atomic mass, many of which can be found in the periodic table to be in a pattern (for example <span><span>Iron </span><span>Cobalt </span><span>Nickel, elements 26, 27, 28)</span></span>
Answer: There are 6.9 mol of 
are required to react completely with 2.30 mol of S.
Explanation:
The given reaction equation is as follows.
Here, 1 mole of S is reaction with 3 moles of which means 1 mole of S requires 3 moles of .
Therefore, moles of required to react completely with 2.30 moles S are calculated as follows.
Thus, we can conclude that there are 6.9 mol of are required to react completely with 2.30 mol of S.
