The standard enthalpy of formation for chlorine is zero but the standard entropy is larger than 0 because it is the elemental state of chlorine.
The standard enthalpy of formation for chlorine is zero because cl2 is the elemental state of chlorine and it does not require any energy for the formation of the standard state of chlorine.
The entropy of any system cannot be negative. It can only be positive or zero.
The entropy of a system will become zero only at a absolute zero temperature.
That's why the entropy of chlorine in elemental state is more than zero because absolutely zero temperature can't be obtained.
To know more about entropy, visit,
brainly.com/question/6364271
#SPJ4
The mass of krypton tetrachloride that can be produced assuming 100% yield is mathematically given as
molar mass=33.29g
<h3>What mass of
krypton tetrachloride can be
produced assuming 100% yield?</h3>
Generally, the equation for ideal gas is mathematically given as
PV=nRT
Therefore
n=(0.50)(15.)/0.082*623
n=0.147mol
Hence for clorine
n=0.441mol
Given the reaction
Kr+2cl2---->KrCL4
Hence
molar mass=225.60*0.147
molar mass=33.29g
Read more about Chemical Reaction
brainly.com/question/11231920
#SPJ1
Answer:
The total heat required is 691,026.36 J
Explanation:
Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L
Where Q: amount of heat, m: mass and L: latent heat
On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).
In this case, the total heat required is calculated as:
- Q for liquid water. This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C
Q= c*m*ΔT

Q=103,763.2 J
- Q for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then
)
Q= m*L

Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)
- Q for temperature change from 100.0
∘
C to 154
∘
C, this is, the sensible heat of steam from 100 °C to 154°C.
Q= c*m*ΔT

Q=25,176.96 J
So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J
<u><em>The total heat required is 691,026.36 J</em></u>