When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
Answer:
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Explanation:
<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ
<u>Explanation:</u>
To calculate the mass of water, we use the equation:

Density of water = 1 g/mL
Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:

To calculate the heat absorbed by the water, we use the equation:

where,
q = heat absorbed
m = mass of water = 1500 g
c = heat capacity of water = 4.186 J/g°C
= change in temperature = 
Putting values in above equation, we get:

Hence, the amount of heat required to warm given amount of water is 470.9 kJ
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