Answer:
=3,723.3 J=3.72 but if it has the option of -3.72 kJ then use that
Explanation:
Use the formula q=m×Cp×delta T
m=1.500 kg=1,500 g
Co=2.52 J/g·k
delta T=0.985k
q=(1,500g)(2.52 J/g·k)(0.985k)
Answer:
The correct answer to the question is Option E (Strongly retained analytes will give broad peaks).
Explanation:
The other options are true because:
A. Initial temp = 50 °C
Final temp = 270 °C
Differences in temp = 270 - 50 = 220°C
Rate = 10 °C/minute.
So, at 10 °C/minute,
total of 220°C /10 °C = number of minutes required to reach the final temp.
220/10 = 22 minutes
B. A column has a minimum and maximum use temperature. Solutes that are already retained would remain stationary while temperatures are low. This would only change if there is an increase in temperature. Heat transfers more energy to the liquid which would make the solute interact with the column phase.
C. Weakly retained solutes may contain larger molecules, will separate by absorbing into the solvent early in separation making the mobile phase separates out into its components on the stationary phase.
D. Retained solute's vapor pressure is higher at higher temperatures making it possible for particle to escape more from the solute when the temperature is high than when it is low.
Answer:
i need a picture of ur question
Explanation:
