If the.pressure exerted by a gas at [math]25^{\circ} \mathrm{C}[/math] in a volume of 0.044 L is 3.81 atm, how many moles of gas are present
        
             
        
        
        
        
Answer:
<h2>377 kPa</h2>
Explanation:
The original pressure can be found by using the formula for Boyle's law which is

where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we're finding the original pressure

150 kPa = 150,000 Pa
We have

We have the final answer as
<h3>377 kPa</h3>
Hope this helps you
 
        
             
        
        
        
Answer:
1) -COOH
2) -NH2
3) hydrogen bonds
4) dispersion forces
5) -CH3
6) hydrogen bonds
7) negative
8) negative
9) positive
Explanation:
Alanine has a <u>-COOH</u> and a <u>-NH2</u> group available to form <u>hydrogen bonds</u> with water molecules.
 Although there are some potential <u>dispersion forces</u> between the terminal <u>-CH3</u> group of alanine and hexane molecules, we expect the <u>hydrogen bonds</u> between alanine and water to be stronger.
Stronger intermolecular attractive forces between alanine and water lead to a more <u>negative ΔHmix</u> and more <u>negative (smaller positive)</u> ΔHsoln for water than for hexane.