Given:
<span> 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
volume of CL2
Solution:
Use the ideal gas law
PV = nRT
V = nRT/P
V = (2.1 moles Cl2) (0.08203 L - atm / mol - K) (273K) / (1 atm)
V = 47 L</span>
Answer:
8.3 kJ
Explanation:
In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:
q for water:
q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g
c: specific heat of water = 4.186 J/gºC
ΔT : change in temperature = 2.06 ºC
so solving for q :
q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J
For calorimeter
q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC
ΔT : change in temperature = 2.06 ºC
q calorimeter = 69.60J x 2.06 ºC = 143.4 J
Total heat released = 8,140 J + 143.4 J = 8,2836 J
Converting into kilojoules by dividing by 1000 we will have answered the question:
8,2836 J x 1 kJ/J = 8.3 kJ
Chlorine is a halogen and all halogens and oxygen, nitrogen and hydrogen are diatomics
Answer:
The correct answer is "Fragment B likely has a higher Guanosine/Citosine content".
Explanation:
Guanosine/Citosine content, or GC content, refers to how many molecules of guanosine and citosine have a DNA fragment, respect to the content of adenine and thymine. The higher the GC content, the higher the temperature needed to denature the fragment of DNA. This happens because guanosine and citosine establish three hydrogen bonds, while adenine and thymine establish two hydrogen bonds when they bind together. Therefore, if fragment A and B are the same length, but at 89 C only fragment A is completely denatured, fragment B likely has a higher GC content.
