Answer:
The answer to your question is Volume = 11.4 L
Explanation:
Data
Volume 1 = V1 = 6 L
Pressure 1 = P1 = 1 atm
Temperature 1 = T1 = 22°C
Volume 2 = V2 = ?
Pressure 2 = 0.45 atm
Temperature 2 = -21°C
Process
1.- Convert temperature (°C) to °K
T1 = 273 + 22 = 295°K
T2 = 273 + (-21) = 252°K
2.- Use the combined gas law to solve this problem
P1V1 / T1 = P2V2 / T2
-Solve for V2
V2 = P1V1T2 / T1P2
-Substitution
V2 = (6)(1)(252) / (295)(0.45)
- Simplification
V2 = 1512 / 132.75
- Result
V2 = 11.38 L
Answer:
The answer will be Ligand A with a dissociation constant (Kd) of
M
Explanation:
When the dissociation constant in the ligand is small (in order of nano) (
) it will be more tied. Due to a dissociation constant measures how much a ligand can be able to be separated from the protein so if the number is small it means that the ligand is highly binded to the protein.
On the other hand, the occupancy percentage of the ligand does not imply binding. Conversely, a High-affinity ligand binding with the proteins implies that a relatively low concentration of a ligand is adequate to occupy the maximum ligand-binding site.
Rubbing Alcohols are chemical compounds because rubbing alcohol itself is a strecture made up of many different chemicals combinding in a compound that make, C3H8O
The volume increases when the balloon temperature increases.
<u>Explanation:</u>
-10 F is converted into Kelvin as 249 K.
0°C is nothing but 0+ 273 = 273 K
And the room temperature is 25°C which is converted into Kelvin as 273 + 25 = 298 K.
249 K is below room temperature.
As per the Charles' law volume and temperature are directly proportional to each other, when the pressure of the gas remains constant.
V ∝ T
As the balloon temperature increases, the volume also increases.
Answer:
The reactants would appear at a higher energy state than the products.
Have a nice day!