Answer:
80.7 L
Explanation:
PV = nRT
P = 1520 mmHg = 2 atm
n = 5 mol
R = 0.08206 (L * atm)/(mol * K)
T = 393.15 K
2 (V) = 5 (0.08206) (393.15)
V ≈ 80.7 L
This is because their tails are hydrophobic and their heads are hydrophilic. Hydrophobic meaning dislikes being near water and hydrophilic meaning likes to be near water. Therefore, they will orientate themselves in such a manner that the heads are facing externally and all the tails are facing together protected by the hydrophilic heads. google a photo of lipid chains in water if you are still confused. I'm not sure if that is what you are asking, but I hope it helps.
The metal properties that make up electrical wires are conductivity and ductility. Conductivity is important for free electrons to flow, hence electricity. Ductility is as well integral for strips to be formed, but with ample strength that does not break the material.
Answer:
See explanation
Explanation:
Hello there!
In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:
![Ksp=[Ag^+][Cl^-]](https://tex.z-dn.net/?f=Ksp%3D%5BAg%5E%2B%5D%5BCl%5E-%5D)
And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

I - 0 0
C - +x +x
E - x x
Which leads to the following modified equilibrium expression:

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.
Regards!
The energy required to raise the temperature of 3 kg of iron from 20° C to 25°C is 6,750 J( Option B)
<u>Explanation:</u>
Given:
Specific Heat capacity of Iron= 0.450 J/ g °C
To Find:
Required Energy to raise the Temperature
Formula:
Amount of energy required is given by the formula,
Q = mC (ΔT)
Solution:
M = mass of the iron in g
So 3 kg = 3000 g
C = specific heat of iron = 0.450 J/ g °C [ from the given table]
ΔT = change in temperature = 25° C - 20°C = 5°C
Plugin the values, we will get,
Q = 3000 g × 0.450 J/ g °C × 5°C
= 6,750 J
So the energy required is 6,750 J.