Answer:
2.445 g
Explanation:
Step 1: Given and required data
- Energy in the form of heat required to boil the water (Q): 5525 J
- Latent heat of vaporization of water (∆H°vap): 2260 J/g
Step 2: Calculate the mass of water
We will use the following expression.
Q = ∆H°vap × m
m = Q / ∆H°vap
m = 5525 J / (2260 J/g)
m = 2.445 g
Answer: The equilibrium constant for the overall reaction is 
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
a) 
![K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
b) 
![K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
For overall reaction on adding a and b we get c
c) 
![K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5E%5Cfrac%7B5%7D%7B2%7D%7D)
![K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_c%3DK_a%5Ctimes%20K_b%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%5Ctimes%20%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
The equilibrium constant for the overall reaction is
Answer: A supersaturated solution will not contain undissolved solute because the undissolved solute will be indicative of saturated solution.
Explanation:
A supersaturated solution is the one that consists of more than the maximum concentration of the solute in the solvent that is being dissolved at a given temperature. A saturated solution is the one in which the maximum concentration of solute has been dissolved in the solvent and no additional solute can be dissolved further.
According to the given statement, a solution with undissolved solute is a saturated solution rather a supersaturated solution.
Answer: 0.52 L of 15 M 
will be used to prepare this amount of 0.52 M base.
Explanation:
But on diluting the number of moles remain same and thus we can use molarity equation.
(to be prepared)
where,
=concentration of stock solution = 15 M
= volume of stock solution = ?
= concentration of solution to be prepared = 0.52 M
= volume of solution to be prepared = 15 L
Thus 0.52 L of 15 M will be used to prepare this amount of 0.52 M base
C plants only have a cell wall, however animal cells have a cell wall and a cell membrane
