Answer:
A.) 4.0
Explanation:
The general equilibrium expression looks like this:
![K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%5BB%5D%5E%7Bb%7D%20%7D)
In this expression,
-----> K = equilibrium constant
-----> uppercase letters = molarity
-----> lowercase letters = balanced equation coefficients
In this case, the molarity's do not need to be raised to any numbers because the coefficients in the balanced equation are all 1. You can find the constant by plugging the given molarities into the equation and simplifying.
<----- Equilibrium expression
<----- Insert molarities
<----- Multiply
<----- Divide
Answer:
The value of
for xylene is 4.309°C/m.
The molar mass of pentane using this data is 73.82 g/mol.
Explanation:

where,
=depression in freezing point
= freezing point constant
we have :
1) freezing point constant for xylene =
=?
Mass of toluene = 0.193 g
Mass of xylene = 2.532 kg = 0.002532 kg ( 1 g =0.001 kg)



The value of
for xylene is 4.309°C/m.
2)
Mass of pentane = 0.123 g
molar mass of pentame= M
Mass of xylene = 2.493 g = 0.002493 kg
Freezing point Constant of xylene = 

M = 73.82 g/mol
The molar mass of pentane using this data is 73.82 g/mol.
Answer:
26.8 °C
Explanation:
Step 1: Given and required data
- Energy transferred to the water (Q): 2100 J
- Initial temperature of the water (T₂): 23.6 °C
- Final temperature of the water (T₁): ?
- Specific heat of water (c): 4.184 J/g.°C
Step 2: Calculate the final temperature of the water
We will use the following expression.
Q = c × m × (T₂ - T₁)
T₂ = 2100 J/(4.184 J/g.°C) × 155 g + 23.6 °C = 26.8 °C