Answer:
27.70 is 
for the reaction.
Explanation:
The equilibrium constant of the reaction = 
The expression of equilibrium constant is given by :
![K_c=\frac{[CO][Br_2]}{[COBr_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5BBr_2%5D%7D%7B%5BCOBr_2%5D%7D)
..[1]
The equilibrium constant expression for above reaction can be written as:
![K_c'=\frac{[COBr_2]^2}{[CO]^2[Br]^2}](https://tex.z-dn.net/?f=K_c%27%3D%5Cfrac%7B%5BCOBr_2%5D%5E2%7D%7B%5BCO%5D%5E2%5BBr%5D%5E2%7D)
( from [1])
27.70 is for the reaction.
I think the correct answer from the choices listed above is option B. The formation of frost on a cold windowpane is an example of condensation. This process involves the phase change from the vapor phase to the liquid phase.
Neither, boiling and melting point per element varies from element to element on the periodic table. Family trends and period trends determine what kind of element each aspect is.
Potassium, aluminum, silicon, magnesium neon
Answer:
0.00676 M
Explanation:
A chemist prepares a solution of calcium bromide by weighing out 0.607g of calcium bromide into a 450ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's calcium bromide solution. Be sure your answer has the correct number of significant digits.
Step 1: Given data
Mass of calcium bromide (solute): 0.607 g
Volume of solution: 450 mL
Step 2: Calculate the moles corresponding to 0.607 g of calcium bromide
The molar mass of CaBr₂ is 199.89 g/mol.
0.607 g × 1 mol/199.89 g = 0.00304 mol
Step 3: Convert the volume of solution to liters
We will use the conversion factor 1 L = 1000 mL.
450 mL × 1 L/1000 mL = 0.450 L
Step 4: Calculate the molar concentration of calcium bromide
The molarity of the solution is:
M = moles of solute / liters of solution
M = 0.00304 mol / 0.450 L
M = 0.00676 M
