It is the smallest unit it is what makes up everything
joke-Don't trust Atoms they make up everything
Answer:
191.6 g of CaCl₂.
Explanation:
What is given?
Mass of HCl = 125.9 g.
Molar mass of CaCl₂ = 110.8 g/mol.
Molar mass of HCl = 36.4 g/mol.
Step-by-step solution:
First, we have to state the chemical equation. Ca(OH)₂ react with HCl to produce CaCl₂:

Now, let's convert 125.9 g of HCl to moles using the given molar mass (remember that the molar mass of a compound can be found using the periodic table). The conversion will look like this:

Let's find how many moles of CaCl₂ are being produced by 3.459 moles of HCl. You can see in the chemical equation that 2 moles of HCl reacted with excess Ca(OH)₂ produces 1 mol of CaCl₂, so we state a rule of three and the calculation is:

The final step is to find the mass of CaCl₂ using the molar mass of CaCl₂. This conversion will look like this:

The answer would be that we're producing a mass of 191.6 g of CaCl₂.
Answer :
The equilibrium concentration of CO is, 0.016 M
The equilibrium concentration of Cl₂ is, 0.034 M
The equilibrium concentration of COCl₂ is, 0.139 M
Explanation :
The given chemical reaction is:

Initial conc. 0.1550 0.173 0
At eqm. (0.1550-x) (0.173-x) x
As we are given:

The expression for equilibrium constant is:
![K_c=\frac{[COCl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCOCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
Now put all the given values in this expression, we get:

x = 0.139 and x = 0.193
We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.139
The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M
The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M
The equilibrium concentration of COCl₂ = x = 0.139 M
Chlorine gas reacts to potassium bromide to form potassium chloride in solution and liquid bromine.
I hope this helps/answers your question! I vaguely remember getting this question before too