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:
(a) Between methanol (CH₃OH) and glycerol (C₃H₅(OH)₃), the substance with the higher surface tension is glycerol (C₃H₅(OH)₃)
(b) Between tetrabromomethane (CBr₄) and chloroform (CHCl₃), the substance with the higher surface tension is chloroform (CHCl₃)
Explanation:
The surface tension of these substances at 20 °C given in mN/m, is as follows:
The surface tension of Methanol is 22.70
The surface tension of Tetrabromomethane is 26.95
The surface tension of Glycerol is 64.00
The surface tension of Chloroform is 27.50
(a) Between methanol (CH₃OH) and glycerol (C₃H₅(OH)₃), the substance with the higher surface tension is glycerol (C₃H₅(OH)₃)
(b) Between tetrabromomethane (CBr₄) and chloroform (CHCl₃), the substance with the higher surface tension is chloroform (CHCl₃)
Answer:

Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
I believe the answer is A the 1st one