As the tone gets higher, the sound waves get closer together
Answer:
(a) ![M=0.257M](https://tex.z-dn.net/?f=M%3D0.257M)
(b) ![M=0.0510M](https://tex.z-dn.net/?f=M%3D0.0510M)
(c) ![M =0.500M](https://tex.z-dn.net/?f=M%20%3D0.500M)
(d) ![M= 0.391M](https://tex.z-dn.net/?f=M%3D%200.391M)
Explanation:
Hello,
In this case, since the molarity or molar concentration of a solution is computed by dividing the moles of solute by the volume of solution in liters, we proceed as follows:
(a) The molar mass of sodium chloride is 58.45 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:
![M=\frac{1.50gNaCl}{0.100L} *\frac{1molNaCl}{58.45gNaCl} =0.257M](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B1.50gNaCl%7D%7B0.100L%7D%20%2A%5Cfrac%7B1molNaCl%7D%7B58.45gNaCl%7D%20%3D0.257M)
(b) The molar of potassium dichromate is 294.2 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:
![M=\frac{1.50gK_2Cr_2O_7}{0.100L} *\frac{1molK_2Cr_2O_7}{294.2gK_2Cr_2O_7} =0.0510M](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B1.50gK_2Cr_2O_7%7D%7B0.100L%7D%20%2A%5Cfrac%7B1molK_2Cr_2O_7%7D%7B294.2gK_2Cr_2O_7%7D%20%3D0.0510M)
(c) The molar of calcium chloride is 111 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:
![M=\frac{5.55gCaCl_2}{0.100L} *\frac{1molCaCl_2}{111gCaCl_2} =0.500M](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B5.55gCaCl_2%7D%7B0.100L%7D%20%2A%5Cfrac%7B1molCaCl_2%7D%7B111gCaCl_2%7D%20%3D0.500M)
(d) The molar of sodium sulfate is 142 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:
![M=\frac{5.55gNa_2SO_4}{0.100L} *\frac{1molNa_2SO_4}{142gNa_2SO_4} = 0.391M](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B5.55gNa_2SO_4%7D%7B0.100L%7D%20%2A%5Cfrac%7B1molNa_2SO_4%7D%7B142gNa_2SO_4%7D%20%3D%200.391M)
Best regards.
I should think because the valence bonds want to escape and bond more whereas the inner layers are for the most part stable
Answer:
The answer is Sodium chloride.
Na is sodium and Cl is chlorine.