Answer:
[tex]h=100.8cm[/tex]
Explanation:
Hello,
In this case, considering the density and mass of both water and heptane we first compute the volume of each one:
[tex]V_{water}=\frac{m_{water}}{\rho _{water}}=\frac{34g}{1.00g/mL}=34mL\\ \\V_{heptane}=\frac{m_{heptane}}{\rho _{heptane}}=\frac{34.6g}{0.684g/mL}=50.6mL\\[/tex]
Now, the total volume is:
[tex]V=50.6mL+34mL=84.6mL[/tex]
Which is equal to:
[tex]V=84.6cm^3[/tex]
Then, by knowing that the volume of a cylinder is πr²h or π(D/2)²h, we solve for the height as follows:
[tex]h=\frac{V}{\pi (D/2)^2} \\\\h=\frac{84.6cm^3}{\pi (3.16cm/2)^2} \\\\h=100.8cm[/tex]
Best regards.
