Answer:
![h=100.8cm](https://img.qammunity.org/2021/formulas/chemistry/college/1hzge1pd6nn32pdu9hhur4k2oynam8he5g.png)
Step-by-step explanation:
Hello,
In this case, considering the density and mass of both water and heptane we first compute the volume of each one:
![V_(water)=(m_(water))/(\rho _(water))=(34g)/(1.00g/mL)=34mL\\ \\V_(heptane)=(m_(heptane))/(\rho _(heptane))=(34.6g)/(0.684g/mL)=50.6mL\\](https://img.qammunity.org/2021/formulas/chemistry/college/nm50hh7g3vd5si91mguhidzgsb6lyhr0dj.png)
Now, the total volume is:
![V=50.6mL+34mL=84.6mL](https://img.qammunity.org/2021/formulas/chemistry/college/klcsqplrc7ptuwhxoa5sljkmxrgno2t366.png)
Which is equal to:
![V=84.6cm^3](https://img.qammunity.org/2021/formulas/chemistry/college/1ndlyqsi37r8681f2jsyu2vui2491sd6gr.png)
Then, by knowing that the volume of a cylinder is πr²h or π(D/2)²h, we solve for the height as follows:
![h=(V)/(\pi (D/2)^2) \\\\h=(84.6cm^3)/(\pi (3.16cm/2)^2) \\\\h=100.8cm](https://img.qammunity.org/2021/formulas/chemistry/college/m1qva0it995xqvvaubdjmgajnjbu2sk9xo.png)
Best regards.