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kenny6666 [7]
3 years ago
15

In looking at the bolded atoms in the answer choices, which pair consists of molecules having the same geometry? view available

hint(s) in looking at the atoms in the answer choices, which pair consists of molecules having the same geometry? ch2o and ch3oh ch2ccl2 and ch2ch2 pcl3 and bf3 co2 and so2
Chemistry
1 answer:
Molodets [167]3 years ago
4 0
The pair which consist of molecules having the same geometry is CH2CCI2 and CH2CH2.
Both of these molecules contain double bonds, which has sp^2 hybridization and they possess a trigonal planar geometry. In trigonal planar geometry, the molecule consist of three equally spaced sp^2 hybrid orbitals, which arranged at angle 120 degree. 
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Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
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Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

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