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Snowcat [4.5K]
1 year ago
7

The mass of ethylene glycol (molar mass 62.1 g / mol), the main component of the antifreeze, must be added to 10.0 L of water to

obtain a solidified solution at -23 ° C. Assume a water density of 1 g / mL.
Chemistry
1 answer:
ser-zykov [4K]1 year ago
7 0
<h2> refer the attachment</h2>

<h2>If it is correct so mark as brainlest answer!</h2>

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What do the following organic chemicals have in common glyphosphate, lindane, 2,4-d, chlordane
otez555 [7]

Glyphosphate, lindane, 2,4-d, chlordane are organic compound that have chlorine in common.

<h3>What is Organic compound?</h3>

Organic compound are group of chemical substance that have carbon and hydrogen atoms in which the bonds are covalently linked together.

Therefore, Glyphosphate, lindane, 2,4-d, chlordane are organic compound that have chlorine in common and these chemicals are use as herbicides that is kill weeds, or to kill insects or pests because of chlorine present in it.

Learn more about organic compounds here.

brainly.com/question/14102593

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The name of the following molecular compound: P203
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B Phosphate trioxide
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Can someone please list one of the groups of representative elements in 1A-7A
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Read 2 more answers
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
8090 [49]

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.

4 0
3 years ago
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