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3 years ago

Solid calcium oxide reacts with sulfur trioxide gas to produce solid calcium sulfate

Chemistry

1 answer:

antiseptic1488 [7]3 years ago

5 0

<u>Answer:</u> The chemical equation for the given reaction is given below.

<u>Explanation:</u>

When solid calcium oxide reacts with sulfur trioxide gas to produce solid calcium sulfate, the chemical equation follows:

$CaO(s)+SO_3(g)\rightarrow CaSO_4(s)$

This is the reaction when a metal oxide reacts with non-metal oxide to produce a salt.

By Stoichiometry of the reaction:

1 mole of solid calcium oxide reacts with 1 mole of sulfur trioxide gas to produce 1 mole of solid calcium sulfate.

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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:

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

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