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3 years ago

What conditions must be met to compress a gas

1 answer:

3 0

In order to compress a gas, high pressure should be met. <span>There are three major groups of compressed gases stored in cylinders: liquefied, non-liquefied and dissolved gases. In each case, the pressure of the gas in the cylinder are high. Hope this helps.</span>

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3 years ago

Why is methane, CH4, a nonpolar compound?

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Each covalent H bond is nonpolar.

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) Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2is 7

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Answer:

Rate of the reaction is 0.2593 M/s

-0.5186 M/s is the rate of the loss of ozone.

Explanation:

The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

$2O_3\rightleftharpoons 3O_2$

Rate of formation of oxygen : $7.78\times 10^{-1} M/s$

Rate of the reaction(R) = $\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}$

$R=\frac{1}{3}\frac{d[O_2]}{dt}$

Rate of formation of oxygen=3 × (R)

$7.78\times 10^{-1} M/s=3\times R$

Rate of the reaction(R): $0.2593 M/s$

Rate of the reaction is 0.2593 M/s

Rate of disappearance of the ozone:

$R=-\frac{1}{2}\frac{d[O_3]}{dt}$

$\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s$

-0.5186 M/s is the rate of the loss of ozone.

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3 years ago

Water is flowing in a pipe with a mass flow rate of 100.0 lb/h (M water H,O=18.016). What is the (i) Molar flow rate of H20 (gmo

Arada [10]

Answer:

i) 0,7 molH20/s

ii)11,2 g O/s

iii)1,4 g H/s

Explanation:

i) To find the molar flow rate of water, we just convert the mass of water to moles of water using its molecular weight(g/mol) and changing to the proper units (lb to grames and hours to seconds):

$100 \frac{lb}{h}*\frac{453,5g}{1 lb}*\frac{1molH20}{18,016g}*\frac{1h}{3600s}=0,7\frac{molH20}{s}$

ii) Now we just consider the oxygen in the water stream (for 1 mole of water there is 1 mole of oxygen):

$100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{1molO}{1molH20}*\frac{16gr}{1molO}*\frac{1h}{3600s}=11,2\frac{gO}{s}$

iii)Just considering the hydrogen in the stream (for 1 mole of water there is 2 moles of hydrogen):

$100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{2molH}{1molH20}*\frac{1gr}{1molH}*\frac{1h}{3600s}=1,4\frac{gH}{s}$

3 0

3 years ago