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3 years ago

Which two characteristics describe all animals

Chemistry

1 answer:

Maru [420]3 years ago

5 0

All animals can be dangerous and they would fight for their family. (This might be wrong)

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Earth is the only planet that has a _____ and a _____.

Serggg [28]

If there are no selections than i would say a thick atmosphere and an unusual large moon.<span />

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3 years ago

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Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

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3 years ago

Spring tides occur when the sun and moon word together to form the highest high tide and the lowest low tide this type of tide o

anyanavicka [17]

It will be B. Full/ new moon

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2 years ago

At standard pressure, during which physical change does the potential energy decrease?

skad [1K]

Gas - > liquid - > solid

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3 years ago

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The reaction AB(aq)→A(g)+B(g) is second order in AB and has a rate constant of 0.0164 M −1 ⋅ s −1 at 25.0 ∘ C . A reaction vesse

uysha [10]

The reaction is second order in AB, so: $v=k[AB]^2$ . In the statement, we obtain that $[AB]=0.104~M$ and, at 25 ºC, $k=0.0164~M^{-1}\cdot s^{-1}$ . Then:

$v=k[AB]^2\\\\
v=0.0164\cdot0.104^2\\\\
v=0.0164\cdot0.010816\\\\
v\approx0.000177=1.77\times10^{-4}~mol/s$

Now, we'll calculate the number of mols of the products in the gas. Using the Ideal Gas Law:

$\bullet~\text{Pressure:}~p=707.3-23.8=683.5~mmHg\\\\
\bullet~\text{Volume:}~V=142~mL=0.142~L\\\\
\bullet~\text{Number of moles:}~n=n_A+n_B\\\\
\bullet~\text{Ideal gas constant:}~R=62.3~L\cdot mmHg\cdot K^{-1}\cdot mol^{-1}\\\\
\bullet~\text{Temperature:}~T=25^oC=25+273~K=298~K\\\\$

$pV=nRT\\\\
683.5\cdot0.142=n\cdot62.3\cdot298\\\\
n=\dfrac{683.5\cdot0.142}{62.3\cdot298}\\\\
n\approx0,0052~mol$

Since each AB molecule forms one of A and one of B, $n_A=n_B$ . Hence: $2n_A\approx0,0052\Longrightarrow n_A=n_B\approx0.0026~mol$ .

We'll consider that in the beginning there was not A or B. So, $\Delta n_A=\Delta n_B=0.0026-0=0.0026~mol$ . Furthermore, since the ratio of AB to A and to B is 1:1, $|\Delta n_{AB}|=|\Delta n_A|=|\Delta n_B|$ .

Calculating the time by the expression of velocity:

$v=\dfrac{|\Delta[AB]|}{\Delta t}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_{AB}|}{V}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_A||}{250~mL}\\\\
1.77\cdot10^{-4}=\dfrac{1}{\Delta t}\cdot\dfrac{0.0026~mol}{0.25~L}\\\\
\Delta t=\dfrac{0.0026}{0.25\cdot1.77\cdot10^{-4}}\\\\
\boxed{\Delta t\approx58.76~s}$

8 0

3 years ago