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2 years ago

14

What is specific heat capacity?

1 answer:

astraxan [27]2 years ago

4 0

Answer:

The energy needed to change the temp of a substance

Explanation:

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(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe

Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration $C_o$ = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

$time (t) = \dfrac{V}{Q}$

$time (t) = \dfrac{200}{0.3}$

$time (t) = 666.66 \ sec$

$time (t) = \dfrac{666.66 }{3600} hrs$

$time (t) = 0.185 hrs$

$\text{Using First Order Reaction:}$

$\dfrac{dc}{dt}=kc$

where;

$t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)$

$0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)$

$0.942 = \Big ( \dfrac{200}{C_e}- 1 \Big)$

$1+ 0.942 = \Big ( \dfrac{200}{C_e} \Big)$

$\dfrac{200}{C_e} = 1.942$

$C_e = \dfrac{200}{1.942}$

$\mathbf{C_e = 102.98 \ mg/l}$

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

$t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)$

$V_{PFR} =0.196(0.663)*0.3*3600$

$V_{PFR} = 140.34 \ m^3$

The volume of the PFR is ≅ 140 m³

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3 years ago

When a given reaction is conducted in a calorimeter, energy is absorbed from the surrounding water and results in a decrease in

rjkz [21]

The potential energy of the products is higher than the potential energy of the reactants.

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3 years ago

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How do we calculate how fast (speed) Earth is rotating on its arist

strojnjashka [21]

Speed= distance/ time

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3 years ago

A sample of nitrogen gas has the temperature drop from 250.°c to 150.°c at constant pressure. what is the final volume if the in

Leviafan [203]

Charles law gives the relationship between temperature of gas and volume of gas.
It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where V- volume , T - temperature and k - constant
$\frac{V1}{T1} = \frac{V2}{T2}$
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
T1 = 250 °C + 273 = 523 K
T2 = 150 °C + 273 = 423 K
Substituting the values in the equation,

$\frac{310 mL}{523 K} = \frac{V}{423K}$
V = 251 mL
the new volume is 251 mL

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using the two ways you have learned about between temperature, pressure and volume, what are two ways you could increase the gas

IrinaVladis [17]

Answer:

I don't know do you know tell me please

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2 years ago