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2 years ago

9

Find the "density" of a copper sample whose mass is 134.4 grams and whose volume is 15.0mL

1 answer:

Sonja [21]2 years ago

3 0

Answer:

8.96 g/mL

Explanation:

density = mass / volume

density = 134.3g / 15.0 mL

density = 8.96 g/mL

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How many grams of water will form if 10.54 g H2 react with 95.10 g O2?

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Answer:

94.0 g H2O

Explanation:

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A postulate is a statement recognized as a truth. The kinetic molecular theory has five postulates. Which of the following is NO

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2nd option is NOT a postulate of kinetic theory of gases

Explanation:

actually particles move in straight random motion and create bombardment with the molecules and container

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Give examples of how land resources are used in a city setting

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Land resources such as gravel and bedrock are used to construct buildings, roads, and sidewalks. Land is also where city structures—such as buildings, roads, and sidewalks—are constructed. People use land resources such as bedrock and aggregate to construct buildings, roads, and sidewalks.

I hope this answer helps

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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

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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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