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4 years ago

11

Alfredo is conducting an experiment on the effects of exercise on concentration. the independent variable is (the) ________ and

the dependent variable is (the) ________. (select only one answer)

2 answers:

nikitadnepr [17]4 years ago

7 0

Answer:

Independent: exercise.

Dependent: concentration.

Explanation:

Hello,

In this case, as the statement says that the experiments is towards the effects of the exercise on concentration it is clear that the independent variable is the exercise as the concentration depends on the performed exercise. Therefore, the dependent variable is the concentration as it changes as a function of the specified exercise.

Best regards.

hodyreva [135]4 years ago

4 0

1. exercise because you are using exercise to affect the amount of concentration. 2. concentration because concentration is what is being measured

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What did the 15th amendment do

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The 15th Amendment to the Constitution granted African American men the right to vote by declaring that the "right of citizens of the United States to vote shall not be denied or abridged by the United States or by any state on account of race, color, or previous condition of servitude."

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3 years ago

What type of reaction is:

nadya68 [22]

Answer:

The answer to your question is Combustion

Explanation:

Chemical reaction

2C₄H₁₀ + 13 O₂ ⇒ 8 CO₂ + 10 H₂O

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3 years ago

(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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