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weqwewe [10]
3 years ago
10

.The arrangement of constituent particles is most ordered in *

Chemistry
1 answer:
Dimas [21]3 years ago
8 0

Answer:

idk srry i tried

Explanation:

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Who pass the grade quarter?.
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Answer:

me

Explanation:

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11.
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D. dishwashing Soap/Liquid
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The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0
3 years ago
7. This part of the brain controls memory, and long term abuse of marijuana can permanently
eduard

Answer:

Hippocampus

Explanation:

The main parts of the brain involved with memory are the amygdala, the hippocampus, the cerebellum, and the prefrontal cortex.

Memory impairment from marijuana use occurs because THC alters how the hippocampus, a brain area responsible for memory formation, processes information.

7 0
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Which of the following equations follows the law of conservation of mass?
Dovator [93]

Answer:

option A = C₂H₄ + 3O₂   →    2CO₂   +  2H₂O

Explanation:

Law of conservation of mass:

This law stated that mass can not be created or destroyed in chemical reaction. It just changed from one to another form.

For example:

C₂H₄   +   3O₂   →    2CO₂   +  2H₂O

28 g   + 96 g    =      88 g  +  36 g

    124  g           =      124 g

This reaction correctly hold the law of conservation of mass.

Other options:

C  + 4H₂  →   CH₄

12 g  + 8g  = 16 g

  20 g  =   16 g

This reaction do not hold the law of conservation of mass.

3H₂O  →   3H₂  +  3O₂

  54 g  =   6 g +  96 g

   54 g =   102 g

This reaction do not hold the law of conservation of mass.

2Na  +  Cl  →  NaCl

46 g  + 35.5 g  =   58.5 g

  81.5 g = 58.5 g

This reaction do not hold the law of conservation of mass.

6 0
3 years ago
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