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2 years ago

QUESTION 1

Chemistry

2 answers:

zhenek [66]2 years ago

6 0

1. Triassic
2. eon > era > period
3. The scientists study fossils and rock layers to find major changes.

otez555 [7]2 years ago

3 0

1. Terrassic: Relating to or denoting the earliest period of the Mesozoic era, between the Permian and Jurassic periods.
2. Eon > Era > Period: An Eon is a large amount of time, An Era is a long and distinct period of history, A length of time.
3. Scientists define the segments of the geological time scale by studying fossil and rock layers for major changes.

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On the pH scale pure water has a pH of A. 0 B. 7 C. 14

konstantin123 [22]

Answer:its b 7

Explanation:

7 is neutral and pure water is the middle man

7 0

3 years ago

Read 2 more answers

You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acetate plus acetate concentration

vovangra [49]

Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM

Given:

Acetic Acid/Sodium Acetate buffer of pH = 5.0

Let HA = acetic acid

A- = sodium acetate

Total concentration [HA] + [A-] = 250 mM ------(1)

pKa(acetic acid) = 4.75

Based on Henderson-Hasselbalch equation

pH = pKa + log[A-]/[HA]

[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77

[A-] = 1.77[HA] -----(2)

From (1) and (2)

[HA] + 1.77[HA] = 250 mM

[HA] = 250/2.77 = 90.25 mM

[A-] = 1.77(90.25) = 159.74 mM

7 0

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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3 years ago

50 POINTS for a legitimate response. Please answer the question, chose the correct unit conversions on the right. Please read th

makkiz [27]

Answer:

0.121 millimeters

5 0

2 years ago

Determine the electron-group arrangement, molecular shape, and ideal bond angle(s) for each of the following:

lubasha [3.4K]

Solution :

a). $SO_3$

This compound is known as sulfur trioxide.

The molecular shape of sulfur trioxide is trigonal planer.

And the bond angle is 120°.

b). $N_2O$

This compound is known as Nitrous oxide. Here, nitrogen is in the center. There is no lone pair around the nitrogen atom and it forms two sigma bonds with the other two atoms.

It is linear in shape.

The bond angle between them is 180°.

c). $CH_2Cl_{2}$

This compound is known as the Dichloromethane.

The molecular shape of the compound is tetrahedral.

The bond angles is 120°

6 0

3 years ago