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2 years ago

In which Colorado Rockies scenario was there more erosion? Why was there more erosion in the scenario?

2 answers:

7 0

There was more erosion in the Colorado Rockies scenario where there was steady rain compared to the scenario in the same area where there was only a drizzle. There is more erosion because the steady rain increased the volume of the river more than the drizzle did. Because there was more water added to the river, the river flowed faster. The faster a river flows, the more sediments it transports or erodes.

DENIUS [597]2 years ago

4 0

Answer:

Explanation:

The Erosion in the Colorado Rookies Scenario that record a ready rain were more than the other scenario in the same area with just a drizzle. The Erosion is much in the Colorado Rockies Scenario that record a steady rain because the volume of the rivers has goes up compare to that of the Scenario with just a drizzle. As a result of the increased in the amount of the rivers bank that is flowing, the river flows more faster which helps in the transportation of the sediments or Erodes.

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How does a conductivity apparatus test whether a solution has ionic or covalent substances in it? It fails to light up except in

larisa [96]

Covalent compounds are those which are made up by sharing of electrons between them and the electronegativity of the elements which form covalent compounds are nearly same that is they are generally non-polar in nature whereas ionic compounds are formed by complete transfer of electrons from one element to the other and thus resulting in formation of ions of opposite charges that is cation (with positive charge) and anion (with negative charge). There is a large difference in electronegativity of the element involved in the formation of ionic compounds. They are held together by electrostatic force of attraction between them.

The conductivity of a substance is determined by the flow of electric charge. When the charged particles move that are present in the ionic compounds only in the conductivity apparatus they complete the circuit by the flow of electrons.

Hence, a conductivity apparatus test whether a solution has ionic or covalent substances in it as ions complete the flow of electrons to form a circuit.

7 0

2 years ago

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The refractive index, n of a polymer was determined as a function of temperature. The results are given in the table. Determine

mylen [45]

Answer:

Explanation:

The given data shows linear trend between refractive index and temperature .

From 20 degree to 80 degree , every increase of 10 degree decreases refractive index by .0015 . This trend breaks at 90 degree beyond which , refractive index falls at greater rate or sharply.

So transition temperature is 90 degree .

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2 years ago

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel in which it partially dissociates at high temperatures. 2NH 3(g)

allochka39001 [22]

Explanation:

system at equilibrium, will the reaction shift towards reactants ~

--?'

2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an

exothermic reaction. Will heating the equilibrium system increase o~e amount of

ammonia produced? . .co:(

3. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). Ifwe use a catalyst, which way will

the reaction shift? ':'\

.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,

4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)

Temperature (K) Kc

300 1.5x104

600 55 k ' pr, cl l<..J~

e- ~ r fee, ct o. ~ 1<

900 3.4 X 10-3

Is the reaction endothermic or exothermic (explain your answer)?

t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\

exothe-rnh't.-- ,.. ..,. (/.., ,~.

5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.

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~ ;)N~~) ~ ~ H ~) ~\-_ == [A!;J:t D~~Jb

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4 0

2 years ago

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a

svetoff [14.1K]

Answer:

The answer to your question is P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

= 373°K

Formula

$(P + \frac{a}{v^{2}} )(v - b) = RT$

Substitution

$(P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)$

Simplify

(P + 0.0094)(22.3829) = 30.586

Solve for P

P + 0.0094 = $\frac{30.586}{22.3829}$

P + 0.0094 = 1.366

P = 1.336 - 0.0094

P = 1.357 atm

7 0

2 years ago

A sample of krypton has a volume of 6.00 L, and the pressure is 0.960 atm. If the final temperature is 55.0°C, the final volume

MissTica

193.38 K was the initial temperature of the krypton.

Explanation:

Data given:

Initial volume of the krypton gas = 6 litres

initial pressure of the krypton gas = 0.960 atm

initial temperature of the krypton gas = ?

final volume of the krypton gas = 7.70 litres

final pressure of the Krypton gas = 1.25 atm

final temperature of the krypton gas = 55 degrees or 273.25+55 = 323.15 K

Applying the Combined Gas Laws:

$\frac{P1V1}{T1} = \frac{P2V2}{T2}$

Rearranging the equation:

T1 = $\frac{P1V1T2}{P2V2}$

Putting the value in the equation:

T1 = $\frac{0.960 X 6 X 323.15}{1.25 X 7.70}$

T1 = 193.38 K

Initial temperature of the krypton gas is 193.78 K

5 0

2 years ago