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prohojiy [21]
3 years ago
11

Which factor most likely facilitates mass movement after heavy rains in a dry region?

Chemistry
2 answers:
melomori [17]3 years ago
4 0

Answer: Option (b) is the correct answer.

Explanation:

The process in which sediment moves downhill is known as mass movement.

Different types of mass movement are landslides, mud slides, slump, creep etc.

Mud flow contains mass of saturated rock particles of all sizes. Mud flow arises due to sudden flood of water or due to heavy rain in a dry region (semi-arid region). Soil and rocks from a large slope area flow along with the flood water and gets washed to a gulch or canyon.

As a result, debris and water moves down canyon and lay out on the gentle slopes below.

Thus, we can conclude that mud flow is most likely facilitates mass movement after heavy rains in a dry region.

kvasek [131]3 years ago
4 0

Answer:

mudflow

Explanation:

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How many moles are present in 45.0 g of lithium oxide (Li2O)?
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Molar mass Li2O = 7.0 x 2 + 16 => 30 g/mol

1 mole Li2O ---------- 30 g
( moles Li2O ) -------- 45.0 g

moles Li2O = 45.0 x 1 / 30

moles Li2O = 45.0 / 30

= 1.50 moles of Li2O 

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Which describes the difference between frogs and bacteria?
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The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,
Masja [62]

<u>Answer:</u> The concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

<u>Explanation:</u>

The given cell is:

Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V ( × 3)

<u>Reduction half reaction:</u> Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V ( × 2)

<u>Net reaction:</u> 3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.50-0=1.50V

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}

where,

E_{cell} = electrode potential of the cell = 1.23 V

E^o_{cell} = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[Au^{3+}]=?M

[H^{+}]=1.0M

Putting values in above equation, we get:

1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})

[Au^{3+}]=1.87\times 10^{-14}M

Hence, the concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

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