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3 years ago

7

The largest pieces being dragged along the streambed are undergoing the process called ________. pushing saltation traction susp

ension solution

1 answer:

lianna [129]3 years ago

8 0

Answer:

saltation

Explanation:

Saltation is the transportation of particles by fluids such as by winds and water. It takes place when loose matter is removed from the sea bed and gets carried by the fluid back to the surface.

Examples include pebbles, sand, and soil. Saltation layers can form during avalanches.

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Minchanka [31]

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2 years ago

Used oil needs to be kept in what kind of containers

Paul [167]

Answer:

Never store used oil in anything other than tanks and storage containers. Used oil may also be stored in units that are permitted to store regulated hazardous waste. Tanks and containers storing used oil do not need to be RCRA permitted, however, as long as they are labeled and in good condition.

8 0

3 years ago

Match each body system with its main organ.

olya-2409 [2.1K]

Answer:

it's in the explanation

Explanation:

Lungs- Respiratory

Heart- Circulatory

Brain- Nervous

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7 0

3 years ago

What would be the pH of an H2SO4 solution if the [H+] = 4.58 x 10-5 moles/liter?

Ghella [55]

Answer: pH of an $H_2SO_4$ solution is 4.34

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$H_2SO_4\rightarrow 2H^++SO_4^{2-}$

Putting in the values:

$pH=-\log[4.58\times 10^{-5}]$

$pH=4.34$

Thus pH of an $H_2SO_4$ solution if the $[H^+]$ is $4.58\times 10^{-5}]$ is 4.34

4 0

3 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

3 years ago