All categories

3 years ago

Which type of weathering helped to form barrier islands

1 answer:

7 0

Barrier islands typically have sand in the beach zone and dune field, and mud in the back-barrier. Overwash deposits sand in the back-barrier.

Barrier islands form in three ways. They can form from spits, from drowned dune ridges or from sand bars. Longshore drift is the movement of sand parallel to the shore caused by the angle of the waves breaking on the beach. ... When a storm such as a hurricane digs an inlet through the spit a barrier island is formed.

You might be interested in

What is the bond order for a single covalent bond?

hram777 [196]

Answer:

We typically represent covalent bonds with a dash ( - ) between the atoms. This indicates a single bond. Ex: Cl - Cl

Single bond, double bond, triple bond.

Explanation:

We call it a single covalent bond because the atoms are sharing a single pair of electrons.

4 0

3 years ago

In a desert, soil containing a mixture of sand and small rocks is exposed to wind erosion. Over time, how would the land surface

amid [387]

Sand dunes would be created due to the mixture falling on each other
x

3 0

2 years ago

Read 2 more answers

12a. How does the model of the volcano provide evidence for the cycling of Earth materials through the rock cycle?

Bond [772]

Answer:

lava is coming from the crust

Explanation:

volcano's explode with lava because the crust pushes it out

6 0

2 years ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

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

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

3 years ago

If the volume occupied by 0.500 mol of nitrogen gas at 0°C is 11.2 L, then the volume occupied by 2.00 mol of nitrogen gas at th

Paraphin [41]

Avagadros law states that volume of gas is directly proportional to number of moles of gas at constant pressure and temperature.
$\frac{V1}{n1} = \frac{V2}{n2}$
where V -volume , n - number of moles
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting these values in the equation
$\frac{11.2 L}{0.500 mol} = \frac{V}{2.00 mol}$
V = 44.8 L
answer is C. 44.8 L

5 0

3 years ago