All categories

3 years ago

Some soils are made mostly of loess. loess is silt-sized sediment deposited by _______.

2 answers:

8 0

There are soils that are made up of loess. Loess is a sediment that is formed from silt deposits due to wind. The composition of the silt is clay and sand. The components are cemented together by calcium carbonate. The silts are blown by wind and accumulate in a singular location.

Angelina_Jolie [31]3 years ago

5 0

The answer to: Some soils are made mostly of loess. Loess is silt-sized sediment deposited by <u>Wind.</u> The answer is wind

You might be interested in

Let’s assume that you put a balloon into the freezer. Initially,the balloon had 3.0liters of gas at a pressure of 400kPa and was

EastWind [94]

Here, we should use combined gas law which can be derived from combined gas law, “PV=nRT”. Rearranging, we can get PV/T=nR. Then we can set the two states in the problem together to get

P1V1/T1 = P2V2/T2

Then just plug in and solve algebraically.

Hope this helps

6 0

3 years ago

The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151

vitfil [10]

Answer:- The natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for $^1^5^1_E_u$ and $^1^5^3_E_u$ as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of $^1^5^1_E_u$ as n then the abundance of $^1^5^3_E_u$ would be 1-n .

Let's plug in the values in the formula:

$151.964=150.919860(n)+152.921243(1-n)$

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

$151.964-152.921243=150.919860n-152.921243n$

$-0.957243=-2.001383n$

negative sign is on both sides so it is canceled:

$0.957243=2.001383n$

$n=\frac{0.957243}{2.001383}$

$n=0.478$

The abundance of $^1^5^1_E_u$ is 0.478 which is 47.8%.

The abundance of $^1^5^3_E_u$ is = $1-0.478$

= 0.522 which is 52.2%

Hence, the natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

3 0

3 years ago

Choose the correct answer to complete the paragraph about the acceptance of the heliocentric model. In the second century BCE, t

PtichkaEL [24]

Answer:epicycles,copermicus, earth

Explanation:

4 0

3 years ago

Read 2 more answers

54,000,000 nanometers equals 0.000000054 megameters.

LiRa [457]

That is true b<span>ecause 1 nanometer = 1.0 × 10^-15 megameters.</span>

3 0

3 years ago

Read 2 more answers

How many Valance electrons are in an atom of phosphorous

densk [106]

Answer:

Five

Explanation:

Valance electrons are the outermost electrons in an atom. There are five electrons in the third layer of the phosphorus atom. So your answer is five valance electrons.

7 0

3 years ago