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3 years ago

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A student makes a list of all the soil layers in a temperate region Which layer would be located farthest from the surface of th

e
Earth?

Obedrock (R)

topsoil (A)

organic layer (0)

parent material (C)

1 answer:

Novay_Z [31]3 years ago

5 0

Answer: Bedrock

Explanation: Bedrock is the lowest layer of rock

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The three naturally occurring isotopes of potassium are 39K, 38.963707u; 40K, 39.963999u; and 41K.The percent natural abundances

sweet-ann [11.9K]

Answer:

The isotopic mass of 41K is 40.9574 amu

Explanation:

Step 1: Data given

The isotopes are:

39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%

40K with an isotopic mass of 39.963999u

41K wit natural abundance of 6.7302 %

Average atomic mass =39.098 amu

Step 2: Calculate natural abundance of 40 K

100 % - 93.2581 % - 6.7302 %

100 % = 0.0117 %

Step 3: Calculate isotopic mass of 41K

39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302

39.098 = 36.33681 + 0.0046758 + X * 2.067302

X = 40.9574 amu

The isotopic mass of 41K is 40.9574 amu

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3 years ago

Which of the following statements BEST describes the function of the large ears of an elephant?

Marina CMI [18]

D one defines its best uwu

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3 years ago

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The chemical equation below shows the formation of aluminum oxide (Al2O3) from aluminum (Al) and oxygen (O2). 4Al 3O2 Right arro

Andrew [12]

Formation reaction is the formation of 1 mole of product from the constituents of the reactant molecules. The mass of oxygen that must react is 182 gm.

<h3>What is mass and molar mass?</h3>

Mass of the substance is the weight while the molar mass of the substance is the addition of the atomic mass of the individual mass of the constituent atoms of the compound or the molecule.

The chemical reaction can be shown as:

$\rm 4 Al + 3 O_{2} \rightarrow 2 Al_{2}O_{3}$

From the reaction, it can be said that 3 moles of oxygen are required to produce 2 moles of aluminium oxide, so x moles of oxygen will be required to produce 3.80 moles of aluminium oxide.

Solving for x:

$\begin{aligned}\rm 2x &= 3.80\times 3\\\\\rm x &= \dfrac{11.4}{2}\\\\&= 5.7\;\rm mol\end{aligned}$

If 1 mol of oxygen is 32 gm then 5.7 moles of oxygen will be 182.4 gm.

Therefore, option D. 182 gm is the mass of oxygen required.

Learn more about moles and molar mass here:

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2 years ago

A liquid may reach its boiling point if

EleoNora [17]

A. heat is added to it

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The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

<u>For $N_2$ :</u>

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

<u>For $H_2$ :</u>

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K$

$V=409.9L$

Therefore, the volume of reactant measured at STP left over is 409.9 L

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3 years ago