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1 year ago

What causes soil damage and loss?

Chemistry

2 answers:

timofeeve [1]1 year ago

8 0

Answer:

Mosty Oil but it depends there is lots of things that can damage soil

Explanation:

SSSSS [86.1K]1 year ago

5 0

Answer:

Sheet erosion by water, wind erosion, and rill erosion

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Aluminium sulfate hydrate al2(so4)3.xh2o contains 13.63% al by mass. calculate x, that is, the number of water molecules associa

Advocard [28]

MAl₂(SO₄)₃·xH₂O:
(mAl×2) + (mS×3) + (mO×12) + (mH₂O×x)
(27×2)+(32×3)+(16×12)+(x×18) = 342 + 18x [g]
mAl₂: 27×2 = 54 [g]

54g ---------- 13,63%
342+18x ---- 100%
0,1363(342+18x) = 54
46,6146 + 2,4534x = 54
2,4534x = 7,3854
x ≈ 3

>>> Al₂(SO₄)₃·3H₂O <<<<
:)

8 0

3 years ago

Part 1. determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. show your work.

d1i1m1o1n [39]

The molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol.
If this sample was placed under extreme pressure, the volume of the sample will decrease.

<h3>How to calculate molar mass?</h3>

The molar mass of a substance can be calculated by first calculating the number of moles using ideal gas law equation:

PV = nRT

Where;

P = pressure
V = volume
T = temperature
R = gas law constant
n = no of moles

0.98 × 1.2 = n × 0.0821 × 287

1.18 = 23.56n

n = 1.18/23.56

n = 0.05moles

mole = mass/molar mass

0.05 = 0.458/mm

molar mass = 0.458/0.05

molar mass = 9.15g/mol

Therefore, the molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol
If this sample was placed under extreme pressure, the volume of the sample will decrease.

Learn more about gas law at: brainly.com/question/12667831

8 0

2 years ago

If more solute can be dissolved in a solvent, the solution is

Mekhanik [1.2K]

Answer: B. Unsaturated

Explanation: in a saturated/super saturated solution, more solute will not be able to dissolve.

8 0

3 years ago

Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)

Blizzard [7]

Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?

8 0

3 years ago

The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

Answer: The molecular formula of the compound is $C_4H_{10}O_4$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

Calculating the molecular formula:

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=36.26g$

Mass of $H_2O=14.85g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

4 years ago