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dolphi86 [110]
2 years ago
14

Commercial agriculture can often lead to water-quality problems. In one to two sentences, explain how two of those problems occu

r.(2 points)
Chemistry
1 answer:
anygoal [31]2 years ago
4 0

Commercial agriculture can often lead to water-quality problems in the following ways:

  • The washing of fertilizers and pesticides into water bodies from farms.

<h3>What is water quality?</h3>

Water quality refers to the state of a water body that encompasses it's physical, chemical and biological characteristics.

The water quality of a water body is crucial to its suitability for domestic or drinking purpose.

Commercial agriculture greatly affects water quality in the following ways:

  • The washing of fertilizers and pesticides into water bodies from farms.

Learn more about water quality at: brainly.com/question/20848502

#SPJ1

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In the reaction n2 + 3h2 ---&gt; 2nh3, how many grams of nh3 are produced if 25.0 g n2 reacts with excess h2? question 10 option
Lisa [10]
Molar mass of N2 = 28

Moles of N2 = 25 / 28 = 0.89

So, moles of NH3 produce = 2 x 0.89 = 1.78 

Note: H2 is in excess. so no need to care about it. 
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3 years ago
Propose a 3-step reaction for the conversion of cyclopentene to propanedioic acid
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Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha
elixir [45]

Answer:

0.269 g

Explanation:

Po_{210} ^{84}  ⟶  Pb_{206} ^{82}+  + He_{4} ^{2}

Data:

Half-life of  Po_{210} ^{84}  (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol  

Time = 338.8 days  

Isotopic masses  

Po_{210} ^{84} = 209.98 g/mol  

Pb_{206} ^{82} = 205.97 g/mol  

Concepts  

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.  

The radioactive decay law is  

N=Noe^(-λt)

Where:  

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days  

λ= radioactive decay constant  

The radioactive constant is related to the half-life by the next equation  

λ= \frac{ln 2}{t(1/2)}

so  

λ= \frac{ln2}{138.4 days}  =0,005008 days^(-1)

No (Atoms of  Po_{210} ^{84}  in t=0)

To get No we need to calculate the number of atoms of  Po_{210} ^{84}   in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of  Po_{210} ^{84}  in the initial sample.  

This is:

\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol} = 0,001599 mol of  PoCl4

This is the number of mol of  Po_{210} ^{84} in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23  

0,001599 mol of  Po_{210} ^{84} * 6.023*10^23 atoms/ mol of  Po_{210} ^{84} = 9.632 *10^20 atoms of  Po_{210} ^{84}

Atoms after 338.8 days

We use the radioactive decay law to get this value  

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of  Po_{210} ^{84} in the sample after 338.8 days has passed  

The number of atoms  Po_{210} ^{84} transformed is equal to the number of atoms of Pb_{206} ^{82}  produced.  

The number of atoms of Po_{210} ^{84} transformed is No - N  

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of Pb_{206} ^{82} produced  

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of Pb_{206} ^{82} ) / ( 6.023*10^23 atoms of Pb_{206} ^{82} / mol of Pb_{206} ^{82} =  0.001306 moles of Pb_{206} ^{82}

This number of moles have a mass of:

(0,001306 moles of Pb_{206} ^{82} )* (205.97 g of Pb_{206} ^{82} /mol of Pb_{206} ^{82} ) = 0.269 g  

3 0
3 years ago
A solution is made by mixing equal masses of methanol, CH 4 O , and ethanol, C 2 H 6 O . Determine the mole fraction of each com
Fynjy0 [20]

Answer:

Mole fraction of CH_4O = 0.58

Mole fraction of C_2H_6O = 0.42

Explanation:

Let the mass of CH_4O and C_2H_6O = x g

Molar mass of CH_4O = 33.035 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles_{CH_4O}= \frac{x\ g}{33.035\ g/mol}

Moles_{CH_4O}=\frac{x}{33.035}\ mol

Molar mass of C_2H_6O = 46.07 g/mol

Thus,

Moles= \frac{x\ g}{46.07\ g/mol}

Moles_{C_2H_6O}=\frac{x}{46.07}\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ CH_4O=\frac {n_{CH_4O}}{n_{CH_4O}+n_{C_2H_6O}}

Mole\ fraction\ of\ CH_4O=\frac{\frac{x}{33.035}}{\frac{x}{33.035}+\frac{x}{46.07}}=0.58

Mole fraction of C_2H_6O = 1 - 0.58 = 0.42

6 0
3 years ago
Which of the following states that an orbital can contain two electrons only if all other orbitals at that sublevel contain at l
Vesnalui [34]
Isn't that Hund's rule????

5 0
2 years ago
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