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forsale [732]
3 years ago
14

The higher the number of living organisms in an area, the ___ fertile the soil. Less, more.

Chemistry
1 answer:
goldenfox [79]3 years ago
4 0
I think it is the more fertile the soil
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Electrolysis is used in the electroplating of metals. The same amount of current is passed through separate aqueous solutions of
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Answer:

CuSO4 cell will have the greatest amount of deposit among all three. The deposit will occur at the cathode

Explanation:

The valence of the elements in this case is as follows -

Cu - 2e-

Sn - 4e-

Cr - 3e-

CuSO4 cell will have the greatest amount of deposit among all three

The atoms of copper metal will deposit at the cathode. At the cathode, the least number of moles of electrons needed .

Hence, more amount of copper  can be extracted out by the electrolyte

 

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Chemistry Questions: Which of the following is a reasonable ground-state electron configuration?
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D. </span><span>-[Kr] 5s24d105p3</span>
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The electron dot diagram for a neutral atom of chlorine (atomic number 17) is shown below.
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A sample that contains only SrCO3 and BaCO3 weighs 0.846 g. When it is dissolved in excess acid, 0.234 g carbon dioxide is liber
In-s [12.5K]

Answer:28.605

Explanation:First, the molar mass of of SrCO3, BaCO3 and CO2 has to be calculated, (using the molar mass of each element Sr = 87.62, Ba = 137.327, C=12.011, O= 16.00)

The molar masses are;

SrCO3 = 87.62 + 12.011 + (3*16) = 147.631g/mol

BaCO3 = 79.904 + 12.011 + (3*16) = 197.34 g/mol

CO2 = 12.011 + (2*16) = 44.011 g/mol

To obtain one of the equations to solve the problem;

The sample is made of SrCO3 and BaCO3 and has a mass of 0.846 g. Representing the mass of SrCO3 as ma and that of BaCO3 as mb. The first equation can be written as:

ma + mb = 0.846g                 (1)

To obtain another equation in order to be able to determine the different percentages of the compounds (SrCO3 and BaCO3) that make of the sample, a relationship can be obtained by determining the relationship between the number of moles of CO2 formed as the mass of the SrCO3 and BaCO3;

The number of moles of CO2 formed = (mass of CO2)/(molar mass) =0.234/44.011 =0.00532moles

CO2 contains 1 mole of carbon (C) so therefore 0.00532 moles of CO2 contains 0.00532 moles of C

The sample produced 0.00532 moles of CO2, therefore the number of moles SrCO3 and BaCO3 that produced this amount can be calculated using the formula;

= (mass )/(molar mass)

No of moles of SrCO3 and BaCO3 will be ma/147.631 and mb/197.34 moles respectively

The total amount of C molecules produced by SrCO3 and BaCO3 will be 0.00532 moles of C

The second equation can be written as

ma/147.631 + mb/197.34= 0.00532          (2)

Solving Equation (1) and (2) simultaneously;

ma = 0.604g; mb = 0.242g

Therefore the percentage of BaCO3   = (mass of BaCO3 )/(mass of sample )*100

                                                         = 0.242/(0.846 )*100

                                                         = 28.605%

5 0
3 years ago
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