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abruzzese [7]
3 years ago
9

Which of these statements best explains why the soil around a volcanic region is fertile?

Chemistry
2 answers:
Akimi4 [234]3 years ago
8 0

Answer:

Option (C)

Explanation:

The volcanic eruptions are usually very destructive but it also has some advantages of which, one is that it makes the soil fertile in the nearby region.

During the time of the volcanic eruption, a large number of pyroclastic materials, blocks and ashes that are comprised of various gases are released, which eventually adds to the soil of the nearby areas. These materials ejected from the volcano are rich in various minerals such as oxygen, silicon, and some amount of CO₂, SO₂, and HCl which facilitates the growth of the soil and increases its productivity.

Thus, the soil around a volcanic region is much fertile.

Hence, the correct answer is option (C).

AVprozaik [17]3 years ago
3 0

Answer:

The correct statement is option c, that is, particles discharged in the air by volcanoes fall to the ground and enrich the soil.  

Explanation:

The eruptions of volcanoes lead to the dispersion of ash over the broader regions surrounding the site of eruption. On the basis of the chemistry of the magma, the ash will be comprising different concentrations of soil nutrients. While the major elements found in the magma are oxygen and silica, the eruptions also lead to the discharging of carbon dioxide, water, hydrogen sulfide, sulfur dioxide, and hydrogen chloride.  

In supplementation, the eruptions also discharge bits of rocks like pyroxene, potolivine, amphibole, feldspar that are in turn enriched with magnesium, iron, and potassium. As an outcome, the areas which comprise huge deposits of the volcanic soil are quite fertile.  

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S8 + 24 F2 ⟶ 8 SF6
Arturiano [62]

Answer:

Theoretical Yield of SF₆ = 2.01 moles

Explanation: If you understand and can apply the methodology below, you will find it applies to ALL chemical reaction stoichiometry problems based on the balanced standard equation; i.e., balanced to smallest whole number coefficients.

Solution 1:

Rule => Convert given mass values to moles, solve problem using coefficient ratios. Finish by converting moles to the objective dimensions.

Given      S₈            +          24F₂            =>    8SF₆

             425g                    229g                      ?

= 425g/256g/mol.      = 226g/38g/mol.

= 1.66 moles S₈          = 6.03 moles F₂ <= Limiting Reactant

<em>Determining Limiting Reactant => Divide moles each reactant by their respective coefficient; the smaller value will always be the limiting reactant. </em>

S₈ = 1.66/1 = 1.66

F₂ = 6.03/24 = 0.25 => F₂ is the limiting reactant

<em>Determining Theoretical Yield:</em>

Note: When working problem do not use the division ratio results for determining limiting reactant. Use the moles F₂ calculated from 229 grams F₂ => 6.03 moles F₂. The division procedure to define the smaller value and limiting reactant is just a quick way to find which reactant controls the extent of reaction.  

Given      S₈            +          24F₂            =>    8SF₆

             425g                    229g                      ?

   = 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈          = 6.03 moles F₂ <= Limiting Reactant

<em>Max #moles SF₆ produced from 6.03 moles F₂ and an excess S₈ </em>

Since coefficient values represent moles, the reaction ratio for the above reaction is 24 moles F₂ to 8 moles SF₆. Such implies that the moles of SF₆ (theoretical) calculated from 6.03 moles of F₂ must be a number less than the 6.03 moles F₂ given. This can be calculated by using a ratio of equation coefficients between 24F₂ and 8SF₆  to make the outcome smaller than 6.03. That is,

moles SF₆ = 8/24 x 6.03 moles = 2.01 moles SF₆ (=> theoretical yield)  

S₈ + 24F₂ => 8SF₆

moles SF₆ = 8/24(6.03) moles = 2.01 moles

You would NOT want to use 24/8(6.03) = 18.1 moles which is a value >> 6.03.        

This analysis works for all reaction stoichiometry problems.

Convert to moles => divide by coefficients for LR => solve by mole mole ratios from balanced reaction and moles of given.    

____________________

Here's another example just for grins ...

             C₂H₆O   +   3O₂     =>     2CO₂    + 3H₂O

Given:    253g          307g               ?               ?

a. Determine Limiting Reactant

b. Determine mass in grams of CO₂ & H₂O produced        

Limiting Reactant

moles  C₂H₆O = 253g/46g/mol = 5.5 moles  => 5.5/1 = 5.5

moles  O₂ = 307g/32g/mol = 9.6 moles         =><em>  9.6/24 = 0.4 ∴ O₂ is L.R.</em>

But the problem is worked using the mole values; NOT the number results used to ID the limiting reactant.  

 C₂H₆O   +       3O₂          =>     2CO₂    + 3H₂O

------------ 9.6 mole (L.R.)              ?               ?

mole yield CO₂ = 2/3(9.6)mole = 6.4 mole  (CO₂ coefficient < O₂ coefficient)

mole yield H₂O = 9.6mole  = 9.6mole (coefficients O₂ & CO₂ are same.)

mole used C₂H₆O = 1/3(9.6)mole = 3.2 mole (coefficient  C₂H₆O < coefficient O₂)

For grams => moles x formula weight (g/mole)

7 0
3 years ago
Francine makes several measurements of the mass of a metal block. The data set is shown in the table below.
BARSIC [14]

Answer:

Mean

Explanation:

The mean of a series of measurements is calculated when a<em>ll the measurements are added up and then divided by the number of measurements taken</em>, as follows:

  • Sum of Measurements = 20.73 + 20.76 + 20.68 + 20.75 = 82.92

As<u> there are 4 measurements</u>, the mean is:

  • Mean = 82.92 / 4 = 20.73
3 0
3 years ago
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