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1 year ago

Ca(s) + O₂(g) + S(s) → CaSO4(s)

Chemistry

1 answer:

san4es73 [151]1 year ago

5 0

The mass of Calcium required to complete this reaction is 4.008 g.

Law of conservation of mass states that In a closed system, mass cannot be produced or destroyed, but it can be changed from one form to another.
The mass of the chemical constituents before a chemical reaction is equal to the mass of the constituents after the reaction.
In several disciplines, including chemistry, mechanics, and fluid dynamics, the idea of mass conservation is widely applied.

In the given reaction mass of product after completion of reaction is 13.614 g that means total mass of constituents before reaction should also be 13.614.

So,

mass of Ca + mass of O₂ + mass of S = mass of CaSO4

Ca + 6.400 g + 3.206 g = 13.614 g

mass of Ca = 13.614 - 9.606 = 4.008 g

Therefore, by law of conservation of mass 4.008 g of Ca is required for the completion of the reaction.

Learn more about mass conservation here:

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[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was con

telo118 [61]

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = $284 \ grams$

solid soil volume = $205 \ cc$

saturated mass soil = $361 \ g$

The weight of the soil after drainage is = $295 \ g$

Water weight for soil saturation = $(361-284) = 77 \ g$

Water volume required for soil saturation = $\frac{77}{1} = 77 \ cc$

Sample volume of water: $= \frac{\text{water density}}{\text{water density input}}$

$= 361- 295 \\\\ = 66 \ cc$

Soil water retained volume = (draining field weight - dry soil weight)

$= 295 - 284 \\\\ = 11 \ cc.$

$\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}$

$= \frac{77}{(205 + 77)} \\\\= \frac{77}{(282)} \\\\ = 27.30 \%$

(Its saturated water volume is equal to the volume of voids)

$\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}$

$= \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23$

$\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}$

$= \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04$

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2 years ago

Calculate AH°rxn from AH°7 values (use table in textbook appendix) a) Cl2 (g) + 2 Na (s) -› 2 NaCl (s) EITHER -411.1kJ/mol b) 2

MrMuchimi

To find AH°rxn, we use the following equation:

What we're going to do is to sum the enthalpy of the products and then substract with the enthalpy of the reactives:

As you can see, we need to multiply by the coefficients of the reaction.

Now, just replace the values of the table:

So the answer is -822.2kJ/mol.

For b:

Now, just replace the values of the table:

The answer for b is -1036kJ/mol.

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kherson [118]

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