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dalvyx [7]
3 years ago
5

How do scientific test help determine the properties of a substance

Chemistry
1 answer:
Nataly_w [17]3 years ago
5 0
Data analysis' of what ever they are studying.  <span />
You might be interested in
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
a student determined that 8.2 milligrams of oxygen is dissolved in a 1000 gram sample of water at 15 degrees Celsius and 1 atm w
melamori03 [73]

Answer: We do not know. We have not been given the solubility of oxygen in water at a given temperature nor have we been given the Henry's laws constant. We also do not know whether you mean 1 atmosphere of air, or 1 atmosphere of oxygen.

8 0
3 years ago
S this statement true or false? All worms belong to the same phylum.
telo118 [61]

Answer:

FALSE.

Explanation:

The right answer is false. There are three different groups of phylum for worms.

Hope this helped!

8 0
3 years ago
Read 2 more answers
The temperature versus time graph of a solid substance absorbing heat is shown. A graph is shown with Temperature followed by de
navik [9.2K]

Answer:

The particles of the liquid slide around faster as the kinetic energy of the particles increases.

Explanation:

After all the bonds in the solid state are broken in part CD, the more free particles in the liquid state gain more kinetic energy with increase in energy supplied.

The increase in kinetic energy is indicated by the temperature increase thus the positive gradient of the part CD.

Kinetic energy means more vibrations thus the particles slide more and more against each other.

7 0
3 years ago
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction:
slamgirl [31]

Answer:

1.5 mole

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

Step 2:

Determination of the masses of Al and Cl2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar Mass of Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 from the balanced equation = 3 x 71 = 213g

From the balanced equation,

54g of Al reacted.

213g of Cl2 reacted

Step 3:

Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

54g of Al reacted with 213g of Cl2.

Therefore, 40.5g of Al will react with = (40.5 x 213)/54 = 159.75g of Cl2.

From the calculations made above, there are leftover of Cl2 as 159.75g reacted out of 212.7g. Therefore, Cl2 is the excess reactant and Al is the limiting reactant.

Step 4:

Determination of the number of mole in 40.5g of Al. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al = 40.5g

Number of mole of Al =?

Number of mole = Mass/Molar Mass

Number of mole of Al = 40.5/27

Number of mole of Al = 1.5 mole

Step 5:

Determination of the number of mole of AlCl3 produced When 40.5 g of Al and 212.7 g of Cl2 combine together. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

From the balanced equation above,

2 moles of Al produced 2 moles of AlCl3.

Therefore, 1.5 mole of Al will also produce 1.5 mole of AlCl3.

From the calculations made above, 1.5 mole of AlCl3 is produced When 40.5 g of Al and 212.7 g of Cl2 combine together.

8 0
3 years ago
Read 2 more answers
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