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Klio2033 [76]
2 years ago
9

3. What's the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen?

Chemistry
1 answer:
Hunter-Best [27]2 years ago
8 0

Answer:

Empirical formula is  Li₂CO₃.

Explanation:

Percentage of oxygen= 65.0%

Percentage of lithium = 18.7%

Percentage of carbon= 16.3%

Empirical formula = ?

Solution:

Number of gram atoms of C = 16.3/12 = 1.4

Number of gram atoms of Li = 18.7/6.94 = 2.7

Number of gram atoms of O = 65.0/ 16 = 4.1

Atomic ratio:

Li              :            C          :    O

2.7/1.4      :       1.4/1.4         :   4.1/1.4

     2          :            1           :     3  

Li : C : O = 2 : 1 : 3

Empirical formula is  Li₂CO₃.

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Explanation:

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2 years ago
A laboratory manual gave precise instructions for carrying out the reaction described by the chemical equation, C2H6O + O2 C2H4O
Alexxandr [17]
C2H6O + O2 ---> C2H4O2 + H2O

using the molar masses:-
24+ 6 + 16 g of C2H6O  produces 24 + 4 + 32 g C2H4O2    (theoretical)
 46 g produces 60g 

60 g C2H4O2 is produced from 46g C2H6O
1g      .     .................................46/60 g 
 700g     .................................    (46/60) * 700  Theoretically

But as the yield is only 7.5% 

the required amount is    ((46/60) * 700 ) / 0.075 =  7155.56 g 

=  7.156 kg to nearest gram.  Answer




8 0
2 years ago
A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.
Sonja [21]

Answer:

Equilibrium constant for PCl_5 is 0.5

Equilibrium constant for decomposition of NO_2 is 1.79 \times 10^{-14}

Explanation:

PCl_5 dissociates as follows:

                    PCl_5 \rightleftharpoons PCl_3+Cl_2

initial          0.72 mol     0         0

at eq.     0.72 - 0.40   0.40      0.40

Expression for the equilibrium constant is as follows:

k=\frac{[PCl_3][Cl_2]}{[PCl_5]}

Substitute the values in the above formula to calculate equilibrium constant as follows:

k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5

Therefore, equilibrium constant for PCl_5 is 0.5

Now calculate the equilibrium constant for decomposition of  NO_2

It is given that 3.3 \times 10^{-3} \% is decomposed.

NO_2 decomposes as follows:

                                  2NO_2 \rightleftharpoons 2NO + O_2

initial                            1.0 M       0           0

at eq. concentration of  NO_2   is:

[NO_2]_{eq}=1-(0.000066) = 0.999934\ M

[NO]_{eq}=6.6 \times 10^{-5}\ M

[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M      

Expression for equilibrium constant is as follows:

K=\frac{[NO]^2[O_2]}{[NO_2]^2}

Substitute the values in the above expression

K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}

Equilibrium constant for decomposition of NO_2 is 1.79 \times 10^{-14}

8 0
2 years ago
When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction, 2 ch3oh(
umka21 [38]
The balanced equation for combustion is as follows;
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
The stoichiometry of CH₃OH to O₂ is 2:3
the limiting reagent is the reactant that is fully consumed during the reaction. The amount of product formed is directly proportional to the amount of limiting reactant produced. The excess reagent is the reactant that is provided in excess and is not fully used up, there will be an amount of this reagent remaining after the reaction.
If methanol is the limiting reactant,
If 2 mol of methanol reacts with 3 moles of O₂
Then 24 mol of methanol reacts with - 3/2 x 24 = 36 mol of O₂ should be present
But only 15 mol of O₂ is present, therefore O₂ is the limiting reactant and methanol is in excess.
3 mol of O₂ reacts with 2 mol of CH₃OH 
then 15 mol of O₂ reacts with 2/3 x 15 = 10 mol of CH₃OH

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How could installing new technology, such as scrubber machines, affect the factories required to install them? Name a positive a
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