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Arisa [49]
3 years ago
6

Both FeO and Fe2O3 contain only iron and oxygen. The mass ratio of oxygen to iron for each compound is given in the following ta

ble:
Compound mass O : mass Fe
FeO 0.2865
Fe2O3 0.4297
Show that these data are consistent with the law of multiple proportions.
Chemistry
2 answers:
FromTheMoon [43]3 years ago
6 0

Answer:

Explanation:

Mass of O / Mass of Fe = .2865/1  = 112 x .2865/112 = 32.088/112

Mass of O / Mass of Fe = .4297 /1 = 112 x .4297 / 112 = 48 / 112

Ratio of mass of O That reacts with constant mass of Fe of 112 g is as follows

32.088 : 48 = 32/16 : 48/16

= 2 : 3

Hence given data are consistent with the law of multiple proportions.

Finger [1]3 years ago
4 0

Answer:

- For FeO:

\frac{O}{Fe}=\frac{16}{55.8}=0.2865

- For Fe₂O₃:

\frac{O}{Fe}=\frac{16*3}{55.8*2}=0.43

Explanation:

Hello,

The law of multiple proportions states that: "if two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers". In this manner, both FeO and Fe₂O₃ are in such way that they have the small whole numbers in their structures, thus, to compute the required ratios we perform the following mathematical relationships:

- For FeO:

\frac{O}{Fe}=\frac{16}{55.8}=0.2865

- For Fe₂O₃:

\frac{O}{Fe}=\frac{16*3}{55.8*2}=0.43

Wherein such values are consistent with the given data.

Best regards.

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The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol
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Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration PCl_3 and Cl_2.

\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}

\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M

\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}

\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M

The given equilibrium reaction is,

                            PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initially                 0.70        0.70              0

At equilibrium    (0.70-x)   (0.70-x)           x

The expression of K_c will be,

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

K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}

Now put all the given values in the above expression, we get:

49=\frac{(x)}{(0.70-x)\times (0.70-x)}

By solving the term x, we get

x=0.59\text{ and }0.83

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of PCl_3 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of Cl_2 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of PCl_5 at equilibrium = x = 0.59 M

Therefore, the concentration of PCl_3 at equilibrium is 0.11 M

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