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Ira Lisetskai [31]
3 years ago
5

A binary compound created by reaction of bismuth and an unknown element E contains 52.07% Bi and 47.93% E by mass. If the formul

a of the compound is Bi2E3, calculate the atomic mass of E.
Chemistry
1 answer:
yulyashka [42]3 years ago
6 0

Answer:

Atomic mass of E is 128.24

Explanation:

  • The percentage composition by mass of an element in a compound is given by dividing the mass of the element by the total mass of the compound and expressing it as a percentage.
  • In this case; the compound Bi₂E₃

Percentage composition of bismuth = 52.07%

Percentage composition of E = 47.93%

Mass Bismuth in the compound is (2×208.9804) = 417.96 g

Therefore,

To calculate the atomic mass of E

52.07% = 417.96 g

47.93% = ?

            = (47.93 × 417.96 ) ÷ 52.07 %

            = 384.729

         E₃ = 384.729

Therefore; E = 384.729 ÷ 3

                     = 128.24  

The atomic mass of E is 128.24

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Explanation:

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3 years ago
In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
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Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

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What volume will 0.405 g of krypton gas occupy at STP?
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Answer:

The answer to your question is V = 0.108 L or 108 ml

Explanation:

Data

Volume = ?

mass = 0.405 g

Temperature = 273°K

Pressure = 1 atm

Process

1.- Convert mass of Kr to moles

                  83.8 g of Kr -------------------- 1 mol

                     0.405 g     -------------------  x

                     x = (0.405 x 1) / 83.8

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