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astraxan [27]
3 years ago
10

when 6.25 grams of pure iron are allowed to react with oxygen , a black oxide forms. if the product weighs 8.15 g, what is the e

mpirical formula of the oxide
Chemistry
1 answer:
erik [133]3 years ago
5 0
We are given with the mass of  pure iron that reacts with oxygen to form an oxide which has a given mass as well. the mass of oxygen reacted is 8.15-6.25 g or 1.9 grams. THen we convert the mass of the reactants to moles. Iron is equal to 0.1119 moles and oxygen is equal to 0.1188. We divide each number to the less amount. Hence iron is 1 and oxygen is approx 1. The empirical formula hence is FeO or ferrous oxide or Iron (II) oxide.
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A sample of pure NH4HS is placed in a sealed 2.0-L container and heated to 550 K at which the equilibrium constant is 3.5 x 10-3
Zolol [24]

Answer:

The mass of NH_{3} in the container is 2.074 gram

Explanation:

Given:

Volume of NH_{4} HS    V = 2 lit

Equilibrium constant k _{eq} = 3.5 \times 10^{-3}

The reaction in which NH_{3} is produced

  NH_{4} HS ⇄ NH_{3} + H_{2}S

Here equal moles of NH_{3} and H_{2}S is formed.

From the formula of equilibrium constant,

  k_{eq} = (NH_{3})(H_{2}S )

   x^{2} = 3.5 \times 10^{-3}

     x = 0.061 M

Above value shows,

   NH_{3} = 0.061 \frac{moles}{L}

So in 2 L no. moles of NH_{3} = 0.061 \times 2 = 0.122 moles.

So mass of 0.122 mole of NH_{3} is = 0.122 \times 17 = 2.074 g

Therefore, the mass of NH_{3} in the container is 2.074 gram

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3 years ago
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ASAP
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5 0
1 year ago
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

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                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

4 0
3 years ago
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