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sladkih [1.3K]
2 years ago
7

What mass of oxygen would be produced upon thermal decomposition of 25 g of potassium chlorate (kclo3)? the molecular weight (mw

) of potassium chlorate is 122.5 g/mol?
Chemistry
1 answer:
blagie [28]2 years ago
4 0
The formula for potassium chlorate is:    2KClO₃   →   2KCl   +   3O₂

If moles =  mass ÷ molar mass  

then moles of KClO₃   =  25 g  ÷  [( 39 × 1 ) + ( 35.5 × 1) + ( 16 × 3)] g/mol
                     
                                   = 25 g  ÷  ( 122.5 g ) g/mol

                                   =  0.2041 mol


Now the mole ratio of   KClO₃    :   O₂     is    2   :   3

∴  if moles of KClO₃   =  0.2041 mol

  then moles of O₂     =  ( 0.2041 mol ÷ 2 ) × 3
          
                                  = 0.3061 mol


Now since mass = moles × molar mass

∴ the mass of oxygen produced  =  0.3061 mol  × ( 16 × 2) g/mol
                                                    =  9.796 g


∴ the grams of oxygen gas produced when 25 grams of potassium chlorate is decomposed is ~ 9.80 g.
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The measured voltage of an electrochemical cell consisting of pure nickel immersed in a solution of Ni2+ ions of unknown concent
Verizon [17]

Answer:

[Ni²⁺] = 1.33 M

Explanation:

To do this, we need to use the Nernst equation which (in standard conditions of temperature)

E = E° - RT/nF lnQ

However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:

E = E° - 0.05916/n logQ

As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:

Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)

With this, Q:

Q = [Cd²⁺] / [Ni²⁺]

Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:

Cd(s) ------------> Cd²⁺ + 2e⁻       E°₁ = 0.40 V

Ni²⁺ + 2e⁻ -------------> Ni(s)        E°₂ = -0.25 V

E° = E°₁ + E°₂

E° = 0.40 - 0.25 = 0.15 V

Now that we have all the data, we can solve for the [Ni²⁺]:

0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])

0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])

-0.017 = -0.02958 log(5/[Ni²⁺])

-0.017/-0.02958 = log(5/[Ni²⁺])

0.5747 = log(5/[Ni²⁺])

10^(0.5747) = 5/[Ni²⁺]

[Ni²⁺] = 5/3.7558

<h2>[Ni²⁺] = 1.33 M</h2>
7 0
3 years ago
Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. CCl4(g) + (1/2)O
madam [21]

The question is incomplete, here is the complete question:

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9 at 1,000 K

Calculate Kc for the reaction 2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

<u>Answer:</u> The value of K_c' for the final reaction is 1.936\times 10^{19}

<u>Explanation:</u>

The given chemical equations follows:

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c

We need to calculate the equilibrium constant for the equation, which is:

2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c'=(K_c)^2

We are given:

K_c=4.4\times 10^9

Putting values in above equation, we get:

K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}

Hence, the value of K_c' for the final reaction is 1.936\times 10^{19}

3 0
3 years ago
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Answer:

Gamma decay/radiation

Explanation:

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4 0
3 years ago
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Hence answer is

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7 0
3 years ago
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3 years ago
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