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Phoenix [80]
3 years ago
7

Balance the following equations and then write the net ionic equations. (Use the lowest possible whole number coefficients. Incl

ude states-of-matter in your answers. Use instead of to balance equations. Write weak electrolytes in their undissociated form. Omit any coefficients of 1).
A. (NH4)3PO4(aq) + Bi(NO3)3(aq) + BiPO4(s) + NH4NO3(aq).
B. AGOH(s) + H2SO4 (aq) + Ag, SO4(s) + H2O.
C. HCIO2 (aq) + Mn(OH)2(s) + Mn(C102)2 (aq) + H2O.
Chemistry
1 answer:
Fiesta28 [93]3 years ago
8 0

Answer:

Explanation:

A )  (NH₄)₃PO₄(aq) + Bi(NO₃)₃(aq) =  BiPO₄(s) + 3 NH₄NO₃(aq).

B₁³⁺ +   PO₄⁻³  =  Bi ( PO₄ )

B )  AgOH(s) + H₂SO₄ (aq)  =  AgSO₄(s) + H₂O.

Ag⁺  + OH⁻ + H⁺ +  2 SO₄⁻² (aq)  =  Ag(SO₄)₂(s) + H₂O.

C )  HCIO₂ (aq) + Mn(OH)₂(s) =   Mn(Cl0₂)₂ (aq) + H₂O.

     H⁺  + OH⁻  =  H₂O .

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Calculate the standard cell potential at 25 ∘c for the reaction x(s)+2y+(aq)→x2+(aq)+2y(s) where δh∘ = -793 kj and δs∘ = -319 j/
QveST [7]
First we will calculate free energy change:
ΔG₀ = ΔH₀ - (T * ΔS₀)
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We know the relation between free energy change and cell potential is:
ΔG₀ = - n F E⁰ where
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3 years ago
A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of N
vaieri [72.5K]

Answer:

Kc = 8.05x10⁻³

Explanation:

This is the equilibrium:

                 2NH₃(g)   ⇄     N₂(g)     +     3H₂(g)

Initially       0.0733

React         0.0733α          α/2                3/2α

Eq     0.0733 - 0.0733α    α/2                0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

α = 0.103 . 2/3 ⇒ 0.0686

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NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

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So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³

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