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Yakvenalex [24]
2 years ago
15

Which contain a common ion that will shift the equilibrium system represented by the equation shown? Select all that apply. MgSO

4 Na2S HNO3 CaCl2.
Chemistry
1 answer:
inn [45]2 years ago
4 0

The compounds in the reaction that shifts the equilibrium with the common ion effect are \rm MgSO_2\;and\;HNO_3.

<h3>What is the common ion effect?</h3>

The equilibrium condition has an equal amount of reactant and products in the reaction.

The addition of ions same as product or reactant results in a change in the equilibrium of condition and shifts the equilibrium in opposite direction.

The given reaction is:

\rm H_2SO_4\;\rightleftharpoons 2H^+\;+\;SO_4^{2-}

The addition of sulfate and hydrogen ions shifts the equilibrium.

Thus, the compounds that shift the equilibrium for the reaction are \rm MgSO_2\;and\;HNO_3.

Learn more about equilibrium, here:

brainly.com/question/4289021

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Why will a reaction occur between calcium oxalate and platinum?
olga2289 [7]

Answer:

Complex formation

Explanation:

At first sight, one may think that a reaction between calcium oxalate and platinum metal is impossible considering the relative positions of the two metals in the electrochemical series.

However, in solution, calcium oxalate dissociates into calcium and oxalate ions. When platinum is added to this solution, a complex is formed between the platinum ions and the oxalate ions. Hence a reaction occurs for this reason.

7 0
3 years ago
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Barium chloride bea CO2 is a binary compound because it contains
Tpy6a [65]

Answer:

Because it contains only two elements.

Explanation:

A binary compound is a compound made up of only two elements. Barium chloride contains:

1. Barium.

2. Chlorine.

Since Barium chloride contains only two elements, it is therefore a binary compound.

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3 years ago
PLEASE HURRY TIME LIMIT WILL GIVE BRAINLIEST A solution is made using 80.1 g of toluene (MM = 92.13 g/mol) and 80.0 g of benzene
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10.9 because of mass= moles/mv it’s just basic chemistry
7 0
3 years ago
The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,
Masja [62]

<u>Answer:</u> The concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

<u>Explanation:</u>

The given cell is:

Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V ( × 3)

<u>Reduction half reaction:</u> Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V ( × 2)

<u>Net reaction:</u> 3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.50-0=1.50V

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}

where,

E_{cell} = electrode potential of the cell = 1.23 V

E^o_{cell} = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[Au^{3+}]=?M

[H^{+}]=1.0M

Putting values in above equation, we get:

1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})

[Au^{3+}]=1.87\times 10^{-14}M

Hence, the concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

7 0
3 years ago
Map
grin007 [14]

Answer: 1650 hope i got it in time.

Explanation:

3 0
2 years ago
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