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ollegr [7]
4 years ago
6

For the chemical reaction , I identify the reactant and the products .

Chemistry
1 answer:
svetlana [45]4 years ago
7 0

The reactant is Mercury (II) Oxide while the products are Mercury and Oxygen separately.

This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.

Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.

Hope this helps!

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Alla [95]

Answer:

Yes

Explanation:

As long as its a solid, you can count it. It will be hard, but possible.

5 0
3 years ago
Zinc reacts with excess hydrochloric acid to produce zinc chloride and hydrogen gas. What happens to the reaction rate if you in
agasfer [191]
The correct response I believe is D. The reaction rate increases because the probability of collisions increases as there are more Zn atoms to react.
4 0
4 years ago
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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
notsponge [240]

<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at 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}=-0.136-(-0.76)=0.624V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

where,

E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

[Sn^{2+}] = ?

Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

3 0
3 years ago
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jekas [21]
D is the correct answer... if u need in depth let me know
3 0
3 years ago
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Neporo4naja [7]

Answer:

The principle of superposition states that the oldest rock units are at rock bottom , and therefore the youngest are at the highest . Based on this, layer C is oldest, followed by B and A. So the full sequence of events is as follows: Layer C formed.

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