A Metal pan will heat up quickly when placed over a gas or electric burner. The Metal pan is a good what...

Chemistry

1 answer:

Alik [6]3 years ago

8 0

Answer:

heating up, lol

Explanation:

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Write a net ionic equation for the reaction that occurs when nickel(ii carbonate and excess hydrobromic acid (aq are combined.

seropon [69]

First, we write the reaction equation:
NiCO₃ + 2HBr → NiBr₂ + H₂CO₃

Now, writing this in ionic form:
NiCO₃ + 2H⁺ + 2Br⁻ → NiBr₂ + 2H⁺ + CO₃⁻²

(NiCO₃ is insoluble so it does not dissociate in to ions very readily)

Overall equation:
NiCO₃ + 2Br⁻ → NiBr₂ + CO₃⁻²

7 0

3 years ago

A voltaic cell that uses the reaction PdCl2−4(aq)+Cd(s)→Pd(s)+4Cl−(aq)+Cd2+(aq) has a measured standard cell potential of +1.03

Ghella [55]

Answer : The balanced two-half reactions will be,

Oxidation half reaction : $Cd(s)\rightarrow Cd^{2+}(aq)+2e^-$

Reduction half reaction : $Pd^{2+}(aq)+2e^-\rightarrow Pd(s)$

Explanation :

Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

The given redox reaction is:

$PdCl_4^{2-}(aq)+Cd(s)\rightarrow Pd(s)+4Cl^-(aq)+Cd^{2+}(aq)$

The balanced two-half reactions will be,

Oxidation half reaction : $Cd(s)\rightarrow Cd^{2+}(aq)+2e^-$

Reduction half reaction : $Pd^{2+}(aq)+2e^-\rightarrow Pd(s)$

7 0

3 years ago

The metals in Groups 1A, 2A, and 3A ____.

choli [55]

All of these metals from cations, so they lost electrons when forming ions. The answer is D.

3 0

3 years ago

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The element iodine would be expected to form covalent bond(s) in order to obey the octet rule. Use the octet rule to predict the

zhannawk [14.2K]

Answer:

HI (hydrogen iodide).

Explanation:

Hello,

In this case, since iodine is in group VIIA, it has seven valence electrons at its outer shell, which means that it only needs one bond to attain the octet. Moreover, since hydrogen has just one electron at its outer shell, one hydrogen turns out enough to obey the octet when hydrogen and iodine bond as shown below:

$H-I$