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marysya [2.9K]
3 years ago
5

Write a balanced half-cell equation for the reaction occurring at the anode. ignore phases in the reaction.

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
7 0

The balanced half-cell equation for the reaction occurring at the anode is  H2 ---> 2H(+) + 2e(-)

E<u>xplanation:</u>

  • The balanced half-cell equation taking place at the anode is explained below
  • The product produced in the reaction in the fuel cell is water.
  • H2 ---> 2H(+) + 2e(-)
  • In the above reaction, the oxidation state of hydrogen switches from 0 to +1.
  • It is becoming oxidized by delivering two electrons at the anode.
  • In the fuel cell, hydrogen molecules get oxidized to hydronium ions.Thus half-reaction is the oxidation reaction.

<u />

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1093.5 rounded into 3 sig figs
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Significant figures means the number of important figures, 0 not included.

This means that the answer is 1094.00

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A hydrogen-like ion is an ion containing only one electron. The energies of the electron in a hydrogen-like ion are given by the
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Answer:

The ionization energy (in kJ/mol) of the helium ion is 21,004.73 kJ/mol .

Explanation:

E_n = -(2.18 10-18 J)\times \frac{Z^2}{n^2}

Z = atomic mass

n = principal quantum number

Energy of the electron in n=1,

E_1= -(2.18 10^{-18} J)\times \frac{4^2}{1^2}=-3.488\times 10^{-17} J

Energy of the electron in n = ∞

E_{\infty}= -(2.18 10^{-18} J)\times \frac{2^2}{\infty ^2}=0 J

Ionization energy of the He^+ ion:

I.E=E_{infty}-E_1=0-(-3.488\times 10^{-17} J)=3.488\times 10^{-17} J

I.E=3.488\times 10^{-20} kJ

To convert in into kj/mol multiply it with N_A=6.022\times 10^{23} mol^{-1}

I.E=3.488\times 10^{-20} kJ\times 6.022\times 10^{23} mol^{-1}=21,004.73kJ/mol

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3 years ago
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Explanation:

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After substantial heating, 6.25 g of iron produced 18.00 g of a compound with chlorine. The empirical formula is:
babunello [35]

Answer:

Option A. FeCl3

Explanation:

The following data were obtained from the question:

Mass of iron (Fe) = 6.25g

Mass of the compound formed = 18g

From the question, we were told that the compound formed contains chlorine. Therefore the mass of chlorine is obtained as follow

Mass of chlorine (Cl) = Mass of compound formed – Mass of iron.

Mass of chlorine (Cl) = 18 – 6.25

Mass of chlorine (Cl) = 11.75g

The compound therefore contains:

Iron (Fe) = 6.25g

Chlorine (Cl) = 11.75g

The empirical formula for the compound can be obtained by doing the following:

Step 1:

Divide by their molar mass

Fe = 6.25/56 = 0.112

Cl = 11.75/35.5 = 0.331

Step 2:

Divide by the smallest

Fe = 0.112/0.112 = 1

Cl = 0.331/0.112 = 3

The empirical formula for the compound is FeCl3

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3 years ago
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Answer:

They do not

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Gold is more advanced

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