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1. 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
2. CH₄ + 2O₂ → CO₂ + 2H₂O
3. Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag
4. MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
5. Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃
1)

2)
CuSO_4+Cu_2Cl_2\neq>

Isotopes of any given factor all incorporate the equal variety of protons, so they have the identical atomic wide variety (for example, the atomic wide variety of helium is usually 2). Isotopes of a given factor include exceptional numbers of neutrons, therefore, special isotopes have special mass numbers.
Answer:
See attached figure.
Explanation:
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In this case, according to the given substances, we recall the concept of Lewis structure as such showing the bonds and valence electrons each atom has in the molecule. Thus, since chlorine atoms have seven valence electrons, carbon atoms four of them, hydrogen atoms have 1 and oxygen atoms 6, we are able to draw such Lewis dot structures, by obeying the octet as shown on the attached figure.
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