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3 years ago

What is the formula for the polyatomic ion and name the following compoundSn(NO2)2

2 answers:

8 0

Ok so i was writing a big àss explanation but i realized im not that good at explaining as i thought i was so plz just watch these vids by tyler dewitt. my awesome teacher recommended these videos and tyler dewitt's also a really good teacher and if you watch his vids you'll really master this topic:

Writing Ionic Formula: https://youtu.be/URc75hoKGLY

Naming Ionic Compounds: https://youtu.be/mKo72RnN37E

Writing Formulas with Polyatomic Ions: https://youtu.be/p9iQ5Qn42DM

Writing Ionic Formulas with Transition Metals: https://youtu.be/H4nfE8gRX7Q

Naming Ionic Compounds with Transition Metals Introduction: https://youtu.be/H4nfE8gRX7Q

*btw the pic i attached will be very useful to you except when youre dealing with transition metals and also these links will also help you. the professor dave explains videos about naming ionic compounds just didnt help me but tyler dewitt really did so if you watch his videos you'll be fine

garri49 [273]3 years ago

3 0

There are 9 polyatomic ions so which one and the compound “Sn(NO2)2 is Tin(ll) Nitrite

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How is the periodic law demonstrated in halogens

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3 years ago

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2 years ago

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a titration of 25.0mL of an x M hcl solution with 0.15M NaOH starts at a burette reading for NaOH of 0.20mL . Thr burette readin

VARVARA [1.3K]

Burette is a very accurate measuring instrument when adding solutions and has a measurement error of 0.05 mL.
Small volumes of solutions can be transferred from the burette at a controllable rate.
In this instance NaOH is in the burette.
Initial reading of NaOH is 0.20 mL
end point is the point at which the chemical reaction reaches completion. In acid base reactions, end point is when all the H⁺ ions have reacted with OH⁻ ions.
final reading of NaOH is 24.10 mL
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3 years ago

Please help me equalize: PbO2 + MnSO4 + HNO3 = HMnO4 + PbSO4 + Pb(NO3)2 + H2O

Feliz [49]

Answer:

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄+ 2H₂O

Explanation:

PbO₂ + MnSO₄ + HNO₃ ⟶ HMnO₄ + PbSO₄ + Pb(NO₃)₂ + H₂O

It will be easiest to balance this equation by the ion-electron method.

1. Write the ionic equation

PbO₂ + Mn²⁺ + SO₄²⁻ + H⁺ + NO₃⁻ ⟶ H⁺ + MnO₄⁻ + Pb²⁺ + SO₄²⁻ + Pb²⁺ + NO₃⁻ + H₂O

2. Eliminate H⁺, H₂O, and spectator ions

PbO₂ + Mn²⁺ ⟶ MnO₄⁻ + Pb²⁺

3. Separate the skeleton equation into two half-reactions.

PbO₂ ⟶ Pb²⁺

Mn²⁺ ⟶ MnO₄⁻

4. Balance all atoms other than H and O

Done

5. Balance O by adding water molecules to the deficient side

PbO₂ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻

6. Balance H by adding H⁺ ions to the deficient side.

PbO₂+ 4H⁺ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺

7. Balance charge by adding electrons to the deficient side.

PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e-

8. Multiply each half-reaction by a number to equalize the electrons transferred.

5 × [PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O]

2 × [Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e⁻]

9. Add the two half-reactions.

5PbO₂+ 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 10H₂O

2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

10. Cancel species that occur on each side of the equation

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

becomes

5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O

11. Add the missing spectator ions

5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O

+ 2SO₄²⁻ + 4NO₃⁻ + 2H⁺ + 2NO₃⁻ +2SO₄²⁻ + 6NO₃⁻ + 2H⁺

becomes

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

12. Check that all atoms are balanced.

$\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{Pb} & 5 & 5\\\text{O} & 36 & 36\\\text{S} & 2 & 2\\\text{H} & 6 & 6\\\text{N} & 6 & 6\\\end{array}$