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3 years ago

What is the hybridization of carbon in nco-?

Chemistry

2 answers:

never [62]3 years ago

5 0

Answer : The hybridization of carbon in $NCO^-$ is, $sp$ hybridized.

Explanation :

As we know that when a carbon attached to the four single sigma bonds then the hybridization will be, $sp^3$ .

When a carbon attached to the three single sigma bonds and one- $\pi$ bond that means a carbon bonded to another atom by the double bond then the hybridization will be, $sp^2$ .

When a carbon attached to the two single sigma bonds and two- $\pi$ bonds that means a carbon bonded to another atom by the triple bond then the hybridization will be, $sp$ .

As per question, the given molecule is, $NCO^-$ . In this molecule, a triple bond is present between the nitrogen and carbon atom and a single bond is present between the carbon and oxygen atom. That means, a carbon bonded to another atom by the triple bond then the hybridization will be, $sp$ .

Hence, the hybridization of carbon in $NCO^-$ is, $sp$ hybridized.

Svet_ta [14]3 years ago

4 0

First let us see what kind of bonds are formed in the compound. By drawing the structure, we see that the kind of bonds are:

N =- triple bond -= C – O

<span>So there is only single bond between C and O therefore the hybridization of C is sp.</span>

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Answer:

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Explanation:

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The nature of zinc powder and Cobalt (II )oxide is heated the following reaction occurs ;

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1) The metal which reduces the other compound is the one higher in the reactivity. So in this case it is $\mathrm{Zn}$ .

2) The substance which brings about reduction while itself getting oxidised (that is losing electrons) is called a reducing agent. Here, $\mathrm{Zn}$ is the reducing agent and reduces Cobalt Oxide to Cobalt while itself getting oxidised to Zinc oxide.

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3 years ago

When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for

Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

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3 years ago

Calculate the weight of 3.491 into 10 to the power 19 molecules of cl2

Rom4ik [11]

Answer:

The mass of 3.491 × 10¹⁹ molecules of Cl₂ of Cl₂ is 4.11 × 10⁻³ grams

Explanation:

The number of particles in one mole of a substance id=s given by the Avogadro's number which is approximately 6.023 × 10²³ particles

Therefore, we have;

One mole of Cl₂ gas, which is a compound, contains 6.023 × 10²³ individual molecules of Cl₂

3.491 × 10¹⁹ molecules of Cl₂ is equivalent to (3.491 × 10¹⁹)/(6.023 × 10²³) = 5.796 × 10⁻⁵ moles of Cl₂

The mass of one mole of Cl₂ = 70.906 g/mol

The mass of 5.796 × 10⁻⁵ moles of Cl₂ = 70.906 × 5.796 × 10^(-5) = 4.11 × 10⁻³ grams

Therefore;

The mass of 3.491 × 10¹⁹ molecules of Cl₂ of Cl₂ = 4.11 × 10⁻³ grams.

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