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3 years ago

Which of the following is the flow of electrons through a wire or a conductor

2 answers:

german3 years ago

3 0

The answer is the direct current may flow in a conductor such as a wire, but can also flow through semiconductors insulators or even through a vacuum as in electron or ion beams.

masha68 [24]3 years ago

3 0

Answer:

Electric current.

Explanation:

Hello,

Electric current is defined to exist when there is a net flow of electric charge through a space, in this case the wire (circuits) or the conductor material. Remember that its SI units are the amperes (A) as well as it is measured using the ammeter

Best regards.

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What function do mirrors serve in reflecting telescopes????

UkoKoshka [18]

Enlarge image i think

4 0

3 years ago

Spread Spectrum (SS) encoding: a) Name four spread spectrum techniques. b) Which one of them was originally devised against eave

azamat

Explanation:

( a )

The four types of spread spectrum techniques are as follows -

1. Direct sequence spread spectrum .

2. frequency hopping spread spectrum .

3. chirp spread spectrum .

4. time hopping spread spectrum .

( b )

The Direct sequence spread spectrum was devised for eavesdropping in the military .

In the field of telecommunications , the Direct sequence spread spectrum , it is the technique of spread spectrum modulation which is used to reduce the overall inference of the signal .

4 0

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.

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

Everything checks. The balanced equation is

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

7 0

3 years ago

X) = 3x - 13 g(x) = 2x2 - 4x-5 {x) = -4-7 6. Find (f- g(x).

Nastasia [14]

Answer:

sorry

Explanation:

don know

8 0

3 years ago

Hich one of the following statements about orbital hybridization is incorrect?

photoshop1234 [79]

Answer : The incorrect statement is, (C) The nitrogen atom in NH₃ is sp² hybridized.

Explanation :

Formula used :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(a) The given molecule is, $CH_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(b) The given molecule is, $CO_2$

$\text{Number of electrons}=\frac{1}{2}\times [4]=2$

The number of electron pair are 2 that means the hybridization will be $sp$ and the electronic geometry of the molecule will be linear.

(c) The given molecule is, $NH_3$

$\text{Number of electrons}=\frac{1}{2}\times [5+3]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are 3 atoms around the central nitrogen atom, the fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be trigonal pyramidal.

(d) sp² hybrid orbitals are coplanar, and at 120° to each other.

The sp² hybrid orbitals has trigonal planar geometry and are in the same plane that means coplanar. Thus, the bond angle is 120° to each other.

(e) sp hybrid orbitals lie at 180° to each other.

The sp hybrid orbitals has linear geometry. Thus, the bond angle is 180° to each other.

Hence, the incorrect statement is, (C) The nitrogen atom in NH₃ is sp² hybridized.

4 0

3 years ago