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liubo4ka [24]
3 years ago
12

A magnet moved near a coil of wire can cause a(n)

Chemistry
2 answers:
Brilliant_brown [7]3 years ago
8 0

Answer:

A magnet moved near a coil of wire can cause a(n)  electric current to flow. (The magnet's magnetic field also moves.)

Explanation:

The answer is A.

Phantasy [73]3 years ago
7 0
When a magnet moves near a coil of wire it can cause an A. electric current
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The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other plati
Alinara [238K]

Here we have to identify the metal powder by the given disposal.

The identification of Zinc can be done by 1 m nitric acid (HNO₃) and Ni(NO₃)₂ which will produce hydrogen gas by reaction and displacement reaction as shown below.

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

The identification of lead can be done by the reaction with 1 m nitric acid (HNO₃) which produces lead nitrate.

The reaction is-

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

The identification of platinum can be done by the reaction with all the given disposal as it will not react with any of the compound.

1. Identification of Zinc (Zn):

(a) Zn metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than zinc.

Zn + NaNO₃ = No reaction

(b) Zn metal will react with 1 m HNO₃ to form hydrogen gas. The reaction is:

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

(c) Zn will react with nickel nitrate [Ni(NO₃)₂] because it may only cause displacement reaction the reduction potential of Zn²⁺/Zn (-0.76) is less than that of Ni²⁺/Ni (-0.23). Thus the reaction will be:

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

2. Identification of lead (Pb):

(a) Pb metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pb.

Pb + NaNO₃ = No reaction

(b) Pb reacts with HNO₃ to form lead nitrate. The reaction is:

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

(c) The standard reduction potential of Pb²⁺/Pb is more than nickel Ni²⁺/Ni thus there will be no reaction between Pb and NI(NO₃)₂.

Pb + Ni(NO₃)₂ = No reaction.

3. Identification of platinum (Pt)

(a) Pt metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pt.

Pt + NaNO₃ = No reaction.

(b) The standard reduction potential of Pt²⁺/Pt is so high (+1.188) thus there will be no reaction with HNO₃.

Pt + HNO₃ = No reaction

(c) The standard reduction potential of Pt²⁺/Pt is more than nickel Ni²⁺/Ni thus there will be no reaction between Pt and Ni(NO₃)₂.

Pt +  Ni(NO₃)₂ = No reaction.    

8 0
3 years ago
There are three bulbs powered by a battery. What will happen if the middle light burns out.
Vadim26 [7]

Answer:

nuclear war ;P

Explanation:

8 0
3 years ago
Need help answering
svp [43]

Answer:

Your answer would be E

Explanation:

4 0
2 years ago
PLEASE HELP
DENIUS [597]

I believe that it is B

3 0
3 years ago
Write the complete ionic equations, spectator ions and net ionic equation for the following.
gizmo_the_mogwai [7]

Answer:

Explanation:

1) ZnBr₂ (aq) + AgNO₃ (aq)

Chemical equation:

 ZnBr₂ (aq) + AgNO₃ (aq)  →Zn(NO₃)₂(aq) + AgBr(s)

Balanced chemical equation:

ZnBr₂ (aq) + 2AgNO₃ (aq)  →Zn(NO₃)₂(aq) + 2AgBr(s)

Ionic equation:

Zn²⁺(aq) + Br₂²⁻ (aq) + 2Ag⁺ (aq)+ 2NO⁻₃ (aq)  → Zn²⁺(aq) +(NO₃)₂²⁻(aq) + 2AgBr(s)

Net ionic equation:

Br₂²⁻ (aq) + 2Ag⁺ (aq)   →    2AgBr(s)

The Zn²⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The AgBr can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

2) HgCl₂ (aq) + KI (aq)  →

Chemical equation:

HgCl₂ (aq) + KI (aq)  → KCl + HgI₂

Balanced chemical equation:

HgCl₂ (aq) + 2KI (aq)  → 2KCl(aq) + HgI₂(s)

Ionic equation:

Hg²⁺(aq)  + Cl₂²⁻  (aq) + 2K⁺(aq) + 2I⁻ (aq)  →  HgI₂ (s) + 2K⁺(aq) + 2Cl⁻ (aq)

Net ionic equation:

Hg²⁺(aq)  + 2I⁻ (aq) →   HgI₂ (s)

The Cl⁻ ((aq)  and K⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The HgI₂ (s) can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

 

3) Ca(OH)₂ (aq) + Na₂SO₄ (aq)

Chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + NaOH(aq)

Balanced chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + 2NaOH(aq)

Ionic equation:

Ca²⁺(aq)  + OH₂²⁻  (aq) + 2Na⁺(aq) + SO₄²⁻ (aq)  →   CaSO₄(s) + 2Na⁺(aq) + 2OH⁻ (aq)

Net ionic equation:

Ca²⁺(aq)   + SO₄²⁻ (aq)  →   CaSO₄(s)

The OH⁻ ((aq)  and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The CaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

4 0
3 years ago
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