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Sonja [21]
3 years ago
12

If a current-carrying wire is in a magnetic field, in what direction will a force be exerted on the wire?

Chemistry
2 answers:
Nataly_w [17]3 years ago
5 0

Answer : Perpendicular to both the magnetic field and the current direction

Explanation :

If a current-carrying wire is in a magnetic field, the magnetic force exerted on the wire is given by :

F=q(v\times B).............(1)

Where

q is the electric charge

v is the velocity with which the particle is moving

B is the magnetic field

The direction of magnetic force is can be find by using right hand rule.

From equation (1), it is clear that the direction of force exerted on the wire is perpendicular to both the magnetic field and the direction of current.

Hence, the correct option is (d).

OverLord2011 [107]3 years ago
4 0
D.Perpendicular to both the magnetic field and the current direction
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∴  R = 8.314 J/K.mol

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4 years ago
The reaction below demonstrates which characteristic of a base?
svlad2 [7]

Answer: option D. the ability of a base to react with a soluble metal salt.


Justification:


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On the other hand, CuSO₄ is a soluble ionic salt which in water will dissociate into its ions according to this other reaction:

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Hence, in solution, the sodium ion (Na⁺) will  react with the metal salt in a double replacement reaction, where the highly reactive sodium ion (Na⁺) will substitute the Cu²⁺ in the CuSO₄ to form the sodium sulfate salt, Na₂SO₄ (water soluble), and the copper(II) hydroxide, Cu(OH)₂ (insoluble).


That is what the given reaction represents:


  CuSO₄ (aq)     +     2NaOH(aq)    →    Cu(OH)₂(s)       +     Na₂SO₄(aq)

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4 years ago
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Andrej [43]
<h3><u>moles of H2SO4</u></h3>

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Note the way this works is to make sure the units are going to give us moles. To check, we do division of the units just like we were dividing two fractions:

(molecules of H2SO4) = (molecules of H2SO4)/1 and so we have 3.4 x 1023/6.022 × 1023 [(molecules of H2SO4)/1]/[(molecules of H2SO4)/(moles of H2SO4)]. Now, invert the denominator and multiply:

<h3 />
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Answer:

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