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Ainat [17]
4 years ago
11

What will happen if an electric current is flowing through a cord and a magnet is brought near the cord?

Chemistry
1 answer:
Nostrana [21]4 years ago
4 0

Answer:

The magnet will exert a force on the electric current

Explanation:

The electric current flowing through the cord will experience magnetic force due to presence of magnetic field created by the magnet brought near it.

The magnitude of the magnetic force experienced by this flowing current is given as;

F = qvB

where;

q is charge of electron (measured in coulomb's, C)

v is the speed of the flowing current (m/s)

B is the magnetic field strength (T)

Therefore, the magnet will exert a force on the electric current

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A sample containing KBr (M 119&amp;/mol) and KNO3 (M 101 g/mol) was analysed by using the Mohr method, A 1.1250-g sample is diss
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Explanation:

(a)   As we know that relation between normality and volume of two solutions is as follows.

               N_{1}V_{1} = N_{2}V_{2}

As the given data is as follows.

             N_{1} = ?,              N_{2} = 0.050 N

              V_{1} = 20 ml,      V_{2} = 16.25 ml

Therefore, putting the given values into the above formula as follows.

                N_{1}V_{1} = N_{2}V_{2}

                N_{1} \times 20 ml = 0.050 N \times 16.25 ml

                N_{1} = 0.0406 N

Hence, normality of solution A is 0.0406 N.

(b)   As gram equivalent weight of KBr is 119 g/mol.

Formula to calculate normality is as follows.

              Normality = \frac{weight}{gm equi. wt} \times \frac{1000}{V_{ml}}

As volume is 200 ml, normality is 0.0406 N, and gram equi. wt is 119 g/mol. Therefore, putting these values into the above formula as follows.

          Normality = \frac{weight}{gm equi. wt} \times \frac{1000}{V_{ml}}      

          0.0406 N  = \frac{weight}{119 g/mol} \times \frac{1000}{200 ml}

              weight = 0.9668 g

As 0.9668 g of KBr is present in 1.1250 g sample. Hence, weight percentage will be calculated as follows.

                   Weight % of KBr = \frac{0.9668 g}{1.1250 g}

                                               = 85.94%

Thus, we can conclude that C% w/w of KBr in the original sample is 85.94%.

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