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3 years ago

When an electric field is applied to a metallic crystal, the movement of electrons is?

1 answer:

4 0

When an electric field is applied to a metallic crystal, the movement of electrons is random. Metals form giant structures in which electrons in the outer shells of the metal atoms are free to move. Hope this answers the question. Have a nice day.

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If the strength of the magnetic field at A is 10 units, the strength of the magnetic field at B is _____. 640 units 6.4 units 2.

dezoksy [38]

Answer : Option C) 2.5 units

Explanation : For calculating the strength of a unknown magnetic field in comparison with the given magnetic field we can use the magnetic formula as;

$\frac{ S_{1}}{S_{2}} = \frac{ (D_{1}^{2}) }{(D_{2}^{2})} .$

On substituting the values we get;

$\frac{10}{x} = \frac{4^{2}}{2^{2}}$ ;

On solving we get the magnetic B field strength as 2.5 units

4 0

2 years ago

Read 2 more answers

Cyclopropane is converted to propene in a first–order process with a rate constant of 5.4 × 10–2 hour–1 . The initial concentrat

Hitman42 [59]

Answer:

The concentration of cyclopropane after 22.0 hour is 0.0457 M.

Explanation:

Conversion of cyclopropane into propene follows first order kinetics.

The integrated rate of first order kinetic is given by :

$[A]=[A_o]\times e^{-kt}$

$[A_o]$ = Initial concentration of reactant

$[A]$ = final concentration of reactant after time t

k = rate constant of the reaction

We have :

Rate constant of the reaction = k = $5.4\times 10^{-2} hour^{-1}$

$[A_o]=0.150 M$

t = 22.0 hour

[A] =?

$[A]=0.150 M\times e^{-5.4\times 10^{-2} hour^{-1}\times 22.hour}$

$[A]=0.0457 M$

The concentration of cyclopropane after 22.0 hour is 0.0457 M.

3 0

3 years ago

For+the+reaction+H2+++I2+-+2HI+the+equilibrium+constant,+kc+is+49+at+a+fixed+temperature.+Two+mole+of+hydrogen+and+two+moles+of+

Sonja [21]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M respectively.

Solution : Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially 0.11 0.11 C

At equilibrium (0.11-x) (0.11-x) (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M = (0.11-x)

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0

3 years ago

You're provided with a bottle labled [CoCl2.6H2O] = 0.056 M in 4.60 M HCl. You heat a small volume of the solution in a hot wate

sweet [91]

Answer: Equilibrium concentration of $[Cl^-]$ at $50^0C$ is 4.538 M

Explanation:

Initial concentration of $CoCl_2$ = 0.056 M

Initial concentration of $Cl^-$ = 4.60 M

The given balanced equilibrium reaction is,

$COCl_2+2Cl^-\rightleftharpoons [CoCl_4]^{2-}+6H_2O$

Initial conc. 0.056 M 4.60 M 0 M 0 M

At eqm. conc. (0.056-x) M (4.60-2x) M (x) M (6x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CoCl_4]^{2-}\times [H_2O]^6}{[CoCl_2]^2\times [Cl^-]^2}$

Given : equilibrium concentration of $[CoCl_4]^{2-}$ =x = 0.031 M

Concentration of $Cl^-$ = (4.60-2x) M = $(4.60-2\times 0.031)$ =4.538 M

Thus equilibrium concentration of $[Cl^-]$ at $50^0C$ is 4.538 M

8 0

3 years ago

how many calories of energy are given off to lower the temperature of 100.0g of iron from 150.0°C to 35.0°C?

jeyben [28]

Answer:

ΔT = Tfinal − Tinitial = 150°C − 35.0°C = 125°C

given the specific heat of iron as 0.108 cal/g·°C

heat=(100.0 g)(0.108 cal /g· °C )(125°C) =

100x 0.108x125= 1350 cal

4 0

3 years ago