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weqwewe [10]
3 years ago
5

Which type of mixture can be separated using magnetism?

Chemistry
1 answer:
Sergeu [11.5K]3 years ago
5 0

Answer: The correct option is heterogeneous mixture whose components are attracted differently to a magnet.

Explanation: There are two types of mixtures:

1) Homogeneous mixtures: In these mixtures, the particles are uniformly distributed throughout the mixture. These particles cannot be separated.

2) Heterogeneous Mixtures: These are the mixtures where the particles are visible separated and are not-uniformly distributed. These particles can be separated easily.

If magnet is used to separate the components of a mixture, the heterogeneous mixtures will only get separated.

To separate the components by a magnet, the components of a mixture should attract the magnet differently. One component should attract the magnet and another should not. Hence, they can be easily separated.

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A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solut
leonid [27]

Answer:

x(t) = −39e

−0.03t + 40.

Explanation:

Let V (t) be the volume of solution (water and

nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid

measured in liters after t minutes, and let c(t) be the concentration (by volume) of

nitric acid in solution after t minutes.

The volume of solution V (t) doesn’t change over time since the inflow and outflow

of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is

c(t) = x(t)

V (t)

=

x(t)

200

.

We model this problem as

dx

dt = I(t) − O(t),

where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,

both measured in liters of nitric acid per minute. The input rate is

I(t) = 6 Lsol.

1 min

·

20 Lnit.

100 Lsol.

=

120 Lnit.

100 min

= 1.2 Lnit./min.

The output rate is

O(t) = (6 Lsol./min)c(t) = 6 Lsol.

1 min

·

x(t) Lnit.

200 Lsol.

=

3x(t) Lnit.

100 min

= 0.03 x(t) Lnit./min.

The equation is then

dx

dt = 1.2 − 0.03x,

or

dx

dt + 0.03x = 1.2, (1)

which is a linear equation. The initial condition condition is found in the following

way:

c(0) = 0.5% = 5 Lnit.

1000 Lsol.

=

x(0) Lnit.

200 Lsol.

.

Thus x(0) = 1.

In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is

µ(t) = exp Z

P(t) dt

= exp

0.03 Z

dt

= e

0.03t

.

The solution is

x(t) = 1

µ(t)

Z

µ(t)Q(t) dt + C

= Ce−0.03t + 1.2e

−0.03t

Z

e

0.03t

dt

= Ce−0.03t +

1.2

0.03

e

−0.03t

e

0.03t

= Ce−0.03t +

1.2

0.03

= Ce−0.03t + 40.

The constant is found using x(t) = 1:

x(0) = Ce−0.03(0) + 40 = C + 40 = 1.

Thus C = −39, and the solution is

x(t) = −39e

−0.03t + 40.

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