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kap26 [50]
3 years ago
12

How many seconds are required to deposit 0.110 grams of magnesium metal from a solution that contains Mg2 ions, if a current of

0.998 A is applied?
Physics
1 answer:
lakkis [162]3 years ago
8 0

Answer: The time required to deposit such amount of Mg is 886secs.

Explanation: According to Faraday Law of Electrolysis, the mass of a substance deposited is directly proportional to quantity of electricity passed.

He also stated that;

96500C(1Farday) of electricity is required to deposit 1 mole of any metal.

For Mg^2+ +2e- ==>Mg

193000C(2Faraday) of electricity is required to deposit 1mole of Magnesium metal (1 mole of Mg=24g)

Which implies;

19300C will liberate 24g

xCwill liberate 0.110g

Where x is the amount of electricity to deposit 0.110g

x = (19300 × 0.110)/24

x = 884.6C

Recall that Q =It

Where Q is the quantity of electricity, I is the current and t is the time taken

884.6= 0.998 × t

t= 884.6/0.998

t= 886.3secs.

Therefore the time require is 886.3s.

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Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
3 years ago
Sue and Jenny kick a soccer ball at exactly the same time. Sue’s foot exerts a force of 52.6 N to the north. Jenny’s foot exerts
Karolina [17]
Using the addition of forces using right angled triangles. The resultant force sqaured. = 112.8 sqaured + 52.6 squared. So resultant force sqaured is 15490.6. So the resultant force is the sqaure root of this which is 124N to 3 significant figures
6 0
3 years ago
Read 2 more answers
What is the kinetic energy of a 5-kg object moving at 7 m/s?
Sunny_sXe [5.5K]
To answer this question, we would need the formula for the kinetic energy which is Kinetic Energy = ½ x m x v^2

Where the following means: m is the mass of the object and v is the velocity

So At 7 m/s: Kinetic Energy = ½ x 5 x 7^2 = 122.5 J is the answer
6 0
3 years ago
In some areas where there are adequate geothermal resources, a pollutant occurs naturally that escapes, making the location unsu
Luba_88 [7]
Hydrogen sulfide gas
7 0
3 years ago
Read 2 more answers
A hockey puck with mass 0.200 kg traveling east at 10.0 m/s strikes a puck with a mass of .250 kg heading north at 20 m/s and st
love history [14]

Answer:

4.44 i + 11.11 j , 11.96 m/s

Explanation:

m1 = 0.2 kg, v1 = 10 m/s along X axis, v1 = 10 i

m2 = 0.25 kg, v2 = 20 m/s along north, v2 = 20 j

After collision they stick together and let they move with velocity v.

Use the conservation of momentum

m1 x v1 + m2 x v2 = (m1 + m2) x v

0.2 x (10 i) + 0.25 x (20 j) = (0.2 + 0.25) x v

2i + 5j = 0.45 v

v = 4.44 i + 11.11 j

magnitude of velocity = \sqrt{(4.44)^{2}+(11.11)^{2}} = 11.96 m/s

Thus, the velocity of 0.250 kg hockey puck is 4.44 i + 11.11 j  metre per second and the magnitude of velocity is 11.96 m/s.

6 0
3 years ago
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