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Mrrafil [7]
3 years ago
9

Calculate the volume in liters of a 29.8 g/dL nickel(II) chloride solution that contains 131. G of nickel(II) chloride . Be sure

your answer has the correct number of significant digits.
Chemistry
1 answer:
ohaa [14]3 years ago
3 0

Answer:

0.44 L.

Explanation:

Density of nickel(II) chloride = 29.8 g/dL.

Mass of nickel(II) chloride = 131 g

Volume of nickel(II) chloride =?

Next, we shall convert 29.8 g/dL to g/L. This can be obtained as follow:

Recall:

1 g/dL = 10 g/L

Therefore,

29.8 g/dL = 29.8 x 10 = 298 g/L

Therefore, 29.8 g/dL is equivalent to 298 g/L.

Finally, we shall determine the volume of nickel(II) chloride as follow:

Density of nickel(II) chloride = 298 g/L

Mass of nickel(II) chloride = 131 g

Volume of nickel(II) chloride =?

Density = mass /volume

298 = 131/Volume

Cross multiply

298 x Volume = 131

Divide both side by 298

Volume = 131/298

Volume = 0.44 L

Therefore, the volume of of nickel(II) chloride is 0.44 L

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Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

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wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

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p = 0.0453 lbs/in³

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b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

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Answer:

10 g

Explanation:

Right from the start, just by inspecting the values given, you can say that the answer will be  

10 g

.

Now, here's what that is the case.

As you know, a substance's specific heat tells you how much heat is needed to increase the temperature of  

1 g

of that substance by  

1

∘

C

.

Water has a specific heat of approximately  

4.18

J

g

∘

C

. This tells you that in order to increase the temperature of  

1 g

of water by  

1

∘

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, you need to provide  

4.18 J

of heat.

Now, how much heat would be required to increase the temperature of  

1 g

of water by  

10

∘

C

?

Well, you'd need  

4.18 J

to increase it by  

1

∘

C

, another  

4.18 J

to increase it by another  

1

∘

C

, and so on. This means that you'd need

4.18 J

×

10

=

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

.

Now look at the value given to you. If you need  

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

, what mass of water would require  

10

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10

∘

C

?

1 g

×

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=

10 g

And that's your answer.

Mathematically, you can calculate this by using the equation

q

=

m

⋅

c

⋅

Δ

T

 

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m

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c

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Δ

T

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Plug in your values to get

418

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⋅

4.18

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(

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m

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Silicon

Germanium

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Lead

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