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tamaranim1 [39]
3 years ago
10

A hydrate is determined to be 45.43% water and 54.57% CoCl. Find the chemical formula and name

Chemistry
1 answer:
patriot [66]3 years ago
5 0

Answer:

Chemical formula is CoCl. 3H₂O

Explanation:

Data Given

Percentage of water = 45.43%

Percentage of CoCl. = 54.57%

Chemical Formula of the hydrates = ?

Solution:

First, find the mass of each of the part ( CoCl and water) in 100 g of the Compound.

Mass of CoCl = 28 + 35.5

Mass of CoCl = 63.5

Mass of H₂O = 18 g

Now find how many moles are there for each element in 100 g of compound

So,

The percentage will be count in grams for 100g in compound

Find the moles in total compounds

Formula Used for CoCl

mole of CoCl = mass of CoCl / Molar mass of CoCl

mole of CoCl = mole of  54.57 g / 63.5 g/mol

mole of CoCl = 0.859

Formula Used for H₂O

mole of H₂O = mass of H₂O/ Molar mass ofH₂O

mole of H₂O = 45.43 g / 18 g/mol

mole of  H₂O = 2.539

Now

To find the Chemical formula

Divide each one by the smallest number of moles

CoCl = 0.859 / 0.859

CoCl = 1

For H₂O

H₂O = 2.539  / 0.859

H₂O = 3

Multiply the mole fraction to a number to get the whole number.

CoCl =  1

H₂O  = 3

So,  

The Chemical formula is CoCl. 3H₂O

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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

a) wet density of the CL sample

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

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

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

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0
3 years ago
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