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cupoosta [38]
2 years ago
12

Determine the transition length at the entrance to a 10 mm tube through which 100 percent glycerol at 60 C is flowing at a veloc

ity of 0.5 m/s. The density of a glycerol is 1240 kg/m Data: The dynamic viscosity is 0.0813 Pa-s. a) 18.13 mm b) 50 mm c) 45.43 mm d) 38.13 mm
Chemistry
1 answer:
Readme [11.4K]2 years ago
5 0

Answer: Option (d) is the correct answer.

Explanation:

The given data is as follows.

Tube diameter d = 10 mm = 0.01 m

Velocity of glycerol, v = 0.5 m/s

Density of glycerol (\rho) = 1240 kg/m3

Dynamic viscosity of glycerol (\mu) = 0.0813 pa.s

Reynolds number (Re) = \rho \times velocity \times \frac{density}{\mu}

                                     = 1240 \times 0.5 \times \frac{0.01}{0.0813}

                                     = 76.26

Therefore, according to Reynolds number we can say that flow is laminar.

                     Lt = 0.05 \times Re \times d

                         = 0.05 \times 76.26 \times 0.01

                         = 0.03813 m

As it is known that 1 m = 1000 mm. Hence, in 0.03813 m will be equal to 0.03813 m \times \frac{1000 mm}{1 m}

                         = 38.13 mm

Thus, we can conclude that the transition length of glycerol is 38.13 mm.

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A 0.4657 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr b
Yanka [14]

Answer:

The mass percentage of bromine in the original compound is 81,12%

Explanation:

<u>Step 1: Calculate moles AgBr</u>

moles AgBr = mass AgBr / molar mass  AgBr

= 0.8878 g / 187.77 g/mol

= 0.00472812 moles AgBr

⇒

Since 1 mol AgBr contains 1 mol Br-

Then the amount of moles Br- in the original sample must also have  been 0.00472812 moles

<u>Step 2:</u> Calculating mass Br-

mass Br- = molar mass Br x moles  Br-

= 79.904 g/mol x 0.00472812 mol

= 0.377796 g Br-

⇒

There were 0.377796 g Br- in the original sample

<u>Step 3:</u> Calculating mass percentage Br-

⇒mass percentage  = actual mass Br- / total mass x 100%

% mass Br = 0.377796 g / 0.4657 g x 100  %

= 81.12%

7 0
2 years ago
Table salt, NaCl, is an example of an amorphous solid.<br> TRUE<br> FALSE
prohojiy [21]
False, it is an example of an Ionic solid
3 0
2 years ago
Read 2 more answers
131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___
inn [45]
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

8 0
1 year ago
Sierra has a special kind of liquid rubber. She knows that ultraviolet light is absorbed by the rubber, X-ray light is transmitt
Grace [21]
Send a picture of the question
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1 year ago
A 2.80 g sample of Al reacts with 4.15 g sample of Cl2 according to the equation shown below.
solong [7]

Answer:

Mass = 5.33 g

Explanation:

Given data:

Mass of Al = 2.80 g

Mass of Cl₂ = 4.15 g

Theoretical yield of AlCl₃ = ?

Solution:

Chemical equation:

2Al  +  3Cl₂        →       2AlCl₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 2.80 g/ 27 g/mol

Number of moles = 0.10 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.15 g/71 g/mol

Number of moles = 0.06 mol

Now we will compare the moles of AlCl₃ with Al and Cl₂.

                    Cl₂           :        AlCl₃

                    3              :          2

                   0.06         :        2/3×0.06 = 0.04

                   Al             :        AlCl₃

                     2            :          2

                   0.10         :        0.10

Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.

Mass of AlCl₃:Theoretical yield

Mass = number of moles ×molar mass

Mass = 0.04 mol × 133.34 g/mol

Mass = 5.33 g

4 0
1 year ago
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