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lidiya [134]
3 years ago
11

Assuming the following materials will not mix,show how the materials would "Stack up" in a graduated cylinder.

Chemistry
1 answer:
mihalych1998 [28]3 years ago
4 0

<u>Answer:</u>

Density is mass per unit volume.

Its formula is  

Density = Mass / Volume

Its units can be kg/L  or kg/(dm^3) or g/mL or g/(cm^3)

Density of various substances are given. The more denser substance goes to the bottom of the graduated cylinder.  

Air on the top then comes  Wood followed by  Corn oil followed by  Ice and then  Water followed by  Glycerin then  Rubber then  Corn syrup and at last  Steel at the bottom  which will more denser than all the substances mentioned above.

The staking up of the substances is in the order mentioned below in the tabular column

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Formula of isocyanic acid is HNCO. It colourless, volatile compound. It is poisonous inorganic compound. Melamine is synthesized from urea. The reaction involves two steps. In first step, urea gets converted to isocyanic acid  which is an intermediate. In the second step, isocyanic acid gives melamine (molecular formula C₃H₆N₆) and carbon dioxide (CO₂). The balanced chemical reaction involved in second step is given below:

6 HNCO → 1 C₃N₃(NH₂)₃ + 3CO₂

The co-efficient of the reaction are: 6, 1, 3

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3 years ago
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Why Are aquifers important to our freshwater supply? 5TH GRADE ANSWER c:
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Explanation:

Aquifers are porous and permeable formations that stores ground water. The ground water system is made up of mostly fresh water.

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An aquifer is able to store this fresh water and it is is good prospect for sourcing ground water.

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Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

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3 years ago
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Answer:  The image from the question has the correct answers.

Explanation:  

As summarized in the attached table.

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