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Mashcka [7]
2 years ago
14

Identify each described physical separation technique.

Chemistry
1 answer:
scZoUnD [109]2 years ago
6 0

Answer:

A. A solid mixture is heated. One component transitions directly between solid and gas ---> Sublimation

B. A liquid is carefully poured out, leaving the solid behind in the container ----> Decantation

C. A solvent is added to dissolve only one mixture component and then the liquids are separated ----> solvent extraction

D. A solid is caught on paper or a membrane while the liquid passes through ---> Filtration using filter paper

Explanation:

A. Sublimation is a process of separating solids which sublimes (solids turning directly to gases) from a a mixture of other solids which do not sublime. When the mixture is heated, the solids which sublime are separated and those which do not are left behind. examples of solids which sublime are naphthalene, Iodine crystals, etc.

B. Decantation is the process of carefully pouring out a liquid from a mixture of it with a solid, leaving the solid behind. Example, a mixture of sand and water where water can be easily decanted, leaving the sand behind.

C. In the process of solvent extraction, a mixture of solids which are soluble in different solvent, one of the components is dissolved in a solvent in which it is soluble while the others are not. The solution is then filtered out and the solid is then separated from the solvent by evaporation.

D. In filtration using a filter paper, the solid-liquid mixture is passed through a filter paper which has pores the size through which liquids can pass through but the solid can not. therefore, the solid remains behind on the paper while the liquid passes through.

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3 years ago
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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is
Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

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7 0
3 years ago
The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat
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First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

                                             =  1 mol * 18 mol/g

                                            = 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

       = 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus 

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

     = 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g 

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

      = 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

                                                  = 1.881 KJ  +6.01 KJ + 4.514 KJ

                                                  = 12.405 KJ


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Answer:

13.5 %

Explanation:

First we<u> calculate the mass of 500 mL of water</u>, using <em>its density</em>:

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