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zepelin [54]
3 years ago
14

Name the separation technique which could be used to separate mud and water without heating​

Chemistry
1 answer:
Westkost [7]3 years ago
8 0

Answer:

Decantation

Explanation:

Decantation is a process to separate

mixtures by removing a liquid layer that

is free of a precipitate, or the solids

deposited from a solution. The purpose

may be to obtain a decant (liquid free

from particulates) or to recover the

precipitate.

Decantation relies on gravity to pull

precipitate out of the solution, so there is

always some loss of product, either from

the precipitate not fully falling out of the

solution or from liquid remaining when

separating it from the solid portion.

The Decanter

A piece of glassware called a decanter is

used to perform decantation. There are

several decanter designs. A simple

version is a wine decanter, which has a

wide body and a narrow neck. When

wine is poured, solids stay in the base of

the decanter.

In the case of wine, the solid is usually

potassium bitartrate crystals. For

chemistry separations, a decanter may

have a stopcock or valve to drain the

precipitate or dense liquid, or it may

have a partition to separate fractions.

Use of Alum and filtration process can also be administered for further purification

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Explain how intermolecular attractions between molecules influence the bulk of properties of a material
irina1246 [14]

Answer:

<em><u>Intermolecular forces determine bulk properties, such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid.</u></em>

5 0
2 years ago
What states of matter were present when pure water reached its boiling point
maxonik [38]
I believe it’s evaporation?
7 0
3 years ago
Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
3 years ago
what are the slope and the y-intercept of the linear function that is represented by the equation y=-10X+1 ?
Vladimir79 [104]

The slope is 10

And the y-intercept is 1

Hope it helps

6 0
3 years ago
Read 2 more answers
A 15.0 L tank of gas is contained at a high pressure of 8.20 x 10^4 torr. The tank is opened and the gas expands into an empty c
Inga [223]

Answer:

20.5torr

Explanation:

Given parameters:

V₁  = 15L

P₁  = 8.2 x 10⁴torr

V₂ = 6 x 10⁴L

Unknown:

P₂  = ?

Solution:

To solve this problem we have to apply the claims of Boyle's law.

Boyle's law is given mathematically as;

            P₁ V₁   = P₂V₂

where P₁ is the initial pressure

          V₁ is the initial volume

           P₂ is final pressure

           V₂ is final volume

   8.2 x 10⁴  x 15  = P₂  x 6 x 10⁴

         P₂  = 20.5torr

5 0
3 years ago
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