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insens350 [35]
3 years ago
10

Which of the following mixture types is characterized by the settling of particles?

Chemistry
1 answer:
torisob [31]3 years ago
8 0
The answer is B. Suspension. Suspension mixtures are composed of two or more materials mixed together wherein the solute particles are usually larger than those found in a solution or colloid. In cases of solid-fluid suspension mixtures, the solid solute particles tend to settle at the bottom of the mixture after some time.
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A 0.250 gram chunk of sodium metal is cautiously dropped into a mixture of 50.0 grams of water and 50.0 grams of ice, both at 0
JulsSmile [24]

Answer:

The ice will not melt, and the temperature will remain at 0°C.

Explanation:

The reaction of the sodium in water is exothermic because heat is being released. In an isolated system, the change in heat must be 0, so the released heat must be absorbed by the ice.

The molar mass of Na is 23 g/mol, so the number of moles that reacted was:

n = 0.250 g/ 23g/mol

n = 0.011 mol

By the reaction:

2 moles ------- -368 kJ

0.011 mol ----- x

By a simple direct three rule:

2x = -4.048

x = -2.024 kJ/mol

So the ice will absorbs 2.024 kJ/mol, which is less than the necessary to melt it (6.02 kJ/mol). Then, the ice will not melt.

The temperature of a pure substance didn't change until all of it has changed of phase, so the temperature must remain at 0°C.

5 0
3 years ago
Calculate the equilibrium constant for the reaction: 2 Cr + 3 Pb2+ ----> 3 Pb + 2 Cr3+ at 25oC. Eocell = 0.61 V
sattari [20]

Answer:

The value is  K  =  8*10^{61}

Explanation:

From the question we are told that

    The equation is  2 Cr  +  3Pb^{2+} \to 3Pb + 2Cr^{3+}

     The  temperature is  T = 25^oC =  298 K   [room  \ temperature ]

     The  emf at standard condition is  E^o_{cell}  =  0.61 \  V

Generally at the cathode

      3Pb^{2+}(aq) + 6 e- --> 3Pb(s)

  At the anode

      2Cr^{3+} + 6e^- \to  2Cr

Generally for an  electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as  

       G =  n*  F *  E^o_{cell}

Here  n  is  the no of electron  with value n = 6

       F  is  the Faraday's constant with value 96487 J/V

  =>   G =  6  * 96487 *  0.61

  =>   G = 3.5 *10^{5} \  J

This Gibbs free energy can also be represented mathematically as

       G =  RTlogK

Here  R  is the cell constant with value 8.314J/K

           K is the equilibrium constant

From above

=>  K  =  antilog^{\frac{G}{ RT} }

Generally  antilog =  2.718

=>K  =  2.718^{\frac{3.5 *10^5}{ 8.314* 298} }

=>   K  =  8*10^{61}

       

         

       

         

6 0
3 years ago
Jenna built the electrical circuit seen here. What would happen to the electrical energy if Jenna removed one of the clips on th
otez555 [7]

Answer: Option (D) is the correct answer.

Explanation:

An electric circuit works well when all the connections are complete but if any of the connections in the circuit is loose or disconnected then it is possible that current will not flow from the circuit.

Therefore, when Jenna removed one of the clips on the battery then circuit becomes incomplete and as a result there will no flow of current.

Thus, we can conclude that the electrical energy would stop because the circuit is incomplete.

6 0
3 years ago
For+the+reaction+H2+++I2+-+2HI+the+equilibrium+constant,+kc+is+49+at+a+fixed+temperature.+Two+mole+of+hydrogen+and+two+moles+of+
Sonja [21]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

    H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0
3 years ago
(Digestion, Intestine, Bile)is secreted by the liver to help in the breaking down of food particles.​
Yuri [45]

Answer:

Bile

Explanation:

Bile is secreted by the liver to help in the breaking down of food particles.​

5 0
1 year ago
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