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ddd [48]
3 years ago
10

3. What does this tell you about the relative tonicity (osmotic pressure) between the contents of the tubing and the solution in

the beaker?
Chemistry
1 answer:
kobusy [5.1K]3 years ago
8 0

Answer:

Please look in the explanation section

Explanation:

Tonicity is defined as the gradient of the osmotic pressure, which is the potential of two solutions that are isolated by a semipermeable membrane. This term is generally used when detailing the cellular response of cells that are submerged in a solution. Tonicity is affected by solutes that are unable to cross the membrane, thereby exerting osmotic pressure. The solutes that cross the membrane do not affect the tone.

On the other hand, the term hypertonic means a higher concentration, defining itself as a solution with a high concentration of solutes that are found outside the cell. If a cell is immersed in this type of solution, the water moves out of the cell to achieve balance between solutes.

The hypotonic term has to do with a lower concentration. These solutions are characterized by having a low concentration of solutes, in this way, water will enter the cell and produce inflammation.

Finally, an isotonic solution is a solution in which the concentration is the same as the concentration of the solutes in other solutions. For example, if the concentration of the water and solute molecules are the same in an external solution as in the cell content. The water molecules spread within the plasma membrane in both directions, and in this way, the cell will neither gain nor lose water.

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Copy the lists of measurements shown below. Pay close attention to the units that follow each number. List 1: 150 mL 11 mL 200 m
stich3 [128]

a) cross out 11mL from list 1, 2 mL from list 2, and 998 cm3 from list 3.

b) circle 200 mL from list 1, 801 mL from list 2, and 1 L from list 3.

7 0
3 years ago
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2. Under a pressure of 95kPa and a temperature of 25 C, a gas occupies 4.0 liters. What would the new
Alekssandra [29.7K]

Answer:

35

Explanation:

multiply and subtract

3 0
3 years ago
The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
Cloud [144]

Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

7 0
3 years ago
Which is an example of chemical weathering?
kobusy [5.1K]

Answer:

b. the layer of oxide formed on statues made of metal

Explanation:

Chemical weathering is a process whereby rocks are disintegrated or weathered causing a chemical reaction thereby leading to change in the parent rock constituents .

In chemical weathering a new substance is usually formed after the weathering takes place.

From the question the layer of statue made by metal will likely form oxides of those metals. Example Silver will react with oxygen to form an entirely new substance like silver(i)oxide.

7 0
3 years ago
A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what
KIM [24]
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water 

Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.

From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water

At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
      = (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)- 
      = 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
       = (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
       = 268.356(T - 27.8)

Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C

Answer: 28.4 °C

3 0
3 years ago
Read 2 more answers
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