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Leno4ka [110]
3 years ago
5

Water will move from a ____________ salt solution to a ____________ salt solution when they are across a differentially permeabl

e membrane.
Chemistry
1 answer:
const2013 [10]3 years ago
6 0
The answers are low concentrated (dilute) and high concentrated respectively. 

As the low concentrated salt solution has a higher water potential than that of the high concentrated salt solution, water molecules will flow from the region of higher water potential to the region of lower water potential, thus from the dilute salt solution to the high concentrated salt solution. This is due to the movement called osmosis. Note that osmosis also requires water to flow through a differentially permeable membrane, which means the membrane can allow certain substances (not all) to go in or out. If the differentially permeable membrane is not present, the movement of water molecules may be regarded as diffusion. 

Therefore, the answers for the blanks are low concentrated and high concentrated.




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<h3><u>What is a Galvanic cell ?</u></h3>

Voltaic or galvanic cells are electrochemical devices that use spontaneous oxidation-reduction events to generate electricity. In order to balance the overall equation and highlight the actual chemical changes, it is frequently advantageous to divide the oxidation-reduction reactions into half-reactions while constructing the equations.

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<h3><u>Oxidation:</u></h3>

The anode is located in one half-cell, which is often shown on the left side of a figure. On the anode, oxidation takes place. In the opposite half-cell, the anode and cathode are linked.

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Please review the attachment
astra-53 [7]

Answer: The correct answer is -297 kJ.

Explanation:

To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:

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Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.

Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:

SO3 —>1/2O2 + SO2 (98 kJ)

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Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.

S + O2 —> SO2

Now, we must add the enthalpies together to get our final answer.

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Hope this helps!

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