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Ostrovityanka [42]
2 years ago
9

b) Explain what happens when a drop of concentrated sugar solution is placed on a rheo leaf peel mounted on a glass slide. Name

this phenomenon. Would the same happen if the rheo leaf was boiled before mounting? Give reason for your answer
Chemistry
1 answer:
user100 [1]2 years ago
6 0

The rheo leaf peel will lose water by osmosis and become flaccid. If the rheo leaf was boiled, this would not happen.

<h3>What is osmosis?</h3>

Across living membranes, water moves from a region of high to a region of low water potential.

Concentrated sugar solutions have low water potential when compared to the cell sap of the cells of rheo leaf peel. Thus, water will move from the rheo leaf peel to the sugar solution.

This will result in the wilting of the leaf peel.

If the leaf peel is boiled, the membranes in the cells die and can no longer function as a selectively permeable interface.

More on osmosis can be found here: brainly.com/question/1799974

#SPJ1

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Answer:

3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.

Explanation:

The molecular mass of P2O5 is

P2 = 2* 31 =           62

O5 = 5 *<u> 16 =         80</u>

Molecular Mass = 142

Set up a Proportion

142 grams P2O5 supplies 62 grams of phosphorus

x    kg P2O5        supplies 1.69 kg of phosphorus

Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.

142/x = 62/1.69            Cross multiply

142 * 1.69 = 62x           combine the left

239.98 = 62x               Divide by 62

239.98/62 = x

3.89 kg of P2O5 must be used.

3 0
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In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691
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Explanation:

Let us assume that the value of K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s

Also at 1500 K, K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s

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Relation between K_{p} and K_{c} is as follows.

                  K_{p} = K_{c}RT

Putting the given values into the above formula as follows.

                  K_{p} = K_{c}RT

         0.003691 = K_{c} \times 8.314 \times 1500

                K_{c} = 2.9 \times 10^{-7}

Also,     K_{c} = \frac{K_{f}}{K_{r}}

or,                K_{f} = K_{c} \times K_{r}

                               = 2.9 \times 10^{-7} \times 400.613

                               = 1.1617 \times 10^{-4} m^{6}/mol^{2}s

Thus, we can conclude that the value of K_{f} is 1.1617 \times 10^{-4} m^{6}/mol^{2}s.

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