When would maintaining high proteinase inhibitor concentrations in plant tissues be problematic?

1 answer:

7 0

Maintaining high proteinase inhibitor concentrations in plant tissues would be problematic when there would be no herbivores present.

Explanation:

Maintaining high proteinase inhibitor concentrations in plant tissues would be problematic when there would be no herbivores present.
Proteinase inhibitors are found in plants belonging to a variety of systematic groups
Proteinase inhibitors (PI) are there present in the plant tissues, especially in seeds
The food chain would become limited to producers, suppliers and decomposers.
This mite is present in agricultural soils and participatesin the process of organic matter for decomposition.
for the above reason it will be exposed to the PI .

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How do you balance this:<br> Br2 + H2O + SO2 = HBr + H2SO4

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Answer:

Br2+ 2H2O + SO2= 2HBr + H2SO4

Let's Compare the left side of the equation to the right side of the equation.

Left: Br= 2, H= 2, S= 1, O = 1+2

Right: Br=1, H= 1+2, S=1, O= 4

We can see that only S is balanced and not the other 3 elements.

I'll try to make each element balance.

For Br; I'll multiply by 2 on the left to make it equal to the right.

For H; Since the 2 for Br on the right affected also H, that H ( for HBr) Already has a 2, but then it adds with the other H2( for H2SO4) to give a total of 4 H on the right side. But then there's only 2 H on the left. so we multiply that 2 by a 2 ( which is written infront of the H2O to give a total of 4 H on the left side.

For O; Because of the 2 infront of the H2O, it affects the O in H2O..so now we have 2 O plus the 2 O ( in SO2) to give a total of 4 O which is equal to the right side.

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What is the best reason for the difference in properties of LiCl and C6H14O?

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Answer: Difference in properties of these two compounds are because one compound is covalent and the other one is ionic.

Explanation: LiCl is an inorganic compound having high electronegativity difference between the two elements and hence, is termed as an ionic compound.

$C_6H_{14}O$ is a organic compound formed by the sharing of electrons between the elements, hence is termed as a covalent compound.