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enot [183]
3 years ago
5

You perform an experiment using two bean seeds planted in identical soil. one seed is kept in a window, where it receives a lot

of light. The other seed is kept in a cupboard and received no light. the seed that gets a lot of light grows very tall and has bright green leaves. The other seed is short and has yellow leaves. What conclusion do you draw?
a) The seeds grew differently because of many different environmental conditions
b) The seeds grew differently because of the different environmental conditions and the amount of air they received
c) The seeds grew differently because of the different environmental conditions and the amount of light they received
d) The seeds had different genes for height and color, making one tall and green and the other short and yellow
Chemistry
1 answer:
My name is Ann [436]3 years ago
7 0

Answer: A

Explanation: because I am your father

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Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC
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Answer:

a. 63.2%

b. 11.7%

c. 73.3%

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e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

%IC = (1 - e^{-0.25*(x_A - x_B)^2}) *100%

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a. x_{Ti} = 1.5

x_{O} = 3.5

%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})*100%

%IC = 63.2%

b. x_{Zn} = 1.6

x_{Te} = 2.1

%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})*100%

%IC = 11.7%

c. x_{Cs} = 0.7

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%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})*100%

%IC = 73.3%

d. x_{In} = 1.7

x_{Sb} = 1.9

%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})*100%

%IC = 0.995 %

e. x_{Mg} = 1.2

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})*100%

%IC = 55.5%

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