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sesenic [268]
2 years ago
12

Simply parmanent tissue bibliography please help it's for my project ​

Chemistry
1 answer:
daser333 [38]2 years ago
6 0

Answer:

Do your project by yourself

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Blue eyes are recessive. Jonathan is hybrid for blue eyes. His wife Carly, has blue eyes. If they have four children, how many w
astraxan [27]

Answer:

possibly two

Explanation:

you said Jonathan had hybrid eyes, did you mean that he was heterozygous? if so 2 out of 4 will have the recessive gene "blue".

3 0
3 years ago
Each of the following values was read on an instrument of measuring device. In each case the last digit was estimated. Tell what
Drupady [299]

Answer:

<h3>160 cm</h3>

Explanation:

6 0
3 years ago
After scientists use a careful process to gather evidence and establish scientific facts, which step most likely occurs next in
oee [108]

Answer:

Conducting Experiments

3 0
2 years ago
Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC
sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

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%

Where x_A - x_B is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

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

x_{Cl} = 3.0

%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%

4 0
3 years ago
Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc
svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

8 0
3 years ago
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