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kykrilka [37]
3 years ago
5

6 tons of pig ovaries were required to extract 13.0 mg of estrone. Assuming complete isolation of the hormones was achieved, cal

culate the mass percentage of estrone in the original tissue.
Chemistry
1 answer:
jasenka [17]3 years ago
8 0
One US ton will be equal to 907.185 kg and 1kg will be equal to 10^6mg.

<span>mass percentage of estrone= </span>estrone mass/<span>tissue mass * 100%
</span>mass percentage of estrone= 13mg/ (6 tons * 907.185 kg /ton * 10^6mg/kg) * 100%
mass percentage of estrone=0.238 *10^{-6} %
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3 years ago
Why is it dangerous to react sodium and potassium with dulute sulphuric acid?​
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5 0
3 years ago
Read 2 more answers
Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol
prohojiy [21]

Hey there!:

From the given data ;

Reaction  volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar  = g/L * mol/g

[E]t  = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence   kcat of   xyzase is  5.7*10¹⁰ s⁻¹


Hope that helps!



4 0
3 years ago
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