2 years ago

How many grams of O2 are required to produce 5 mol CO2

Chemistry

1 answer:

maxonik [38]2 years ago

7 0

I did the math i got 220 grams

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The turnover number is the maximum substrate quantitiy converted to product per enzyme and per second. It can be calculate as follows:

$k_c = \frac{V_{max}}{[E]}$ with $[E]$ active enzyme concentration.

In this case we have Vmax an data to calculate $[E]$

$moles of E=\frac{10^{-9}g}{29600 g/mol}=3.38*10^{-14}$

$[E]=\frac{3.38*10^{-14}}{0.01 L}=3.38*10^{-12} M$

Now $k_c=\frac{6.8*10^-10}{3.38*10^{-12}}=201.28 min^{-1}$

it is not like any option.

If we assume that $V_{max}$ have the non usual units of $mol L^{-1} min^{-1}$ and it is $6.8*10^-10 mol min^{-1}$

So we need divide by the moles of E (in place of [E])

Now $k_c=\frac{6.8*10^-10}{3.38*10^{-14}}=20128 min^{-1} = 335 s{-1}$

(pass from $min^{-1}$ to $s^{-1}$ dividing by 60)

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