You are carrying out an experiment in the lab to study the glycolysis pathway. To do this, a liver extract (which is capable of
carrying out all of the normal metabolic reactions) is incubated with 14C-labeled glucose at carbon 1. Based on your knowledge of glycolysis, where does the 14C-label end up in pyruvate?
1 answer:
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<u>Answer:</u> The empirical formula for the given compound is
<u>Explanation:</u>
We are given:
Percentage of H = 5.80 %
Percentage of O = 23.02 %
Percentage of N = 20.16 %
Percentage of Cl = 51.02 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of H = 5.80 g
Mass of O = 23.02 g
Mass of N = 20.16 g
Mass of Cl = 51.02 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Hydrogen =
Moles of Oxygen =
Moles of Nitrogen =
Moles of Chlorine =
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.
For Hydrogen =
For Oxygen =
For Nitrogen =
For Chlorine =
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of H : O : N : Cl = 4 : 1 : 1 : 1
Hence, the empirical formula for the given compound is
Answer:
Explanation:
Hello.
In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:
Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:
Thus, the total number of molecules turns out:
Regards.