Answer:
Explanation:
The first step is the <u>calculation of the moles</u> of 
and 
, so:
Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:
Now we can <u>convert the grams</u> of O to moles, so:
The next step is to divide all the mol values by the <u>smallest one</u>:
Therefore the formula is
Answer:Consider a sample that is a mixture composed of biphenyl, benzoic acid, and benzyl alcohol. The sample is spotted on a TLC ... Predict the relative Rf values for the three components in the sample. Hint: See Table 19.3. ... The sample is spotted on a TLC plate and developed in solvent mixture. We are going to predict the ...
Explanation:MORE POWER
233.856 , sorry if i’m wrong
Answer is: mass of the ore is 8.54kg.<span>
</span>ω(Ca₃(PO₄)₂ - calcium phosphate) = 58.6% ÷ 100% = 0.586.
m(P) = 1.00 kg · 1000 g/kg.
m(P) = 1000 g.
In one molecule of calcium phosphate there are two phosphorus atoms:
M(Ca₃(PO₄)₂) = 310.18 g/mol.
M(P) = 30.97 g/mol.
For one kilogram of phosphorus, we need:
M(Ca₃(PO₄)₂) : 2M(P) = m(Ca₃(PO₄)₂) : m(P).
310.18 g/mol : 61.94 g/mol = m(Ca₃(PO₄)₂) : 1000 g.
m(Ca₃(PO₄)₂) = 5007.75 g ÷ 1000 g/kg = 5.007 kg.
Mass of ore find from proportion:
m(Ca₃(PO₄)₂) : m(ore) = 56% : 100%.
m(ore) = 100% · 5.007 kg ÷ 58.6%.
m(ore) = 8.54kg.
Answer:
The Phosphorylated glucose(glucose +inorganic phosphate), with the energy supplied from ATP hydrolysis formed glucose 6- phosphate, which is later converted to 2 molecules of fructose 6-phosphate- this is phosphorylation.And represented the fate of glucose -6-phosphate.
The fructose 6-phosphate are converted to triose phosphate- which is a 2-molecules of 3C compound. The latter is oxidized by NAD→ NADH+ to form intermediates in the glycolytic pathways .
These intermediates are converted to ribose 5-phosphates in the presence of transketolase and transaldolase enzymes.And they are finally converted to pyruvate in the glycolytic pathway with the production of 2ATPs per molecule of glucose.
Basically the phosphate pathway reaction is very slow due to enzyme catalysis.
