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Ronch [10]
3 years ago
13

Why are low sulfur fuels an advantage to the environment

Chemistry
1 answer:
ipn [44]3 years ago
3 0

Answer: Low sulphur fuels reduce tailpipe CO, HC, and NOx emissions from catalyst-equipped gasoline vehicles

Explanation:

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g How many turns through the citric acid cycle are required to fully oxidize all of the acetly-coA that result from 1 molecule o
faust18 [17]

Answer:

The two molecules of acetyl-CoA that are produced from a molecule of glucose goes through two turn in the citric acid cycle, one for each molecule of acetyl-CoA.

Explanation:

Glycolysis the process by which a molecule of glucose is broken down in a series of steps to yield two molecules of pyruvate. The overall equation for  the reactions of glycolsis is given below:

Glucose + 2NAD+ ----> 2 Pyruvate + 2NADH + 2H⁺

Each of the two pyruvate molecules produced from glucose breakdown is further oxidized to two molecules of acetyl-CoA and CO₂ each.

2 Pyruvate ----> 2 AcetylCoA + 2CO₂

Each of the acetyl-CoA molecule then enters the citric acid cycle for its oxidation. In each turn of the cycle, one acetyl group enters as acetyl-CoA and two molecules of CO₂ leave.

7 0
3 years ago
Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
What purpose do each of the following play in the isolation and purification of ethyl 3- hydroxybutanoate?
Shalnov [3]

Explanation:

-Filter help — delete some big unreacted, undesirable species (norit is probably from what you are sorting through, its only carbon which cleans up things)

— extract with DCM because you are probably in an aqueous phase, and some butanoate is in it

- Anhydrous sodium absorbs excess of  water (dries the material)

-evaporation in the hood to clear the DCM and crystallize the product.

8 0
3 years ago
,SJhfgaiuehrg;oiaerhgoaw<br> ?
Snezhnost [94]

Answer:

I don't understand anything but thanks for the points

please say in English. I mean say properly the question.......

thank you

6 0
2 years ago
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
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